Proof. We start with the ‘only if’ part of the claim. Let $\phi\in\mathcal{A}_{\Phi_0}$ such that $V^{\Phi_0,\phi}_T\geq X$ . Choose $\mathbb{Q}\in \mathcal{Q}$ and $M\in\mathcal{M}^{\mathbb{Q}}_{[0,T]}$ . From (2.1) and the Cauchy–Schwarz inequality, it follows that

$$\sqrt{\int_{0}^T S^2_t \,{\textrm{d}} t} \sqrt{\int_{0}^T \phi^2_t \,{\textrm{d}} t}-\dfrac{\Lambda}{2}\int_{0}^T \phi^2_t \,{\textrm{d}} t\geq X+\Phi_0S_0,$$

and hence, by exploiting the behaviour of the (random) parabola

$$x\rightarrow x \sqrt{\int_{0}^T S^2_t \,{\textrm{d}} t} -\dfrac{\Lambda}{2} x^2,$$

we get

$$\int_{0}^T \phi^2_t \,{\textrm{d}} t\leq c\biggl(1+\max({-}X,0)+\int_{0}^T S^2_t \,{\textrm{d}} t\biggr)$$

for some constant $c>0$ . This, together with Assumption 2.1, (2.5), and the well-known inequality ${\textrm{e}}^x+y\log y\geq xy$ , $x\in\mathbb{R}$ , $y>0$ , yields

$$\mathbb{E}_{\mathbb{Q}}\biggl[\int_{0}^T \phi^2_t \,{\textrm{d}} t \biggr]<\infty.$$

Hence

$$\mathbb{E}_{\mathbb{Q}}\biggl[\Phi_0M_0+\int_{0}^T M_t\phi_t \,{\textrm{d}} t\biggr]=0,$$

so from (2.1) and the simple inequality

$$x y-\dfrac{\Lambda}{2}x^2\leq \dfrac{y^2}{2\Lambda},\quad x,y\in\mathbb{R}$$

we obtain

\begin{align*}\mathbb{E}_{\mathbb{Q}}[X]&\leq\mathbb{E}_{\mathbb{Q}}\biggl[\Phi_0 (M_0-S_0)+\int_{0}^T \phi_t (M_t-S_t) \,{\textrm{d}} t- \dfrac{\Lambda}{2}\int_{0}^T \phi^2_t \,{\textrm{d}} t\biggr]\\&\leq \mathbb{E}_{\mathbb{Q}}\biggl[\Phi_0 (M_0-S_0)+\dfrac{1}{2\Lambda}\int_{0}^T |M_t-S_t|^2 \,{\textrm{d}} t\biggr],\end{align*}

and the result follows.

Next, we prove the ‘if’ part of the claim. Assume by contradiction that this part does not hold true. Namely, there exists X which satisfies (2.5)–(2.6) and there is no $\phi\in\mathcal{A}_{\Phi_0}$ such that $V^{\Phi_0,\phi}_T\geq X$ .

Using the same arguments as in the proof of Proposition 3.5 in [Reference Guasoni and Rásonyi14], it follows that the set

$$\Upsilon\,:\!=\, \bigl(\bigl\{V^{\Phi_0,\phi}_T\,:\, \phi\in\mathcal{A}_{\Phi_0}\bigr\}-L^0_{+}(\mathbb{P})\bigr)\cap L^1(\mathbb{P})$$

is convex and closed in $L^1(\mathbb{P})$ . Observe that from Assumption 2.1 and (2.5)–(2.6) (take $\mathbb{Q}=\mathbb{P}$ and $M\equiv 0$ in (2.6)) it follows that $X\in L^1(\mathbb{P})$ . Since there is no $\phi\in\mathcal{A}_{\Phi_0}$ such that $V^{\Phi_0,\phi}_T\geq X$ , we get $X\in L^1(\mathbb{P})\setminus\Upsilon$ .

Thus, by the Hahn–Banach separation theorem, we can find $Z\in L^{\infty}\setminus\{0\}$ such that

$$\mathbb{E}_{\mathbb{P}}[Z X]>\sup_{\upsilon\in \Upsilon}\mathbb{E}_{\mathbb{P}}[Z \upsilon].$$

Since

$$\bigl(V^{\Phi_0,\hat\phi}_T-L^0_{+}(\mathbb{P})\bigr)\cap L^1(\mathbb{P})\subset\Upsilon\quad \text{for } \hat\phi\equiv -\dfrac{\Phi_0}{T},$$

we must have $Z\geq 0$ . Moreover, from (2.1) we have

$$V^{\Phi_0,\phi}_T\leq -\Phi_0S_0+\dfrac{1}{2\Lambda}\int_{0}^T S^2_t \,{\textrm{d}} t\quad \text{for all $\phi$,}$$

so from Assumption 2.1 it follows that there exists $\epsilon>0$ such that

$$\mathbb{E}_{\mathbb{P}}[(\epsilon+Z) X]>\sup_{\upsilon\in \Upsilon}\mathbb{E}_{\mathbb{P}}[(\epsilon+Z) \upsilon].$$

We conclude that for the probability measure $\mathbb{Q}$ given by

$$\dfrac{{\textrm{d}}\mathbb{Q}}{{\textrm{d}}\mathbb{P}}\,:\!=\, \dfrac{\epsilon+ Z}{\epsilon +\mathbb{E}_{\mathbb{P}}[Z]}$$

we have $\mathbb{Q}\in \mathcal{Q}$ and

(2.7) \begin{equation}\mathbb{E}_{\mathbb{Q}}[X]>\sup_{\phi\in\mathcal{A}_{\Phi_0}}\mathbb{E}_{\mathbb{Q}}\bigl[V^{\Phi_0,\phi}_T\bigr],\end{equation}

where we set

$$\mathbb{E}_{\mathbb{Q}}\bigl[V^{\Phi_0,\phi}_T\bigr]\,:\!=\, -\infty$$

if $V^{\Phi_0,\phi}_T\notin L^1(\mathbb{Q})$ .

Next, fix $n\in\mathbb{N}$ and introduce the set

$$\mathcal{B}_n\,:\!=\, \bigl\{\phi\in L^2({\textrm{d}} t\otimes\mathbb{Q})\,:\, \|\phi\|_{L^2({\textrm{d}} t\otimes\mathbb{Q})}\leq n\bigr\}.$$

We argue that for any $n\in\mathbb{N}$

(2.8) \begin{align}\sup_{\phi\in\mathcal{A}_{\Phi_0}}\mathbb{E}_{\mathbb{Q}}\bigl[V^{\Phi_0,\phi}_T\bigr]&\geq\sup_{\phi\in\mathcal{A}_{\Phi_0} \cap \mathcal{B}_n}\mathbb{E}_{\mathbb{Q}}\bigl[V^{\Phi_0,\phi}_T\bigr]\notag \\&=\sup_{\phi\in \mathcal{B}_n}\inf_{M\in\mathcal{M}^{\mathbb{Q}}_{[0,T]}}\mathbb{E}_{\mathbb{Q}}\biggl[V^{\Phi_0,\phi}_T+M_T\biggl(\Phi_0+\int_{0}^T\phi_t \,{\textrm{d}} t\biggr)\biggr]\notag \\&=\inf_{M\in\mathcal{M}^{\mathbb{Q}}_{[0,T]}}\sup_{\phi\in \mathcal{B}_n}\mathbb{E}_{\mathbb{Q}}\biggl[V^{\Phi_0,\phi}_T+M_T\biggl(\Phi_0+\int_{0}^T\phi_t \,{\textrm{d}} t\biggr)\biggr].\end{align}

Indeed, the inequality is obvious. The first equality follows from the fact that if $\phi\in \mathcal{B}_n\setminus \mathcal{A}_{\Phi_0}$ , then

$$\inf_{M\in\mathcal{M}^{\mathbb{Q}}_{[0,T]}}\mathbb{E}_{\mathbb{Q}}\biggl[M_T\biggl(\Phi_0+\int_{0}^T\phi_t \,{\textrm{d}} t\biggr)\biggr]=-\infty.$$

For the last equality in (2.8) we apply a minimax theorem. Consider the vector space $\mathcal{M}^{\mathbb{Q}}_{[0,T]}$ with the $L^2({\textrm{d}} t\otimes\mathbb{Q})$ norm and the set $\mathcal{B}_n$ with the weak topology which corresponds to $L^2({\textrm{d}} t\otimes\mathbb{Q})$ . Then both of these sets are convex subsets of topological vector spaces and the latter set is even compact. Moreover,

$$(\phi,M)\rightarrow \mathbb{E}_{\mathbb{Q}}\biggl[V^{\Phi_0,\phi}_T+M_T\biggl(\Phi_0+\int_{0}^T\phi_t \,{\textrm{d}} t\biggr)\biggr]$$

is upper semi-continuous and concave in $\phi$ and convex (indeed affine) in M. We can thus apply Theorem 4.2 in [Reference Sion16] to obtain the second equality.

Next, choose $M\in\mathcal{M}^{\mathbb{Q}}_{[0,T]}$ and introduce the process $\phi$ (which depends on n and M):

$$\phi_t\,:\!=\, \dfrac{n(M_t-S_t)}{\max\bigl(\Lambda n,\|M-S\|_{{L^2({\textrm{d}} t\otimes\mathbb{Q})}}\bigr)}, \quad t\in [0,T].$$

Observe that $\phi\in \mathcal{B}_n$ and simple computations give

\begin{align*}&\mathbb{E}_{\mathbb{Q}}\biggl[V^{\Phi_0,\phi}_T+M_T\biggl(\Phi_0+\int_{0}^T\phi_t \,{\textrm{d}} t\biggr)\biggr]\\&\quad =\mathbb{E}_{\mathbb{Q}}\biggl[\Phi_0 (M_0-S_0)+\int_{0}^T \phi_t (M_t-S_t) \,{\textrm{d}} t- \dfrac{\Lambda}{2}\int_{0}^T \phi^2_t \,{\textrm{d}} t\biggr]\\&\quad =\mathbb{E}_{\mathbb{Q}}[\Phi_0 (M_0-S_0)]+G_n\bigl( \|M-S\|_{L^2({\textrm{d}} t\otimes\mathbb{Q})}\bigr),\end{align*}

where

$$G_n(x) = \begin{cases} \dfrac{x^2}{2\Lambda} & \text{if $ x<\Lambda n$,}\\[7pt] n x-\dfrac{\Lambda}{2} n^2 & \text{otherwise}.\end{cases}$$

Since $n\in\mathbb{N}$ and $M\in\mathcal{M}^{\mathbb{Q}}_{[0,T]}$ were arbitrary, from (2.7)–(2.8) we conclude that there exists a sequence of martingales $M^n\in\mathcal{M}^{\mathbb{Q}}_{[0,T]}$ , $n\in\mathbb{N}$ such that

(2.9) \begin{equation}\mathbb{E}_{\mathbb{Q}}[X]> \sup_{n\in\mathbb{N}} \bigl\{\mathbb{E}_{\mathbb{Q}}\bigl[\Phi_0 \bigl(M^n_0-S_0\bigr)\bigr]+G_n\bigl(\|M^n-S\|_{L^2({\textrm{d}} t\otimes\mathbb{Q})}\bigr)\bigr\}.\end{equation}

From the fact that ${{{\textrm{d}}\mathbb{Q}}/{{\textrm{d}}\mathbb{P}}}$ is bounded we have $\mathbb{E}_{\mathbb{Q}}[X]<\infty$ , so $\sup_{n\in\mathbb{N}}\|M^n-S\|_{L^2({\textrm{d}} t\otimes\mathbb{Q})}<\infty$ . Thus from (2.9) we obtain that for any $k>({{1}/{\Lambda}})\sup_{n\in\mathbb{N}}\|M^n-S\|_{L^2({\textrm{d}} t\otimes\mathbb{Q})}$ ,

$$\mathbb{E}_{\mathbb{Q}}[X]>\mathbb{E}_{\mathbb{Q}}\biggl[\Phi_0 \bigl(M^k_0-S_0\bigr)+\dfrac{1}{2\Lambda}\int_{0}^T |M^k_t-S_t|^2 \,{\textrm{d}} t\biggr],$$

which is a contradiction to (2.6). This completes the proof.

Proof. Let C denote the set of all pairs (Z, Y) such that $Z>0$ is a random variable satisfying $\mathbb{E}_{\mathbb{P}}[Z]=1$ , $\mathbb{E}_{\mathbb{P}}[ Z\log Z]<\infty$ , and $Y=(Y_t)_{0\leq t< T}$ is a $\mathbb{P}$ -martingale satisfying

$$ \mathbb{E}_{\mathbb{P}}\biggl[\int_{0}^T \dfrac{Y^2_t}{\mathbb{E}_{\mathbb{P}}[Z\mid \mathcal{F}_t]}\,{\textrm{d}} t\biggr]<\infty. $$

Note that the function $(z,y)\rightarrow y^2/z$ is convex on $\mathbb{R}_{++}\times\mathbb{R}$ , so C is a convex set. Define a map $\Psi\,:\, C\rightarrow \mathbb{R}$ by

$$\Psi(Z,Y)\,:\!=\, \mathbb{E}_{\mathbb{P}}\biggl[\dfrac{1}{\alpha} Z\log Z+\Phi_0 (Y_0-S_0 Z)+\dfrac{Z}{2\Lambda}\int_{0}^T\biggl(\dfrac{Y_t}{\mathbb{E}_{\mathbb{P}}[Z\mid \mathcal{F}_t]}-S_t\biggr)^2 \,{\textrm{d}} t\biggr].$$

Observe that there is a bijection

$$(Y,Z)\in C\leftrightarrow \Bigl(\mathbb{Q}\in\mathcal{Q},M\in\mathcal{M}^{\mathbb{Q}}_{[0,T)}\Bigr) $$

given by

$$Z\,:\!=\, \dfrac{{\textrm{d}}\mathbb{Q}}{{\textrm{d}}\mathbb{P}}\quad \text{and}\quad M_t\,:\!=\, \dfrac{Y_t}{\mathbb{E}_{\mathbb{P}}[Z\mid \mathcal{F}_t]},\quad t\in [0,T), $$

and for this bijection we have

$$\Psi(Z,Y)=\mathbb{E}_{\mathbb{Q}}\biggl[\dfrac{1}{\alpha}\log\biggl(\dfrac{{\textrm{d}}\mathbb{Q}}{{\textrm{d}}\mathbb{P}}\biggr)+\Phi_0 (M_0-S_0)+\dfrac{1}{2\Lambda}\int_{0}^T |M_t-S_t|^2 \,{\textrm{d}} t\biggr].$$

Thus, in order to prove the lemma, it is sufficient to show that there exists a unique minimizer for $\Psi\,:\, C\rightarrow \mathbb{R}$ . Note that the convexity of the map $(z,y)\rightarrow y^2/z$ implies the convexity of $\Psi$ . From the strict convexity of the functions $z\rightarrow z\log z$ and the map $y\rightarrow y^2$ , it follows that $\Psi$ is strictly convex, and hence the uniqueness of a minimizer is immediate. It remains to prove the existence of a minimizer.

Let $(Z^n,Y^n)\in C$ , $n\in\mathbb{N}$ be a sequence such that

(2.10) \begin{equation}\lim_{n\rightarrow\infty}\Psi(Z^n,Y^n)=\inf_{(Z,Y)\in C}\Psi(Z,Y).\end{equation}

Assumption 2.1 and (2.10) yield that without loss of generality we can assume that

$$\sup_{n\in\mathbb{N}}\mathbb{E}_{\mathbb{P}} [ Z^n\log {Z^n}]<\infty\quad \text{and}\quad \sup_{n\in\mathbb{N}}\mathbb{E}_{\mathbb{P}}\biggl[\int_{0}^T \dfrac{|Y^n_t|^2} {\mathbb{E}_{\mathbb{P}}[Z^n\mid \mathcal{F}_t]}\,{\textrm{d}} t\biggr]<\infty.$$

Thus the de la Vallée–Poussin criterion ensures that $Z^n$ , $n\in\mathbb{N}$ are uniformly integrable. Let us argue that for any $s<T$ the random variables $Y^n_s$ , $n\in\mathbb{N}$ are uniformly integrable

Fix $s<T$ . From the Jensen inequality and fact that $Y^n$ is a martingale, it follows that for any given n the function

$$t\rightarrow\mathbb{E}_{\mathbb{P}}\biggl[\dfrac{|Y^n_t|^2}{\mathbb{E}_{\mathbb{P}}[Z^n\mid \mathcal{F}_t]}\biggr]$$

is non-decreasing, and hence

$$\sup_{n\in\mathbb{N}}\mathbb{E}_{\mathbb{P}}\biggl[ \dfrac{|Y^n_s|^2}{\mathbb{E}_{\mathbb{P}}[Z^n\mid \mathcal{F}_s]}\biggr]<\infty.$$

This, together with the inequality $\sup_{n\in\mathbb{N}}\mathbb{E}_{\mathbb{P}} [ Z^n\log {Z^n}]<\infty$ , gives $\sup_{n\in\mathbb{N}}\mathbb{E}_{\mathbb{P}}[ g|Y^n_s|]<\infty$ , where

$$g(y)\,:\!=\, \inf_{z>0} \biggl\{\dfrac{y^2}{z}+z\log z\biggr\},\quad y>0.$$

For a given $y>0$ , the function $z\rightarrow {{y^2}/{z}}+z\log z$ is convex and attains its minimum at the unique $z=z(y)$ which satisfies $1+\log z={{y^2}/{z^2}}$ . Obviously $\lim_{y\rightarrow\infty}z(y)=\infty$ and $y=z(y)\sqrt{1+\log\!(z(y))}$ . Thus

$$\lim_{y\rightarrow\infty}\dfrac{g(y)}{y}\geq \lim_{z\rightarrow\infty} \dfrac{z\log z}{z\sqrt{1+\log z }}=\infty.$$

Hence, from the de la Vallée–Poussin criterion, we conclude that $Y^n_s$ , $n\in\mathbb{N}$ are uniformly integrable.

Next, define the sequence of random variables $(H_n)_{n\in\mathbb{N}}\in L^1({\textrm{d}} t\otimes \mathbb{P}, [0,2T]\times \Omega)$ by

$$H^n(t,\omega) \,:\!=\, \begin{cases} Y^n_t(\omega) & \text{if $ t<T$,}\\ Z^n(\omega) & \text{otherwise.}\end{cases}$$

Observe that the relations

$$ \sup_{n\in\mathbb{N}}\mathbb{E}_{\mathbb{P}} [ Z^n\log {Z^n}]<\infty,\quad \sup_{n\in\mathbb{N}}\mathbb{E}_{\mathbb{P}}\biggl[\int_{0}^T \dfrac{|Y^n_t|^2} {\mathbb{E}_{\mathbb{P}}[Z^n\mid \mathcal{F}_t]}\,{\textrm{d}} t\biggr]<\infty,\quad \lim_{y\rightarrow\infty}\frac{g(y)}{y}=\infty$$

yield that $(H^n)_{n\in\mathbb{N}}$ is a bounded sequence in $L^1({\textrm{d}} t\otimes \mathbb{P}, [0, 2T]\times \Omega)$ . Hence, from the well-known Komlós argument (see Theorem 1.3 in [Reference Delbaen and Schachermayer6]), there exists a sequence $(\hat H^n)\in \operatorname{conv}(H^n,H^{n+1}\cdots)$ , $n\in\mathbb{N}$ such that $(\hat H^n)_{n\in\mathbb{N}}$ converge in probability ( ${\textrm{d}} t\otimes\mathbb{P}$ ) to some $\hat H \in L^1({\textrm{d}} t\otimes \mathbb{P}, [0, 2T]\times \Omega)$ . From the bounded convergence theorem, $\arctan(\hat H^n)\rightarrow\arctan(H)$ in $L^1({\textrm{d}} t\otimes \mathbb{P}, [0, 2T]\times \Omega)$ . Thus, from Fubini’s theorem, we obtain that there exists a dense set $\mathcal{I}\subset [0,2T]$ such that for any $t\in \mathcal{I}$ , $(\hat H^n_t)_{n\in\mathbb{N}}$ converge in probability to $\hat H_t$ . Choose a countable subset $\mathcal{J}\subset \mathcal{I}\cap [0,T)$ of the form $\mathcal{J}=\{t_1<t_2<\cdots\}$ such that $\lim_{n\rightarrow \infty} t_n=T$ .

We conclude that there exist convex combinations (the same combinations as for $H^n$ ) $(\hat Z^n,\hat Y^n)\in \operatorname{conv}((Z^{n},Y^{n}),(Z^{n+1},Y^{n+1})\cdots)$ , $n\in\mathbb{N}$ , such that the sequence $(\hat Z^n)_{n\in\mathbb{N}}$ converges in probability to some $\hat Z$ , and for any $m\in\mathbb{N}$ the sequence $(\hat Y^n_{t_m})_{n\in\mathbb{N}}$ converges in probability to some $U_m$ . From the uniform integrability of the sequences $(Z^n)_{n\in\mathbb{N}}$ and $(Y^n_{t_m})_{n\in\mathbb{N}}$ , $m\in\mathbb{N}$ , we conclude that

(2.11) \begin{equation}\hat Z^n\rightarrow \hat Z \quad \text{in $ L^1(\mathbb{P})$,}\end{equation}

and for any $m\in\mathbb{N}$

(2.12) \begin{equation}\hat Y^n_{t_m}\rightarrow U_m\quad \text{in $ L^1(\mathbb{P})$.}\end{equation}

Notice that (2.11) implies $\mathbb{E}_{\mathbb{P}}[\hat Z]=1$ . Moreover, the function $x\rightarrow x\log x$ , $x>0$ is bounded from below, so from the Fatou lemma and the convexity of the function $x\rightarrow x\log x$ we get $\mathbb{E}_{\mathbb{P}}[ \hat Z\log {\hat Z}]\leq \sup_{n\in\mathbb{N}}\mathbb{E}_{\mathbb{P}} [ Z^n\log {Z^n}]<\infty$ .

Next, define the process $\hat Y=(\hat Y_t)_{0\leq t<T}$ by

$$\hat Y_t\,:\!=\, \sum_{m=1}^{\infty}\mathbf 1_{\{t\in [t_{m-1},t_{m})\}} \mathbb{E}_{\mathbb{P}}[U_m\mid \mathcal{F}_t], $$

where we set $t_0\,:\!=\, 0$ . Clearly, for any n the process $\hat Y^n=\hat Y^n_{[0,T)}$ is a martingale (convex combination of martingales), so from (2.12) we obtain that $\hat Y=\hat Y_{[0,T)}$ is a martingale and

(2.13) \begin{equation}\hat Y^n_t\rightarrow \hat Y_t \quad \text{in $ L^1(\mathbb{P})$}\quad \text{for all $t\in [0,T]$.}\end{equation}

From (2.11) we get

(2.14) \begin{equation}\mathbb{E}_{\mathbb{P}}[\hat Z^n\mid \mathcal{F}_t]\rightarrow \mathbb{E}_{\mathbb{P}}[\hat Z\mid \mathcal{F}_t] \quad \text{in $ L^1(\mathbb{P})$} \quad \text{for all $t\in [0,T]$.}\end{equation}

By combining the Fatou lemma, the convexity of $\Psi$ , (2.10), and (2.13)–(2.14), we obtain

$$\Psi(\hat Z,\hat Y)\leq \lim_{n\rightarrow\infty}\Psi(\hat Z^n,\hat Y^n)=\inf_{(Z,Y)\in C}\Psi(Z,Y).$$

A priori it might happen that ${\textrm{d}} t\otimes \mathbb{P}(\mathbb{E}_{\mathbb{P}}[\hat Z\mid \mathcal{F}_t]=0)>0$ , so we need to be careful with the definition of $\Psi(\hat Z,\hat Y)$ . From (2.13)–(2.14) it follows that we have the convergence in probability $\mathbb{E}_{\mathbb{P}}[\hat Z^n\mid \mathcal{F}_{\cdot}]\rightarrow \mathbb{E}_{\mathbb{P}}[\hat Z\mid \mathcal{F}_{\cdot}]$ and $\hat Y^n\rightarrow \hat Y$ with respect to the product measure ${\textrm{d}} t\otimes\mathbb{P}$ . Hence, by taking a subsequence (which for simplicity we still denote by n), we can assume that $\mathbb{E}_{\mathbb{P}}[\hat Z^n\mid \mathcal{F}_{\cdot}]\rightarrow \mathbb{E}_{\mathbb{P}}[\hat Z\mid \mathcal{F}_{\cdot}]$ and $\hat Y^n\rightarrow \hat Y$ ${\textrm{d}} t\otimes\mathbb{P}$ a.s. Since $\lim_{n\rightarrow\infty}\Psi(\hat Z^n,\hat Y^n)<\infty$ , from the Fatou lemma it follows that

$$\mathbb{E}_{\mathbb{P}}\biggl[\int_{0}^T\biggl(\lim\inf_{n\rightarrow \infty}\dfrac{|\hat Y^n_t|^2}{\mathbb{E}_{\mathbb{P}}[\hat Z^n\mid \mathcal{F}_t]}\biggr)\,{\textrm{d}} t \biggr]<\infty.$$

In particular,

$$\lim\inf_{n\rightarrow \infty}\dfrac{|\hat Y^n_t|^2}{\mathbb{E}_{\mathbb{P}}[\hat Z^n\mid \mathcal{F}_t]}<\infty\quad {\textrm{d}} t\otimes \text{${\mathbb{P}}$ a.s.}$$

This together with the above convergence of the sequences $(\mathbb{E}_{\mathbb{P}}[\hat Z^n\mid \mathcal{F}_{\cdot}])_{n\in\mathbb{N}}$ and $(\hat Y^n)_{n\in\mathbb{N}}$ yields the implication $\mathbb{E}_{\mathbb{P}}[\hat Z\mid \mathcal{F}_t]=0\Rightarrow\hat Y_t=0$ ${\textrm{d}} t\otimes\mathbb{P}$ a.s. Thus we set

$$\dfrac{\hat Y_t}{\mathbb{E}_{\mathbb{P}}[\hat Z\mid \mathcal{F}_t]}\,:\!=\, 0$$

if $\mathbb{E}_{\mathbb{P}}[\hat Z\mid \mathcal{F}_t]=0$ .

Finally, in order to complete the proof it remains to show that $\hat Z>0$ a.s. To this end, define the function $f\,:\, [0,1]\rightarrow\mathbb{R}$ by $f(\alpha)\,:\!=\, \Psi(\alpha +(1-\alpha)\hat Z,\hat Y)$ , $\alpha\in [0,1].$ From the convexity of $\Psi$ it follows that f is convex. The inequality $\Psi(\hat Z,\hat Y)\leq\inf_{( Z, Y)\in C}\Psi( Z, Y)$ yields that the right-hand derivative $f^{\prime}(0{+})\geq 0$ . Moreover, from the monotone (derivative of a convex function) convergence theorem it follows that we can interchange derivative and expectation. Thus

\begin{align*} 0&\leq f^{\prime}(0{+})\\ &= \mathbb{E}_{\mathbb{P}}\biggl[\dfrac{1}{\alpha}(1-\hat Z)\log\hat Z-\Phi_0S_0(1-\hat Z)\biggr]\\&\quad +\dfrac{1}{2\Lambda}\mathbb{E}_{\mathbb{P}}\biggl[(1-\hat Z)\int_{0}^TS^2_t \,{\textrm{d}} t+\int_{0}^T\mathbf{1}_{\{\mathbb{E}_{\mathbb{P}}[\hat Z\mid \mathcal{F}_t]>0\}}\biggl(\dfrac{\hat Y^2_t}{\mathbb{E}_{\mathbb{P}}[\hat Z\mid \mathcal{F}_t]}-\dfrac{\hat Y^2_t}{\mathbb{E}^2_{\mathbb{P}}[\hat Z\mid \mathcal{F}_t]}\biggr)\,{\textrm{d}} t\biggr].\end{align*}

We conclude that $\mathbb{E}_{\mathbb{P}}[\!\log\hat Z]>-\infty$ and complete the proof.

Proof. Let $\bigl(\hat{\mathbb{Q}}\in\mathcal{Q},\hat M\in\mathcal{M}^{\hat{\mathbb{Q}}}_{[0,T]}\bigr)$ be the minimizer from Lemma 2.2. Denote

$$D\,:\!=\,\mathbb{E}_{\hat{\mathbb{Q}}}\biggl[\dfrac{1}{\alpha}\log\biggl(\dfrac{{\textrm{d}}\hat{\mathbb{Q}}}{{\textrm{d}}\mathbb{P}}\biggr)+\Phi_0(\hat M_0-S_0)+\dfrac{1}{2\Lambda}\int_{0}^T|\hat {M}_t-S_t|^2 \,{\textrm{d}} t\biggr].$$

Let us show that there exists $\hat\phi\in\mathcal{A}_{\Phi_0}$ such that

(2.15) \begin{equation}V^{\Phi_0,\hat\phi}_T\geq D-\dfrac{1}{\alpha} \log\biggl(\dfrac{{\textrm{d}}\hat{\mathbb{Q}}}{{\textrm{d}}\mathbb{P}}\biggr).\end{equation}

We apply Lemma 2.1 for

$$X\,:\!=\, D-\dfrac{1}{\alpha} \log\biggl(\dfrac{{\textrm{d}}\hat{\mathbb{Q}}}{{\textrm{d}}\mathbb{P}}\biggr).$$

Clearly X satisfies (2.5), so we need to show that for any $\mathbb{Q}\in\mathcal{Q}$ and $M\in \mathcal{M}^{\mathbb{Q}}_{[0,T]}$ we have

(2.16) \begin{equation}\mathbb{E}_{\mathbb{Q}}\biggl[\dfrac{1}{\alpha} \log\biggl(\dfrac{{\textrm{d}}\hat{\mathbb{Q}}}{{\textrm{d}}\mathbb{P}}\biggr)+\Phi_0(M_0-S_0)+\dfrac{1}{2\Lambda}\int_{0}^T |M_t-S_t|^2 \,{\textrm{d}} t\biggr]\geq D.\end{equation}

Choose $\mathbb{Q}\in\mathcal{Q}$ and $M\in \mathcal{M}^{\mathbb{Q}}_{[0,T]}$ . Define $(Z,Y),(\hat Z,\hat Y)\in C$ by

$$Z=\dfrac{{\textrm{d}}\mathbb{Q}}{{\textrm{d}}\mathbb{P}},\quad \hat Z=\dfrac{{\textrm{d}}\hat{\mathbb{Q}}}{{\textrm{d}}\mathbb{P}},\quad Y_t=M_t\mathbb{E}_{\mathbb{P}}[Z\mid \mathcal{F}_t]\quad \text{and}\quad \hat Y_t=\hat M_t\mathbb{E}_{\mathbb{P}}[\hat Z\mid \mathcal{F}_t],\quad t<T.$$

Define the convex function $h\,:\, [0,1]\rightarrow\mathbb{R}_{+}$ by

$$h(\alpha)\,:\!=\, \Psi(\alpha Z+(1-\alpha)\hat Z,\alpha Y+(1-\alpha)\hat Y),\quad \alpha\in [0,1].$$

The function h attains its minimum at $\alpha=0$ , so $h^{\prime}(0{+})\geq 0$ . Again, the monotone convergence theorem allows us to interchange derivative and expectation. Thus

(2.17) \begin{align} 0&\leq h^{\prime}(0{+}) \notag \\ &= \mathbb{E}_{\mathbb{P}}\biggl[\dfrac{1}{\alpha}(Z-\hat Z)\log\hat Z+\Phi_0(Y_0-\hat Y_0)-\Phi_0 S_0(Z_0-\hat Z_0)\biggr]\notag \\&\quad +\dfrac{1}{2\Lambda}\mathbb{E}_{\mathbb{P}}\biggl[\int_{0}^T((Z-\hat Z) S^2_t-2(Y_t-\hat Y_t)S_t)\,{\textrm{d}} t\biggr]\notag \\&\quad +\dfrac{1}{2\Lambda}\mathbb{E}_{\mathbb{P}}\biggl[\dfrac{1}{2\Lambda}\int_{0}^T\dfrac{2 \hat Y_t(Y_t-\hat Y_t)\mathbb{E}_{\mathbb{P}}[\hat Z\mid \mathcal{F}_t]-\mathbb{E}_{\mathbb{P}}[Z-\hat Z\mid \mathcal{F}_t]\hat Y^2_t}{\mathbb{E}^2_{\mathbb{P}}[\hat Z\mid \mathcal{F}_t]}\biggr].\end{align}

Observe that for any $t<T$

$$2Y_t\hat Y_t \mathbb{E}_{\mathbb{P}}[\hat Z\mid \mathcal{F}_t]\leq \mathbb{E}_{\mathbb{P}}[Z\mid \mathcal{F}_t]\hat Y^2_t+\dfrac{Y^2_t \mathbb{E}^2_{\mathbb{P}}[\hat Z\mid \mathcal{F}_t]}{\mathbb{E}_{\mathbb{P}}[Z\mid \mathcal{F}_t]}.$$

This together with (2.17) gives

\begin{align*}0&\leq \mathbb{E}_{\mathbb{P}}\biggl[\dfrac{1}{\alpha}(Z-\hat Z)\log\hat Z+\Phi_0(Y_0-\hat Y_0)-\Phi_0 S_0(Z_0-\hat Z_0)\biggr]\\&\quad +\dfrac{1}{2\Lambda}\mathbb{E}_{\mathbb{P}}\biggl[\int_{0}^T((Z-\hat Z)S^2_t-2(Y_t-\hat Y_t)S_t)\,{\textrm{d}} t\biggr]\\&\quad +\dfrac{1}{2\Lambda}\mathbb{E}_{\mathbb{P}}\biggl[\int_{0}^T\biggl(\dfrac{Y^2_t}{\mathbb{E}_{\mathbb{P}}[Z\mid \mathcal{F}_t]}-\dfrac{\hat Y^2_t}{\mathbb{E}_{\mathbb{P}}[\hat Z\mid \mathcal{F}_t]}\biggr)\,{\textrm{d}} t\biggr],\end{align*}

which is exactly (2.16). We conclude that (2.15) holds true, and thus

(2.18) \begin{equation}\mathbb{E}_{\mathbb{P}}\bigl[{\textrm{e}}^{-\alpha V^{\Phi_0,\hat\phi}_T}\bigr]\leq {\textrm{e}}^{-\alpha D}.\end{equation}

We arrive at the final step of the proof. Choose $\phi\in\mathcal{A}_{\Phi_0}$ . Without loss of generality, assume that

$$\mathbb{E}_{\mathbb{P}}\bigl[{\textrm{e}}^{-\alpha V^{\Phi_0,\phi}_T}\bigr]<\infty,$$

and hence arguments similar to those in the proof of Lemma 2.1 yield

$$\mathbb{E}_{\hat{\mathbb{Q}}}\biggl[\int_{0}^T\phi^2_t \,{\textrm{d}} t \biggr]<\infty.$$

Let us argue that for any $\gamma>0$

(2.19) \begin{align}&\mathbb{E}_{\mathbb{P}}\Bigl[{\textrm{e}}^{-\alpha V^{\Phi_0,\phi}_T}\Bigr]\notag \\ &\quad \geq\alpha\gamma\mathbb{E}_{\hat {\mathbb{Q}}}\biggl[\Phi_0 S_0+\int_{0}^T S_t\phi_t \,{\textrm{d}} t+\dfrac{\Lambda}{2}\int_{0}^T \phi^2_t \,{\textrm{d}} t\biggr]-\mathbb{E}_{\mathbb{P}}\big[\gamma \hat Z(\!\log\!(\gamma \hat Z)-1)\big]\notag \\ &\quad =\alpha\gamma\mathbb{E}_{\hat {\mathbb{Q}}}\biggl[\Phi_0 (S_0-\hat M_0)+\int_{0}^T (S_t-\hat M_t)\phi_t \,{\textrm{d}} t+\dfrac{\Lambda}{2}\int_{0}^T \phi^2_t \,{\textrm{d}} t\biggr]-\gamma(\!\log \gamma-1)-\gamma \mathbb{E}_{\hat {\mathbb{Q}}}[\!\log \hat Z]\notag \\ &\quad \geq\alpha \gamma\mathbb{E}_{\hat{\mathbb{Q}}}\biggl[ \Phi_0(S_0-\hat M_0)-\dfrac{1}{2\Lambda}\int_{0}^T|\hat M_t-S_t|^2 \,{\textrm{d}} t\biggr]-\gamma(\!\log \gamma-1)-\gamma\mathbb{E}_{\hat{\mathbb{Q}}}[\!\log \hat Z].\end{align}

Indeed, the first inequality follows from the simple inequality ${\textrm{e}}^x\geq xy-y(\!\log y-1)$ , $x\in\mathbb{R}$ , $y>0$ . The equality is due to

$$\mathbb{E}_{\hat {\hat{\mathbb{Q}}}}\biggl[\Phi_0 \hat M_0+\int_{0}^T \hat M_t \phi_t \,{\textrm{d}} t\biggr]=0;$$

for this we need the bound $\mathbb{E}_{\hat{\mathbb{Q}}}\bigl[\int_{0}^T\phi^2_t \,{\textrm{d}} t \bigr]<\infty$ . The last inequality follows from the maximization of the quadratic pattern in $\phi$ .

Optimizing (2.19) in $\gamma>0$ , we arrive at

(2.20) \begin{equation}\mathbb{E}_{\mathbb{P}}\bigl[{\textrm{e}}^{-\alpha V^{\Phi_0,\phi}_T}\bigr]\geq {\textrm{e}}^{-\alpha D}.\end{equation}

Since $\phi\in\mathcal{A}_{\Phi_0}$ was arbitrary, from (2.18), (2.20) and the fact that $\bigl(\hat{\mathbb{Q}}\in\mathcal{Q},\hat M\in\mathcal{M}^{\hat{\mathbb{Q}}}_{[0,T]}\bigr)$ is the minimizer from Lemma 2.2, we obtain (2.3). Moreover, note that there is an equality in (2.19) if and only if

$$\phi=\dfrac{\hat M-S}{\Lambda}\quad {\textrm{d}} t\otimes\text{$\mathbb{P}$ a.s.}$$

This yields (2.4) and completes the proof.

Proof. First, from the assumption $\kappa\in (0,{{1}/{(2\alpha\sigma^2 T)}})$ , it follows that $\mathbb{E}_{\mathbb{P}}\big[{\textrm{e}}^{\alpha\mathcal{X}}\big]<\infty$ . Thus we define the probability measure $\tilde{\mathbb{P}}$ by

$$\dfrac{{\textrm{d}}\tilde{\mathbb{P}}}{{\textrm{d}}\mathbb{P}}\,:\!=\, \dfrac{{\textrm{e}}^{\alpha \mathcal{X}}}{\mathbb{E}_{\mathbb{P}}\big[{\textrm{e}}^{\alpha\mathcal{X}}\big]}.$$

Observe that for any probability measure $\mathbb{Q}\sim\mathbb{P}$ we have

$$\mathbb{E}_{\mathbb{Q}}\biggl[\!\log\biggl(\dfrac{{\textrm{d}}\mathbb{Q}}{{\textrm{d}}\tilde{\mathbb{P}}}\biggr)\biggr]=\mathbb{E}_{\mathbb{Q}}\biggl[\!\log\biggl(\dfrac{{\textrm{d}}\mathbb{Q}}{{\textrm{d}}\mathbb{P}}\biggr)-\alpha \mathcal{X}\biggr]+\alpha \log\!\big(\mathbb{E}_{\mathbb{P}}\big[{\textrm{e}}^{\alpha\mathcal{X}}\big]\big).$$

From Hölder’s inequality and Assumption 2.1 it follows that there exists $b>0$ such that

$$\mathbb{E}_{\tilde{\mathbb{P}}}\biggl[\exp\biggl(b\sup_{0 \leq t\leq T} S^2_t\biggr)\biggr]<\infty.$$

Hence, using Theorem 2.1 for the probability measure $\tilde{\mathbb{P}}$ , we obtain

(3.3) \begin{align}&\min_{\phi\in\mathcal{A}_{\Phi_0}}\biggl\{\dfrac{1}{\alpha}\log\mathbb{E}_{\mathbb{P}}\bigl[\exp\bigl(\alpha\bigl(\mathcal{X}- V^{\Phi_0,\phi}_T\bigr)\bigr) \bigr]\biggr\}\notag\\&\quad = \sup_{\mathbb{Q}\in\mathcal{Q}}\sup_{M\in\mathcal{M}^{\mathbb{Q}}_{[0,T)}}\mathbb{E}_{\mathbb{Q}}\biggl[\mathcal{X}-\dfrac{1}{\alpha}\log\biggl(\dfrac{{\textrm{d}}\mathbb{Q}}{{\textrm{d}}\mathbb{P}}\biggr)-\Phi_0(M_0-S_0)-\dfrac{1}{2\Lambda}\int_{0}^T |M_t-S_t|^2 \,{\textrm{d}} t\biggr].\end{align}

Moreover, there exists a unique maximizer $\bigl(\hat{\mathbb{Q}}\in\mathcal{Q},\hat M\in \mathcal{M}^{\hat {\mathbb{Q}}}_{[0,T)}\bigr)$ for the right-hand side of (3.3), and the process given by (2.4) is the unique utility-based optimal hedging strategy.

Observe that by the Markov property of Brownian motion, in order to prove Theorem 3.1 it is sufficient to establish (3.2) for $t=0$ . Thus, in view of (2.4), it remains to establish that

(3.4) \begin{equation}\dfrac{\hat M_0-S_0}{\Lambda}=\dfrac{\big(2\kappa S_0+{{\mu}/{\big(\alpha\sigma^2\big)}}\big)\tanh\!\big(\sqrt\rho T/2\big)-\big(\!\coth\!\big(\sqrt\rho T\big)-2\Lambda\sqrt{\rho}\kappa\big)\Phi_0}{\frac{1}{\sqrt\rho}-4\kappa\Lambda \tanh\!\big(\sqrt\rho T/2\big)}.\end{equation}

To this end, let $\mathbb{Q}\in\mathcal{Q}$ and $M\in\mathcal{M}^{\mathbb{Q}}_{[0,T)}$ . From the Girsanov theorem it follows that there exists a progressively measurable process $\theta\in L^2({\textrm{d}} t\otimes\mathbb{Q})$ such that

$$\mathbb{E}_{\mathbb{Q}}[\!\log\!({\textrm{d}}\mathbb{Q}/{\textrm{d}}\mathbb{P})]=\mathbb{E}_{\mathbb{Q}}\biggl[\int_0^T\theta^2_s\,{\textrm{d}} s\biggr]/2<\infty\quad \text{and}\quad W^{\mathbb{Q}}_t\,:\!=\, W_t-\int_{0}^t\theta_s \,{\textrm{d}} s,\quad t\in [0,T]$$

is a $\mathbb{Q}$ -Brownian motion. By applying the martingale representation theorem, there exists a process $\gamma=(\gamma_t)_{0\leq t< T}$ such that

(3.5) \begin{equation} M_t=M_0+\sigma\int_{0}^t \gamma_s \,{\textrm{d}} W^{\mathbb{Q}}_s, \quad {\textrm{d}} t\otimes\text{$\mathbb{P}$ a.s.} \end{equation}

Moreover, by applying the martingale representation theorem for $\theta_t$ , $t\in [0,T]$ , we conclude that there exist a deterministic function $a_t$ , $t\in [0,T]$ and a jointly measurable process $\beta_{t,s}$ , $0\leq s\leq t\leq T$ such that $\beta_{t,s}$ is $\mathcal{F}_{t\wedge s}$ measurable and

(3.6) \begin{equation} \theta_t=a_t+\int_{0}^t \beta_{t,s}\,{\textrm{d}} W^{\mathbb{Q}}_s, \quad {\textrm{d}} t\otimes\text{$\mathbb{P}$ a.s.} \end{equation}

Set

$$ \nu_t\,:\!=\, \mu t+\sigma\int_{0}^t a_s \,{\textrm{d}} s,\quad t\in [0,T]\quad \text{and}\quad l_{t,s}\,:\!=\, \int_{s}^t \beta_{u,s}\,{\textrm{d}} u,\quad 0\leq s\leq t\leq T. $$

From Fubini’s theorem, (3.1), and (3.6),

(3.7) \begin{equation} S_t=S_0+\nu_t+\sigma \int_{0}^t (1+l_{t,s})\,{\textrm{d}} W^{\mathbb{Q}}_s, \quad t\in [0,T]. \end{equation}

Given the probability measure $\mathbb{Q}$ , we are looking for a martingale $\tilde M\in\mathcal{M}^{\mathbb{Q}}_{[0,T)}$ which maximizes the right-hand side of (3.3). By combining (3.5), (3.7) and applying the Itô isometry and Fubini’s theorem, we obtain

\begin{align*}&\mathbb{E}_{{\mathbb{Q}}}\biggl[\Phi_0( M_0-S_0)+\dfrac{1}{2\Lambda}\int_{0}^T|{M}_t-S_t|^2 \,{\textrm{d}} t\biggr]\\&\quad =\Phi_0(M_0-S_0)+\dfrac{1}{2\Lambda}\int_{0}^T (M_0-S_0-\nu_t )^2 \,{\textrm{d}} t+\dfrac{\sigma^2}{2\Lambda}\mathbb{E}_{\mathbb{Q}}\biggl[\int_{0}^T \int_{s}^T (\gamma_s-1- l_{t,s})^2 \,{\textrm{d}} t \,{\textrm{d}} s\biggr].\end{align*}

Given a and $\beta$ , we are looking for $\hat M_0$ and $\hat\gamma$ which minimize the above right-hand side. Observe that the right-hand side is a quadratic function in $M_0$ and $\gamma_s$ , $s\in [0,T)$ . Hence we obtain that the minimizer is unique and given by

(3.8) \begin{equation}\hat M_0=S_0+\dfrac{1}{T}\int_{0}^T \nu_t \,{\textrm{d}} t-\dfrac{\Phi_0\Lambda}{T}\end{equation}

and

(3.9) \begin{equation}\hat\gamma_s=1+\dfrac{1}{T-s}\int_{s}^T l_{t,s} \,{\textrm{d}} t, \quad s<T.\end{equation}

Finally, we compute the optimal $\nu$ . From the Itô isometry, Fubini’s theorem, and (3.6)–(3.7), we have

$$\mathbb{E}_{\mathbb{Q}}[S^2_T]=(S_0+\nu_T)^2+\sigma^2 \mathbb{E}_{\mathbb{Q}}\biggl[\int_{0}^T\int_{s}^T (1+l_{t,s})^2 \,{\textrm{d}} t \,{\textrm{d}} s\biggr]$$

and

$$\mathbb{E}_{\mathbb{Q}}\biggl[\!\log\biggl(\dfrac{{\textrm{d}}\mathbb{Q}}{{\textrm{d}}\mathbb{P}}\biggr)\biggr]=\dfrac{1}{2}\mathbb{E}_{\mathbb{Q}}\biggl[\int_0^T\theta^2_s\,{\textrm{d}} s\biggr]=\dfrac{1}{2}\biggl(\int_{0}^Ta^2_t \,{\textrm{d}} t+\mathbb{E}_{\mathbb{Q}}\biggl[\int_{0}^{T}\int_{s}^T \beta^2_{t,s} \,{\textrm{d}} t \,{\textrm{d}} s\biggr]\biggr).$$

These equalities together with (3.8)–(3.9) give

(3.10) \begin{align}&\sup_{M\in\mathcal{M}^{\mathbb{Q}}_{[0,T)}}\mathbb{E}_{\mathbb{Q}}\biggl[\mathcal{X}-\dfrac{1}{\alpha}\log\biggl(\dfrac{{\textrm{d}}\mathbb{Q}}{{\textrm{d}}\mathbb{P}}\biggr)-\Phi_0( M_0-S_0)-\dfrac{1}{2\Lambda}\int_{0}^T | M_t-S_t|^2 \,{\textrm{d}} t\biggr]\notag \\&\quad =\mathbb{E}_{\mathbb{Q}}\biggl[\mathcal{X}-\dfrac{1}{\alpha}\log\biggl(\dfrac{{\textrm{d}}\mathbb{Q}}{{\textrm{d}}\mathbb{P}}\biggr)-\Phi_0( \hat M_0-S_0)-\dfrac{1}{2\Lambda}\int_{0}^T |\hat M_t-S_t|^2 \,{\textrm{d}} t\biggr]\notag \\&\quad =I+\mathbb{E}_{\mathbb{Q}} \biggl[\int_{0}^T J_s \,{\textrm{d}} s\biggr],\end{align}

where

$$I=\kappa(S_0+\nu_T)^2-\dfrac{1}{2\alpha}\int_{0}^Ta^2_t \,{\textrm{d}} t+\dfrac{1}{2\Lambda}\biggl(\dfrac{1}{T}\biggl(\Phi_0\Lambda-\int_{0}^T\nu_t \,{\textrm{d}} t\biggr)^2-\int_{0}^T \nu^2_t \,{\textrm{d}} t\biggr)$$

and

$$J_s=\kappa\sigma^2\int_{s}^T (1+l_{t,s})^2 \,{\textrm{d}} t-\dfrac{1}{2\alpha}\int_{s}^T\beta^2_{t,s} \,{\textrm{d}} t+\dfrac{1}{2\Lambda}\biggl(\dfrac{1}{T-s}\biggl(\int_{s}^T l_{t,s} \,{\textrm{d}} t\biggr)^2-\int_{s}^T l^2_{t,s} \,{\textrm{d}} t\biggr).$$

From Proposition 4.1 we conclude that the optimal $\nu$ satisfies (4.2). Hence from (3.8) we obtain (3.4) and complete the proof.

Proof. The proof will be done in two steps. First we will solve the optimization problem (4.1) for the case where $\delta_T$ and $\int_{0}^T\delta_t \,{\textrm{d}} t$ are given. Then we will find the optimal $\delta_T$ and $\int_{0}^T\delta_t \,{\textrm{d}} t$ .

Thus, for any x, y, let $\Gamma_{x,y}\subset \Gamma$ be the set of all functions $\delta\in\Gamma$ that satisfy $\delta_T=x$ and $\int_{0}^T\delta_t \,{\textrm{d}} t=y$ . Consider the minimization problem

$$\min_{\delta\in \Gamma_{x,y}} \int_{0}^T H(\dot \delta_t,\delta_t)\,{\textrm{d}} t,$$

where

$$H(u,v)\,:\!=\, \dfrac{1}{2\alpha\sigma^2}(u-\mu)^2+\dfrac{1}{2\Lambda} v^2\quad \text{for $u,v\in\mathbb{R}$.}$$

This optimization problem is convex, and thus it has a unique solution which has to satisfy the Euler–Lagrange equation (for details see [Reference Gelfand and Fomin13])

$$\dfrac{{\textrm{d}}}{{\textrm{d}} t}\dfrac{\partial H}{\partial \dot \delta_t}=\lambda+\dfrac{{\textrm{d}}}{{\textrm{d}} t}\dfrac{\partial H}{\partial \delta_t}$$

for some constant $\lambda>0$ (Lagrange multiplier due to the constraint $\int_{0}^T \delta_t \,{\textrm{d}} t=y$ ). Thus the optimizer solves the ODE $\ddot{\delta}_t-\rho \delta\equiv \operatorname{const.}$ (recall the risk–liquidity ratio $\rho={{\alpha\sigma^2}/{\Lambda}}$ ). From the standard theory it follows that

(4.3) \begin{equation}\delta_t=c_1\sinh\!\big(\sqrt\rho t\big)+c_2\sinh\!\big(\sqrt\rho(T-t)\big)+c_3\end{equation}

for some constants $c_1,c_2,c_3$ . From the three constraints $\delta_0=0$ , $\delta_T=x$ , and $\int_{0}^T \delta_t \,{\textrm{d}} t=y$ we obtain

(4.4) \begin{equation} c_1=\dfrac{x-c_3}{\sinh\!\big(\sqrt\rho T\big)}, \quad c_2=-\dfrac{c_3}{\sinh\!\big(\sqrt\rho T\big)}, \quad c_3=\dfrac{\sqrt\rho y-x\tanh\!\big(\sqrt\rho T/2\big)}{\sqrt\rho T-2\tanh\!\big(\sqrt\rho T/2\big)}.\end{equation}

We argue that

(4.5) \begin{align} &\rho\int_{0}^T \delta^2_t \,{\textrm{d}} t+\int_{0}^T \dot{\delta}^2_t\,{\textrm{d}} t \notag \\ &\quad =\rho\int_{0}^T ((\delta_t-c_3)+c_3)^2 \,{\textrm{d}} t+\int_{0}^T \dot{\delta}^2_t\,{\textrm{d}} t\notag \\ &\quad =\dfrac{\sqrt\rho}{2}\bigl(c^2_1+c^2_2\bigr)\sinh\!\big(2\sqrt\rho T\big)-2c_1c_2\sqrt\rho\sinh\!\big(\sqrt\rho T\big)-\rho c^2_3 T+2\rho c_3 y\notag \\ &\quad =\sqrt{\rho}x^2\coth\!\big(\sqrt\rho T\big)+2\sqrt\rho c_1 c_2\sinh\!\big(\sqrt\rho T\big)\big(\cosh\big(\sqrt\rho T\big)-1\big)-\rho c^2_3 T+2\rho c_3 y\notag \\ &\quad =\sqrt{\rho}x^2\coth\!\big(\sqrt\rho T\big)+\big(2\sqrt\rho \tanh\!\big(\sqrt \rho T/2\big)-\rho T\big)c^2_3 +2\big(\rho y-\sqrt\rho \tanh\!\big(\sqrt \rho T/2\big)x\big)c_3\notag \\ &\quad =\sqrt{\rho}\biggl(x^2\coth\!\big(\sqrt\rho T\big)+\dfrac{\big(x\tanh\!\big(\sqrt\rho T/2\big)-\sqrt\rho y\big)^2}{\sqrt\rho T-2\tanh\!\big(\sqrt\rho T/2\big)}\biggr).\end{align}

Indeed, the first equality is obvious. The second equality follows from (4.3) and simple computations. The third equality is due to

$$c_1-c_2=\frac{x}{\sinh\!\big(\sqrt\rho T\big)}.$$

The fourth equality is due to

$$c_1c_2=\dfrac{c^2_3-xc_3}{\sinh^2\!\big(\sqrt\rho T\big)}.$$

The last equality follows from substituting $c_3$ .

By combining (4.5) and the simple equality

$$\dfrac{1}{2\alpha\sigma^2}\int_{0}^T (\dot \delta_t-\mu)^2 \,{\textrm{d}} t +\dfrac{1}{2\Lambda}\int_{0}^T \dot \delta_t^2 \,{\textrm{d}} t=\dfrac{\mu^2 T}{2\alpha\sigma^2}-\dfrac{\mu x}{\alpha\sigma^2}+\dfrac{1}{2\alpha\sigma^2}\biggl(\rho\int_{0}^T \delta^2_t \,{\textrm{d}} t+\int_{0}^T \dot{\delta}^2_t\,{\textrm{d}} t\biggr),$$

we obtain

(4.6) \begin{align}&\min_{\delta\in \Gamma_{x,y}} \int_{0}^T H(\dot \delta_t,\delta_t)\,{\textrm{d}} t\notag \\&\quad =\dfrac{\mu^2 T}{2\alpha\sigma^2}-\dfrac{\mu x}{\alpha\sigma^2}+\dfrac{1}{2\Lambda\sqrt\rho}\biggl(x^2\coth\!\big(\sqrt\rho T\big)+\dfrac{\big(x\tanh\!\big(\sqrt\rho T/2\big)-\sqrt\rho y\big)^2}{\sqrt\rho T-2\tanh\!\big(\sqrt\rho T/2\big)}\biggr).\end{align}

We arrive at the final step of the proof. In view of (4.6), the optimization problem (4.1) is reduced to finding $x\,:\!=\, \delta_T$ and $y\,:\!=\, \int_{0}^T\delta_t \,{\textrm{d}} t$ , which maximize the quadratic form

$$A x^2+B y^2+2C xy + \eta x+\theta y,$$

where

\begin{align*}A&\,:\!=\, \kappa-\dfrac{1}{2\Lambda\sqrt\rho}\biggl(\!\coth\!\big(\sqrt\rho T\big)+\dfrac{\tanh^2\big(\sqrt\rho T/2\big)}{\sqrt\rho T-2 \tanh\!\big(\sqrt\rho T/2\big)}\biggr)\\&=-\dfrac{\sqrt\rho T\coth\!\big(\sqrt\rho T\big)+4\Lambda\sqrt\rho\kappa \tanh\!\big(\sqrt\rho T/2\big)-1-2\Lambda\rho T\kappa}{{2\Lambda\sqrt\rho\big(\sqrt\rho T-2 \tanh\!\big(\sqrt\rho T/2\big)\big)}},\\B&\,:\!=\, \dfrac{1}{2\Lambda T}-\dfrac{\sqrt\rho}{2\Lambda\big(\sqrt\rho T-2 \tanh\!\big(\sqrt\rho T/2\big)\big)}=-\dfrac{\tanh\!\big(\sqrt\rho T/2\big)}{\Lambda T \big(\sqrt\rho T-2 \tanh\!\big(\sqrt\rho T/2\big)\big)},\\C&\,:\!=\, \dfrac{\tanh\!\big(\sqrt\rho T/2\big)}{2\Lambda\big(\sqrt\rho T-2 \tanh\!\big(\sqrt\rho T/2\big)\big)}, \quad \eta\,:\!=\, 2\kappa S_0+\dfrac{\mu}{\alpha\sigma^2} \quad \text{and} \quad \theta\,:\!=\, -\dfrac{\Phi_0}{T}.\end{align*}

Simple computations give

\begin{align*}AB-C^2&=-\dfrac{\kappa\tanh\!\big(\sqrt\rho T/2\big)}{\Lambda T\big(\sqrt\rho T-2 \tanh\!\big(\sqrt\rho T/2\big)\big)}\\&\quad +\dfrac{\coth\!\big(\sqrt\rho T\big)\tanh\!\big(\sqrt\rho T/2\big)}{2\sqrt\rho \Lambda^2 T \big(\sqrt\rho T-2 \tanh\!\big(\sqrt\rho T/2\big)\big)}-\dfrac{\tanh^2\big(\sqrt\rho T/2\big)}{4\sqrt\rho \Lambda^2 T \big(\sqrt\rho T-2 \tanh\!\big(\sqrt\rho T/2\big)\big)}\\&=\dfrac{{{1}/{(4\sqrt\rho\Lambda)}}-\kappa\tanh\!\big(\sqrt\rho T/2\big)}{\Lambda T\big(\sqrt\rho T-2 \tanh\!\big(\sqrt\rho T/2\big)\big)}.\end{align*}

From the inequality $z>\tanh\!(z)z$ , $z>0$ and the assumption $\kappa\in (0,{{1}/{(2\alpha\sigma^2 T)}})$ , we obtain $B<0$ and $AB-C^2>0$ . Thus the above quadratic form has a unique maximizer

$$(\bar x,\bar y)\,:\!=\, \dfrac{1}{2(AB-C^2)}(C\theta-B\eta,C\eta-A\theta).$$

We conclude that the optimization problem (4.1) has a unique solution which is given by (4.3)–(4.4) for $(x,y)\,:\!=\, (\bar x,\bar y)$ . Moreover, direct computations yield

$$\dfrac{\bar y}{T}=\dfrac{\big(2\kappa S_0+{{\mu}/{\big(\alpha\sigma^2\big)}}\big)\tanh\!\big(\sqrt{\rho} T/2\big)- \big(\!\coth\!\big(\sqrt\rho T\big)-2\Lambda\sqrt\rho\kappa\big)\Phi_0}{\dfrac{1}{\sqrt{\rho}\Lambda}-4 \kappa \tanh\!\big(\sqrt{\rho} T/2\big)}+\dfrac{\Phi_0\Lambda}{T},$$

and (4.2) follows.