2. Setup and main result

On a complete filtered probability space $(\Omega, \mathcal F, \mathbb P, (\mathcal F_t)_{0\leq t\lt\infty})$ , let X be the solution of a stochastic differential equation

(1) \begin{equation} {\mathrm{d}} X_t = r X_t\,{\mathrm{d}} t + \sigma(X_t)X_t\,{\mathrm{d}} W_t\end{equation}

with initial condition $x_0\in(0,\infty)$ . Here, $r\gt0$ is the constant interest rate and $\sigma(\!\cdot\!)\;:\;(0,\infty)\to(0,\infty)$ is a given function. We will assume that the local volatility function $\sigma(\!\cdot\!)$ is continuous and bounded at infinity so that (Reference Alòs and García-Lorite1) has a unique strong solution up to the first hitting time of 0; we also assume that X is regular so that any point $y\in(0,\infty)$ can be reached in finite time, and we let 0 be an absorbing state so that if X reaches 0 in finite time, then it remains at 0 at all times.

For any given strike price $K\geq 0$ , consider the price $P(K) \;:\!=\; \sup_{\tau}\mathbb E[{\mathrm{e}}^{-r\tau}(K-X_\tau)^+]$ of a perpetual American put option with strike K, where the supremum is taken over all stopping times. Note here that the discounting rate r is the same as in the drift of X in (Reference Alòs and García-Lorite1), so by standard arbitrage theory, P(K) is the unique arbitrage-free price of the put option. The following result is immediate (and well known, cf. [Reference Ekström and Hobson8]).

2.1. Constant volatility and implied volatility

In the special case of constant volatility, $\sigma(\!\cdot\!)\equiv \gamma\gt0$ , it is well known that $P_{\gamma}(K)\;:\!=\;P(K)$ is given by

(2) \begin{equation} P_\gamma(K)=\left\{ \begin{array}{l@{\quad}l} (K-z(K))({z_\gamma(K)}/{x_0})^{2r/\gamma^2}, & K\lt\hat K_\gamma , \\ K-x_0, & K\geq \hat K_\gamma, \end{array}\right.\end{equation}

where $z_\gamma(K) \;:\!=\; {2rK}/({2r+\gamma^2})$ and $\hat K_\gamma \;:\!=\; {(2r+\gamma^2)K}/{2r}$ . Moreover,

\[\tau_{z_\gamma(K)}\;:\!=\;\inf\{t\geq 0 \;:\; X_t\leq z_\gamma(K)\}\]

is an optimal stopping time, so that $P_\gamma(K) = \mathbb E[{\mathrm{e}}^{-r\tau_{z_\gamma(K)}}(K-X_{\tau_{z_\gamma(K)}})^+]$ . Using the explicit formula (Reference Alvarez2), it is straightforward to check that the price is monotonically increasing in volatility: if $P_{\gamma_1}(K)$ is the price corresponding to a volatility $\gamma_1\gt0$ and $P_{\gamma_2}(K)$ is the price corresponding to a volatility $\gamma_2\gt0$ , then $\gamma_1\leq \gamma_2\implies P_{\gamma_1}(\!\cdot\!)\leq P_{\gamma_2}(\!\cdot\!)$ . Furthermore, if $P_{\gamma_1}(K)\gt (K-x_0)^+$ , then $P_{\gamma_2}(K)\gt P_{\gamma_1}(K)$ for $\gamma_2\gt\gamma_1$ , i.e. the price is strictly increasing in volatility in the continuation region.

Remark 1. Monotonicity of perpetual options with respect to volatility holds true for a much larger class of local volatility models; see [Reference Alvarez2] and [Reference Ekström7].

2.2. Main result

Now assume that a volatility function $\sigma(\!\cdot\!)$ is given, and let $P(\!\cdot\!)$ be the corresponding option price; recall that $P(\!\cdot\!)$ has the properties specified in Lemma 1. We write

\[\hat K\;:\!=\;\inf\{K\geq 0 \;:\; P(K)=(K-x_0)^+\}.\]

Then $\hat K\in(x_0,\infty]$ , $P(K)\gt (K-x_0)^+$ for $K\lt\hat K$ , and $P(K)=(K-x_0)^+$ for $K\geq \hat K$ . By the strict monotonicity of the function $P_\gamma (K)$ with respect to $\gamma$ (recall that $P_\gamma(K)$ is the price of a put option if the volatility is a constant $\gamma$ ), for each $K\in(0,\hat K)$ there exists a unique $\gamma=\gamma(K)\gt0$ such that $P(K)=P_\gamma(K)$ . This value is referred to as the implied volatility.

Remark 2. In contrast to the standard notion of implied volatility (defined by inverting the Black–Scholes formula for options with a finite horizon), which is well defined for all strikes, the implied volatility $\gamma(\!\cdot\!)$ is uniquely determined only for $K\lt\hat K$ .

We now present our main result.

Theorem 1. Let a volatility function $\sigma(\!\cdot\!)$ be given, let $P(\!\cdot\!)$ be the corresponding option value, and let $\hat K=\inf\{K\geq 0 \;:\; P(K)=(K-x_0)^+\}$ .

1. (i) If $\sigma(\!\cdot\!)\;:\; (0,\infty)\to(0,\infty)$ is non-increasing, then the implied volatility $\gamma(\!\cdot\!)$ is non-increasing on $(0,\hat K)$ .

2. (ii) If $\sigma(\!\cdot\!)\;:\; (0,\infty)\to(0,\infty)$ is non-decreasing, then the implied volatility $\gamma(\!\cdot\!)$ is non-decreasing on $(0,\hat K)$ .

A geometric approach to implied volatility is provided in Section 3, and the proof of Theorem 1 is then given in Section 4.

3. A geometric approach to option pricing and implied volatility

While the classical method to determine the price P (along with an optimal stopping time) would be to make a Markovian embedding by varying the initial point $x_0$ and then study an associated free-boundary problem, it is more convenient for our purposes to follow a geometric approach as presented in [Reference Ekström and Hobson8]. As we will see, this geometric approach is well aligned with the inversion of the pricing function in (Reference Alvarez2) that is needed to calculate the implied volatility.

Let $\varphi\;:\;(0,\infty)\to(0,\infty)$ be a non-increasing solution of

(3) \begin{equation} \frac{\sigma^2(x)x^2}{2}\varphi''(x)+rx\varphi'(x)-r\varphi(x)=0.\end{equation}

This function is unique up to multiplication with a positive constant, cf. [Reference Borodin and Salminen5, pp. 18, 19]; we will for simplicity (and without loss of generality) assume that $\varphi(x_0)=1$ . It is straightforward to check that

\begin{equation*} \varphi(x) = Dx\int_x^\infty\frac{1}{y^2} \exp\!\bigg\{{-}\int_{x_0}^y\frac{2r}{z\sigma^2(z)}\,{\mathrm{d}} z\bigg\}\,{\mathrm{d}} y,\end{equation*}

where

\[D=\frac{1}{x_0\int_{x_0}^\infty y^{-2} \exp\!\big\{{-}\int_{x_0}^y{2r}/({z\sigma^2(z)})\,{\mathrm{d}} z\big\}\,{\mathrm{d}} y}.\]

Moreover, straightforward differentiation yields

\[\varphi''(x)=\frac{2rD}{\sigma^2(x)x^2}\exp\!\bigg\{{-}\int_{x_0}^x\frac{2r}{z\sigma^2(z)}\,{\mathrm{d}} z\bigg\}\gt0,\]

so $\varphi$ is strictly convex. Furthermore, denoting by $\tau_z\;:\!=\;\inf\{t\geq 0\;:\; X_t\leq z\}$ the first passage time below a level $z\in(0,\infty)$ , we have, for any $x\geq z$ , $\mathbb E_x[{\mathrm{e}}^{-r\tau_z}] = {\varphi(x)}/{\varphi(z)}$ . In particular, using $\varphi(x_0)=1$ ,

(4) \begin{equation} \mathbb E_{x_0}[{\mathrm{e}}^{-r\tau_z}] = \frac{1}{\varphi(z)}\end{equation}

for $z\in(0,x_0]$ . Moreover, the above formulae also extend to $z=0$ as $\varphi(0)=\infty$ in the case when 0 is unattainable.

The following result was provided in [Reference Ekström and Hobson8]; for completeness of the presentation, we include its proof.

Proposition 1. The value function $P\;:\;[0,\infty)\to[0,\infty)$ satisfies

\begin{equation*} P(K)=\sup_{z\leq K\wedge x_0}\frac{K-z}{\varphi(z)}. \end{equation*}

Proof. Denote the required right-hand side by

\[\hat P(K)\;:\!=\;\sup_{z\leq K\wedge x_0}\frac{K-z}{\varphi(z)}.\]

We clearly have

\begin{equation*} P(K) = \sup_{\tau}\mathbb E[{\mathrm{e}}^{-r\tau}(K-X_\tau)^+] \geq \sup_{z\leq K\wedge x_0}\mathbb E[{\mathrm{e}}^{-r\tau_z}(K-X_{\tau_z})^+] = \sup_{z\leq K\wedge x_0} \frac{K-z}{\varphi(z)} = \hat P(K), \end{equation*}

where the second equality uses that $X_{\tau_z}=z$ on $\{\tau_z \lt\infty\}$ , $\mathbb P$ -almost surely, and (Reference Berestycki, Busca and Florent4).

For the reverse inequality, first assume that $K\leq K'\;:\!=\;-x_0/\varphi'(x_0)$ . For such K, we have

\[\hat P(K)=\sup_{z}\frac{K-z}{\varphi(z)},\]

so that $\hat P(K)\varphi(z)\geq (K-z)^+$ . Consequently,

\begin{equation*} P(K) = \sup_{\tau}\mathbb E[{\mathrm{e}}^{-r\tau}(K-X_\tau)^+] \leq \hat P(K)\sup_\tau\mathbb E[{\mathrm{e}}^{-r\tau}\varphi(X_\tau)] \leq \hat P(K)\varphi(x_0) = \hat P(K), \end{equation*}

where the second inequality uses that $\{{\mathrm{e}}^{-rt}\varphi(X_t),\,0 \leq t \leq \infty\}$ is a supermartingale.

Finally, the above argument gives that $P(K')=(K'-x_0)$ ; hence $P(K)=(K-x_0) \leq \hat P(K)$ for all $K\geq K'$ by Lemma 1, which completes the proof.

Remark 3. For a given K, the convexity of $\varphi$ yields the existence of a unique straight line through the point (K, 0) that is tangent to the graph $\{(z,\varphi(z)), z\leq K\wedge x_0\}$ . Denoting by $(z(K),\varphi(z(K)))$ the tangent point (which is unique by the strict convexity of $\varphi$ ), the price P(K) is given by $P(K)=(K-z(K))/\varphi(z(K))$ . Thus, P(K) equals the negative reciprocal of the slope of this tangent line, cf. Figure 1.

Figure 1. For a given $K \leq \hat K$ , the put price P(K) equals the negative reciprocal of the slope of the tangent line to $\varphi$ which passes through (K, 0).

Moreover, z(K) is non-decreasing in K, and it follows from the proof of Proposition 1 that $\tau_{z(K)}$ is an optimal stopping time.

We now return to the problem of determining the implied volatility, assuming that market prices are calculated using a local volatility function $\sigma(\!\cdot\!)$ with corresponding function $\varphi(\!\cdot\!)$ as above. For a given $K\gt0$ , let $l_K$ be the unique straight line through (K, 0) that is tangent to $\{(z,\,\varphi(z)),z\leq K\wedge x_0\}$ . Denoting the tangent point by $(z(K),\varphi(z(K)))$ , we have

\[l_K(z)=-\frac{\varphi(z(K))}{K-z(K)}(z-K).\]

To describe how to determine the implied volatility in the local volatility model $\sigma(\!\cdot\!)$ , note that, for a given constant volatility $\gamma\gt0$ , the decreasing positive solution of $\frac12\gamma^2x^2f^{\prime\prime}(x)+rxf^{\prime}(x)-rf(x)=0$ on $(0,\infty)$ , also imposing the normalizing condition $f(x_0)=1$ , is given by

(5) \begin{equation} f(x)\;:\!=\;f_\gamma(x)\;:\!=\;\bigg(\frac{x}{x_0}\bigg)^{-2r/\gamma^2}.\end{equation}

Clearly, on $(0,x_0)$ , $f_\gamma$ is strictly decreasing in $\gamma$ . Consequently, given a strike price $K\gt0$ and the corresponding tangent line $l_K(\!\cdot\!)$ (constructed from $\sigma(\!\cdot\!)$ and $\varphi(\!\cdot\!)$ as above) with $l_K(x_0)\lt 1$ (equivalently, $z(K)\lt x_0$ ), there exists a unique $\gamma=\gamma(K)$ such that $l(\!\cdot\!)$ is also a tangent line to $f_\gamma(\!\cdot\!)$ . By Proposition 1, $\gamma(K)$ is the implied volatility of the option.

4. Monotone local volatility functions

In this section we provide a result regarding the function $\varphi$ corresponding to non-increasing local volatility functions (Proposition 2). For its proof we will use a monotonicity result (Lemma 2) with respect to the local volatility functions; its proof is adapted from a similar situation in [Reference Ekström7]. Using Proposition 2, the proof of Theorem 1 then follows.

Lemma 2. Assume that $\sigma_i\;:\;(0,\infty)\to(0,\infty)$ , $i=1,2$ , satisfy $\sigma_1(\!\cdot\!)\leq \sigma_2(\!\cdot\!)$ . Let $h_i$ , $i=1,2$ , be the solutions of

\begin{align*}\left\{ \begin{array}{l@{\quad}l} \displaystyle \frac12\sigma_i^2(x)x^2h''_{\!\!\!i}(x) + rxh'_{\!\!i}(x) - rh_i(x) = 0, & x\in(a,b), \\[10pt] h_i(a) = A,\ h_i(b) = B, \end{array}\right.\end{align*}

for $0\lt a\lt b\lt\infty$ and $A,B\geq 0$ . If ${A}/{a} \geq {B}/{b}$ , then $h_1(\!\cdot\!)\leq h_2(\!\cdot\!)$ on [a, b].

Proof. Denoting by $\varphi_1$ the decreasing fundamental solution corresponding to $\sigma_1$ , the general solution of the ordinary differential equation for $h_1$ is $h_1(x)= Cx + D \varphi_1(x)$ , where C and D are constants. Imposing the boundary conditions $h_1(a)=A$ and $h_2(b)=B$ yields

\[D=\frac{bA-aB}{b\varphi_1(a)-a\varphi_1(b)},\]

which is positive by assumption. Consequently, the function $h_1$ is convex.

Next, for $x\in(a,b)$ , let $X^2$ be the solution of ${\mathrm{d}} X^2_t = rX^2_t\,{\mathrm{d}} t + \sigma_2(X^2_t)X^2_t\,{\mathrm{d}} W_t$ with $X^2_0=x$ . By Itô’s formula, the process $Y_t \;:\!=\; {\mathrm{e}}^{-rt}h_1(X_t^2)$ satisfies

\begin{align*} dY_t & = {\mathrm{e}}^{-rt}\Big(\frac12\sigma^2_2(X^2_t)(X^2_t)^2 h''_{\!\!\!l}(X_t^2) + rX^2_th'_{\!\!l}(X^2_t) - rh_1(X_t^2)\Big)\,{\mathrm{d}} t + {\mathrm{e}}^{-rt}\sigma^2(X^2_t)X^2_th'_{\!\!l}(X^2_t)\,{\mathrm{d}} W_t \\[5pt] & = {\mathrm{e}}^{-rt}\frac12(\sigma^2_2(X^2_t)-\sigma^2_1(X^2_t)) (X^2_t)^2h''_{\!\!\!l}(X_t^2)\,{\mathrm{d}} t + {\mathrm{e}}^{-rt}\sigma^2(X^2_t)X^2_th'_{\!\!l}(X^2_t)\,{\mathrm{d}} W_t \end{align*}

for $t\leq \tau_{ab}\;:\!=\;\inf\{s\geq 0\;:\; X^2_s\notin(a,b)\}$ . Since $0\lt\sigma_1\leq\sigma_2$ and $h_1$ is convex, $Y_t$ is a submartingale. Similarly, the process ${\mathrm{e}}^{-rt}h_2(X_t^2)$ is a martingale. Consequently, optional sampling gives

\[h_1(x)\leq \mathbb E_x\big[{\mathrm{e}}^{-r\tau_{ab}}h_1(X^2_{\tau_{ab}})\big] = \mathbb E_x\big[{\mathrm{e}}^{-r\tau_{ab}}h_2(X^2_{\tau_{ab}})\big] = h_2(x),\]

which completes the proof.

First, assume that $\gamma\leq\sigma(x_0)$ . Then $\gamma\leq \sigma(x)$ for all $x\in[a,x_0]$ , and Lemma 2 immediately yields $f_\gamma(\!\cdot\!)\leq \varphi(\!\cdot\!)$ on $[a,x_0]$ .

Second, assume that $\sigma(a)\leq \gamma$ . If $\gamma=\sigma(x_0)$ , then $\sigma(\!\cdot\!)\equiv \gamma$ on $[a,x_0]$ , and $\varphi(\!\cdot\!)\equiv f_\gamma(\!\cdot\!)$ on $[a,x_0]$ . Therefore, we may assume that $\gamma\gt\sigma(x_0)$ . Note that $\sigma(\!\cdot\!)\leq \gamma$ on $[a,\infty)$ , so Lemma 2 and Remark 5 give $\varphi(\!\cdot\!)\leq f_\gamma(\!\cdot\!)$ on $[a,\infty)$ . Since $f_\gamma(x_0)=1=\varphi(x_0)$ and $f_\gamma(\!\cdot\!)\geq \varphi(\!\cdot\!)$ on $[a,\infty)$ , we then must have $f^\prime_\gamma(x_0)=\varphi'(x_0)$ and $f^{\prime\prime}_\gamma(x_0)\geq \varphi''(x_0)$ . Therefore,

\begin{align*} 0 = \frac12\sigma^2(x_0)x_0^2\varphi''(x_0)+rx_0\varphi'(x_0)-r\varphi(x_0) & \leq \frac12\sigma^2(x_0)x_0^2f^{\prime\prime}_\gamma(x_0)+rx_0f^{\prime}_\gamma(x_0)-rf_\gamma(x_0) \\[5pt] & \lt \frac12\gamma^2x_0^2f^{\prime\prime}_\gamma(x_0)+rx_0f^{\prime}_\gamma(x_0)-rf_\gamma(x_0) = 0, \end{align*}

which shows that the point $a\in(0,x_0)$ with $f_\gamma(a)=\varphi(a)$ cannot exist.

Third, assume that $\sigma(x_0) \lt \gamma \lt \sigma(a)$ . We first show that $f_\gamma(\!\cdot\!)\lt\varphi(\!\cdot\!)$ in a left-neighborhood $(x_0-\varepsilon,x_0)$ of $x_0$ (for some $\varepsilon\gt0$ ). To see this, note that $\sigma(\!\cdot\!)\leq \gamma$ on $(x_0,\infty)$ , so Lemma 2 yields $\varphi(\!\cdot\!)\leq f_\gamma(\!\cdot\!)$ on $(x_0,\infty)$ . Since $\varphi(x_0)=1=f_\gamma(x_0)$ , we have $\varphi'(x_0)\leq f^{\prime}_\gamma(x_0)$ ; consequently,

\begin{align*} 0 = \frac12\sigma^2(x_0)x_0^2\varphi''(x_0)+rx_0\varphi'(x_0)-r\varphi(x_0) & \lt \frac12\gamma^2x_0^2\varphi''(x_0)+rx_0f^{\prime}_\gamma(x_0)-rf_\gamma(x_0) \\[5pt] & = \frac12\gamma^2x_0^2(\varphi''(x_0)-f^{\prime\prime}_\gamma(x_0)), \end{align*}

so $\varphi''(x_0)\gt f^{\prime\prime}_\gamma(x_0)$ . This inequality, together with $\varphi(x_0)=f_\gamma(x_0)$ and $\varphi'(x_0)\leq f^{\prime}_\gamma(x_0)$ , imply that $f_\gamma(\!\cdot\!)\lt\varphi(\!\cdot\!)$ in a left-neighborhood of $x_0$ .

If the set $\{x\in(a,x_0)\;:\; f_\gamma(x)\gt\varphi(x)\}$ is empty, then $f_\gamma(\!\cdot\!)\leq \varphi(\!\cdot\!)$ on $[a,x_0]$ and there is nothing to prove. Instead, assume that $f_\gamma(x)\gt\varphi(x)$ for some $x\in(a,x_0)$ , and let $c\;:\!=\;\sup\{x\in(a,x_0)\;:\; f_\gamma(x)\gt\varphi(x)\}$ . Since $f_\gamma(\!\cdot\!)\lt\varphi(\!\cdot\!)$ in a left-neighborhood of $x_0$ , we then have $c\in(a,x_0)$ . Also note that, by continuity, $f_\gamma(c)=\varphi(c)$ .

Now, if $\gamma\geq\sigma(c)$ , then $\gamma\geq \sigma(\!\cdot\!)$ on $(c,x_0)$ . Thus, Lemma 2 gives $\varphi(\!\cdot\!)\leq f_\gamma(\!\cdot\!)$ on $(c,x_0)$ , which contradicts that $f_\gamma(\!\cdot\!)\lt\varphi(\!\cdot\!)$ in a left-neighborhood of $x_0$ . On the other hand, if $\gamma\lt\sigma(c)$ , then $\gamma\leq\sigma(\!\cdot\!)$ on (a,c), and Lemma 2 yields $f_\gamma(\!\cdot\!)\leq \varphi(\!\cdot\!)$ on (a, c). Since we also have $f_\gamma(\!\cdot\!)\leq \varphi(\!\cdot\!)$ on $(c,x_0)$ by the definition of c, this shows that $f_\gamma(\!\cdot\!)\leq \varphi(\!\cdot\!)$ on $(a,x_0)$ .

Next, assume that $\sigma(\!\cdot\!)$ is non-decreasing. Again, three cases are considered, namely $\sigma(x_0)\leq \gamma$ , $\gamma\leq \sigma(a)$ , and $\sigma(a)\lt\gamma\lt\sigma(x_0)$ .

First, if $\sigma(x_0)\leq \gamma$ , then $\gamma\geq \sigma(\!\cdot\!)$ on $[a,x_0]$ , and Lemma 2 yields $f_\gamma(\!\cdot\!)\geq\varphi$ on $[a,x_0]$ .

Second, if $\gamma\leq \sigma(a)$ , then $\gamma\leq \sigma(\!\cdot\!)$ on $[a,\infty)$ , so Lemma 2 implies that $f_\gamma(\!\cdot\!)\leq \varphi(\!\cdot\!)$ on $[a,\infty)$ . If $\gamma=\sigma(a)=\sigma(x_0)$ , then $f_\gamma(\!\cdot\!)= \varphi(\!\cdot\!)$ on $[a,x_0]$ , and there is nothing to prove. Thus, we may assume that $\gamma\leq \sigma(a)\lt\sigma(x_0)$ ; in this case, however, $f_\gamma(x_0)=1=\varphi(x_0)$ and $f_\gamma(\!\cdot\!)\leq\varphi(\!\cdot\!)$ imply that

\begin{align*} 0 = \frac12\sigma^2(x_0)x_0^2\varphi''(x_0)+rx_0\varphi'(x_0)-r\varphi(x_0) & \geq \frac12\sigma^2(x_0)x_0^2f^{\prime\prime}_\gamma(x_0)+rx_0f^{\prime}_\gamma(x_0)-rf_\gamma(x_0) \\[5pt] & \gt \frac12\gamma^2x_0^2f^{\prime\prime}_\gamma(x_0)+rx_0f^{\prime}_\gamma(x_0)-rf_\gamma(x_0)=0, \end{align*}

which is impossible.

Third, assume that $\sigma(a)\lt\gamma\lt\sigma(x_0)$ . We then have $\gamma\leq \sigma(\!\cdot\!)$ on $[x_0,\infty)$ , so $f_\gamma(\!\cdot\!)\leq\varphi(\!\cdot\!)$ on $[x_0,\infty)$ . Since $f_\gamma(x_0)=1=\varphi(x_0)$ , this implies that $f^{\prime}_\gamma(x_0)\leq \varphi'(x_0)$ , so

\begin{align*} 0 = \frac12\sigma^2(x_0)x_0^2\varphi''(x_0)+rx_0\varphi'(x_0)-r\varphi(x_0) & \gt \frac12\gamma^2x_0^2\varphi''(x_0)+rx_0f^{\prime}_\gamma(x_0)-rf_\gamma(x_0) \\[5pt] & = \frac12\gamma^2x_0^2(\varphi''(x_0)-f^{\prime\prime}_\gamma(x_0)). \end{align*}

Therefore, $f^{\prime\prime}_\gamma(x_0)\gt\varphi''(x_0)$ , and $f_\gamma(\!\cdot\!)\gt\varphi(\!\cdot\!)$ in a left-neighborhood $(x_0-\varepsilon,x_0)$ of $x_0$ .

Now, assume that there exists at least one point $x\in(a,x_0)$ with $f_\gamma(x)\lt\varphi(x)$ , and define $c\;:\!=\;\sup\{x\in(a,x_0)\;:\; f_\gamma(x)\lt\varphi(x)\}$ . Since $f_\gamma(\!\cdot\!)\gt\varphi(\!\cdot\!)$ in a left-neighborhood of $x_0$ , we then have $c\in(a,x_0)$ . If $\gamma\leq \sigma(c)$ , then $\gamma\leq \sigma(\!\cdot\!)$ on $[c,x_0]$ . Lemma 2 then yields $f_\gamma(\!\cdot\!)\leq \varphi(\!\cdot\!)$ on $[c,x_0]$ , which contradicts $f_\gamma(\!\cdot\!)\gt\varphi(\!\cdot\!)$ in a left-neighborhood of $x_0$ . On the other hand, if $\gamma \gt \sigma(c)$ , then Lemma 2 yields $f_\gamma(\!\cdot\!)\geq \varphi(\!\cdot\!)$ on [a, c] and, by the definition of c, we also find that $f_\gamma(\!\cdot\!)\geq \varphi(\!\cdot\!)$ on $[a,x_0]$ .

Next, assume that $\sigma(\!\cdot\!)$ is non-decreasing, and that $z(K)\lt z_\gamma(K)$ . Then, by convexity, there exists a unique $a\in(z(K), z_\gamma(K))$ with $\varphi(a)=f_\gamma(a)$ (where $\gamma\;:\!=\;\gamma(K)$ is again the implied volatility). Proposition 2 then gives $f_\gamma\geq \varphi$ on $[a,x_0]$ , which is a contradiction of $f_\gamma(z_\gamma(K))=l_K(z(K)) \lt\varphi(z_\gamma(K))$ .

Proof of Theorem 1. For $i=1,2$ , let $K_i$ be given strike prices with $K_1\lt K_2$ , denote by $(z(K_i),\varphi(z(K_i)))$ the tangent point of the tangent line $l_{K_i}$ of $\varphi$ through the point $(K_i,0)$ , and let $\gamma_i\;:\!=\;\gamma(K_i)$ be the corresponding implied volatilities. Then the lines

\[l_{K_i}(z)=-\frac{\varphi(z(K_i))}{K_i-z(K_i)}(z-K_i), \qquad i=1,2,\]

are tangent lines to the functions

\[f_{\gamma_i}(z)=\bigg(\frac{z}{x_0}\bigg)^{-2r/\gamma_i^2},\]

respectively, with unique tangent points with z-coordinates $z'_{\!\!i}\;:\!=\;z_{\gamma_i}(K_i)$ , for which $f_{\gamma_i}(z'_{\!\!i})=l_{K_i}(z'_{\!\!i})$ . We denote by $\hat z$ the z-coordinate of the intersection point between $l_{K_1}$ and $l_{K_2}$ . For a graphical illustration, see Figure 2.

Figure 2. The function $\varphi$ , together with two strike prices $K_1\lt K_2$ , the corresponding tangent lines $l_{K_1}$ and $l_{K_2}$ , and their intersection point $\hat z$ .

Now assume that $\sigma(\!\cdot\!)$ is non-increasing. By Proposition 3, we then have $z(K_i)\leq z'_{\!\!i}$ , $i=1,2$ . If $z'_{\!\!1}\gt\hat z$ , then automatically $\gamma_1\geq\gamma_2$ since $f_\gamma$ is decreasing in $\gamma$ on $(0,x_0)$ . On the other hand, if $z'_{\!\!1}\in[z(K_1),\hat z]$ then, by convexity, there is a unique point $a\in[z(K_1),z'_{\!\!1}]$ such that $f_{\gamma_1}(a)=\varphi(a)$ . By Proposition 2, we then have $f_{\gamma_1}(\!\cdot\!)\leq\varphi(\!\cdot\!)$ on $[a,x_0]$ . Consequently, at $z(K_2)$ we have $f_{\gamma_1}(z(K_2))\leq \varphi(z(K_2))=l_{K_2}(z(K_2))$ . Since $f_\gamma$ is decreasing in $\gamma$ on $(0,x_0)$ , it follows that $\gamma_1\geq \gamma_2$ .

Next, assume that $\sigma(\!\cdot\!)$ is non-decreasing. Then $z'_{\!\!1}\leq z(K_1)$ by Proposition 3, so by convexity there exists a point $a\in [z'_{\!\!1},z(K_1)]$ with $f_1(a)=\varphi(a)$ . By Proposition 2, we must have $f_{\gamma_1}(\!\cdot\!)\geq\varphi(\!\cdot\!)\geq l_{K_2}(\!\cdot\!)$ on $[a,x_0]$ , and it follows that $\gamma_1\leq \gamma_2$ .