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Iosif Pinelis
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$\newcommand{\de}{\delta}$ The notion of the generalized gradient, as defined in Clarke's paper linked in your question, is applicable only to Lipschitz functions. In general, depending on your measure space, your function $f$ will not be Lipschitz, because the evaluation functional $L^2(\tau)\ni x\mapsto\de_s(x):=x(s)$ for $s\in T$ will not be Lipschitz in general. Therefore, the generalized gradient of your function $f$ will be undefined in general. In particular, it will be undefined if your measure $\tau$ is non-atomic.


However, we have $\|x\|_2\ge\tau(\{r\})^{1/2}|x(r)|$ for all $r\in T$. So, if $\tau(\{r\})>0$, then the evaluation functional $\de_r$ is bounded and hence Lipschitz. So, assuming now that $\tau(\{s\})>0$ and $\tau(\{t\})>0$, we see that your nonlinear function(al) \begin{equation*} f=\min(\de_s,b\de_t) \end{equation*} is Lipschitz. (I am assuming $a=1$, without loss of generality.) Moreover, then the generalized (upper) directional derivative of $f$ at $x\in L^2(\tau)$ in the direction $v\in L^2(\tau)$ is \begin{multline*}\tag{1} f^0(x;v):=\limsup_{y\to x,h\downarrow0}\frac{f(y+hv)-f(y)}h \\ =\left\{\begin{aligned} v(s)&\text{ if }x(s)<bx(t), \\ bv(t)&\text{ if }x(s)>bx(t), \\ \max[v(s),bv(t)] &\text{ if }x(s)=bx(t); \end{aligned}\right. \end{multline*}\begin{multline*} f^0(x;v):=\limsup_{y\to x,h\downarrow0}\frac{f(y+hv)-f(y)}h \\ =\left\{\begin{aligned} v(s)&\text{ if }x(s)<bx(t), \\ bv(t)&\text{ if }x(s)>bx(t), \\ \max[v(s),bv(t)] &\text{ if }x(s)=bx(t); \end{aligned}\right. \tag{1} \end{multline*} see details on this at the end of this answer; here I am using the (somewhat strange to me) notation $f^0(x;v)$ from the linked paper by Clarke.

So, the generalized gradient is $\{\de_s\}$ if $x(s)<bx(t)$ and $\{b\de_t\}$ if $x(s)>bx(t)$.

Consider now the case $x(s)=bx(t)$:

Consider first the subcase $s\ne t$. We have $$\max(v_s,bv_t)\ge Av_s+Bv_t\quad\text{for all real }v_s,v_t $$ iff $0\le A\le1$ and $B=(1-A)b$. So, if $x(s)=bx(t)$ and $s\ne t$, then the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(s)+(1-A)bv(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[0,1]$.

If $s=t$, $\tau(\{t\})>0$, and $x(t)=bx(t)$ (that is, either $b=1$ or $x(t)=0$), then $f^0(x;v)=\max[v(t),bv(t)]$ and the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[\min(1,b),\max(1,b)]$.

Details on (1): If for a function $x\in L^2(\tau)$ with $\tau(\{s\})>0$ and $\tau(\{t\})>0$ we have $x(s)<bx(t)$, and a function $y\in L^2(\tau)$ is close enough to $x$, and real $h>0$ is small enough, then $y(s)+hv(s)<by(t)+bhv(t)$ and $y(s)<by(t)$, so that $f(y+hv)-f(y)=y(s)+hv(s)-y(s)=hv(s)$, and hence $f^0(x;v)=v(s)$, in the case $x(s)<bx(t)$. Similarly, $f^0(x;v)=bv(t)$ in the case $x(s)>bx(t)$.

To complete the proof of (1), it remains to consider the case $x(s)=bx(t)$. Then the function $x\in L^2(\tau)$ can be however closely approximated by functions $\check x$ and $\hat x$ in $L^2(\tau)$ such that $\check x(s)<b\check x(t)$ and $\hat x(s)>b\hat x(t)$, so that, by what has just been shown, $f^0(\check x;v)=v(s)$ and $f^0(\hat x;v)=bv(t)$. Letting now $\check x\to x$ and $\hat x\to x$ and using the upper semincontinuity of $f^0(\cdot;\cdot)$ (which is item 4 of the list on page 54 in the linked paper by Clarke), we conclude that \begin{equation*} f^0(x;v)\ge\max[v(s),bv(t)]. \tag{2} \end{equation*} On the other hand, for any $y\in L^2(\tau)$ and real $h>0$, \begin{equation*} f(y+hv)-f(y)=\min[y(s)+hv(s),by(t)+bhv(t)]-\min[y(s),by(t)] \le\max[hv(s),bhv(t)], \end{equation*}\begin{multline*} f(y+hv)-f(y)=\min[y(s)+hv(s),by(t)+bhv(t)]-\min[y(s),by(t)] \\ \le\max[hv(s),bhv(t)], \end{multline*} by the inequality $\min[z_1,w_1]-\min[z_2,w_2]\le\max[z_1-z_2,w_1-w_2]$ for real $z_1,w_1,z_2,w_2$. So, \begin{equation*} f^0(x;v)\le\max[v(s),bv(t)]. \tag{3} \end{equation*} Finally, by (2) and (3), we get $f^0(x;v)=\max[v(s),bv(t)]$, in the case $x(s)=bx(t)$.

$\newcommand{\de}{\delta}$ The notion of the generalized gradient, as defined in Clarke's paper linked in your question, is applicable only to Lipschitz functions. In general, depending on your measure space, your function $f$ will not be Lipschitz, because the evaluation functional $L^2(\tau)\ni x\mapsto\de_s(x):=x(s)$ for $s\in T$ will not be Lipschitz in general. Therefore, the generalized gradient of your function $f$ will be undefined in general. In particular, it will be undefined if your measure $\tau$ is non-atomic.


However, we have $\|x\|_2\ge\tau(\{r\})^{1/2}|x(r)|$ for all $r\in T$. So, if $\tau(\{r\})>0$, then the evaluation functional $\de_r$ is bounded and hence Lipschitz. So, assuming now that $\tau(\{s\})>0$ and $\tau(\{t\})>0$, we see that your nonlinear function(al) \begin{equation*} f=\min(\de_s,b\de_t) \end{equation*} is Lipschitz. (I am assuming $a=1$, without loss of generality.) Moreover, then the generalized (upper) directional derivative of $f$ at $x\in L^2(\tau)$ in the direction $v\in L^2(\tau)$ is \begin{multline*}\tag{1} f^0(x;v):=\limsup_{y\to x,h\downarrow0}\frac{f(y+hv)-f(y)}h \\ =\left\{\begin{aligned} v(s)&\text{ if }x(s)<bx(t), \\ bv(t)&\text{ if }x(s)>bx(t), \\ \max[v(s),bv(t)] &\text{ if }x(s)=bx(t); \end{aligned}\right. \end{multline*} see details on this at the end of this answer; here I am using the (somewhat strange to me) notation $f^0(x;v)$ from the linked paper by Clarke.

So, the generalized gradient is $\{\de_s\}$ if $x(s)<bx(t)$ and $\{b\de_t\}$ if $x(s)>bx(t)$.

Consider now the case $x(s)=bx(t)$:

Consider first the subcase $s\ne t$. We have $$\max(v_s,bv_t)\ge Av_s+Bv_t\quad\text{for all real }v_s,v_t $$ iff $0\le A\le1$ and $B=(1-A)b$. So, if $x(s)=bx(t)$ and $s\ne t$, then the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(s)+(1-A)bv(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[0,1]$.

If $s=t$, $\tau(\{t\})>0$, and $x(t)=bx(t)$ (that is, either $b=1$ or $x(t)=0$), then $f^0(x;v)=\max[v(t),bv(t)]$ and the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[\min(1,b),\max(1,b)]$.

Details on (1): If for a function $x\in L^2(\tau)$ with $\tau(\{s\})>0$ and $\tau(\{t\})>0$ we have $x(s)<bx(t)$, and a function $y\in L^2(\tau)$ is close enough to $x$, and real $h>0$ is small enough, then $y(s)+hv(s)<by(t)+bhv(t)$ and $y(s)<by(t)$, so that $f(y+hv)-f(y)=y(s)+hv(s)-y(s)=hv(s)$, and hence $f^0(x;v)=v(s)$, in the case $x(s)<bx(t)$. Similarly, $f^0(x;v)=bv(t)$ in the case $x(s)>bx(t)$.

To complete the proof of (1), it remains to consider the case $x(s)=bx(t)$. Then the function $x\in L^2(\tau)$ can be however closely approximated by functions $\check x$ and $\hat x$ in $L^2(\tau)$ such that $\check x(s)<b\check x(t)$ and $\hat x(s)>b\hat x(t)$, so that, by what has just been shown, $f^0(\check x;v)=v(s)$ and $f^0(\hat x;v)=bv(t)$. Letting now $\check x\to x$ and $\hat x\to x$ and using the upper semincontinuity of $f^0(\cdot;\cdot)$ (which is item 4 of the list on page 54 in the linked paper by Clarke), we conclude that \begin{equation*} f^0(x;v)\ge\max[v(s),bv(t)]. \tag{2} \end{equation*} On the other hand, for any $y\in L^2(\tau)$ and real $h>0$, \begin{equation*} f(y+hv)-f(y)=\min[y(s)+hv(s),by(t)+bhv(t)]-\min[y(s),by(t)] \le\max[hv(s),bhv(t)], \end{equation*} by the inequality $\min[z_1,w_1]-\min[z_2,w_2]\le\max[z_1-z_2,w_1-w_2]$ for real $z_1,w_1,z_2,w_2$. So, \begin{equation*} f^0(x;v)\le\max[v(s),bv(t)]. \tag{3} \end{equation*} Finally, by (2) and (3), we get $f^0(x;v)=\max[v(s),bv(t)]$, in the case $x(s)=bx(t)$.

$\newcommand{\de}{\delta}$ The notion of the generalized gradient, as defined in Clarke's paper linked in your question, is applicable only to Lipschitz functions. In general, depending on your measure space, your function $f$ will not be Lipschitz, because the evaluation functional $L^2(\tau)\ni x\mapsto\de_s(x):=x(s)$ for $s\in T$ will not be Lipschitz in general. Therefore, the generalized gradient of your function $f$ will be undefined in general. In particular, it will be undefined if your measure $\tau$ is non-atomic.


However, we have $\|x\|_2\ge\tau(\{r\})^{1/2}|x(r)|$ for all $r\in T$. So, if $\tau(\{r\})>0$, then the evaluation functional $\de_r$ is bounded and hence Lipschitz. So, assuming now that $\tau(\{s\})>0$ and $\tau(\{t\})>0$, we see that your nonlinear function(al) \begin{equation*} f=\min(\de_s,b\de_t) \end{equation*} is Lipschitz. (I am assuming $a=1$, without loss of generality.) Moreover, then the generalized (upper) directional derivative of $f$ at $x\in L^2(\tau)$ in the direction $v\in L^2(\tau)$ is \begin{multline*} f^0(x;v):=\limsup_{y\to x,h\downarrow0}\frac{f(y+hv)-f(y)}h \\ =\left\{\begin{aligned} v(s)&\text{ if }x(s)<bx(t), \\ bv(t)&\text{ if }x(s)>bx(t), \\ \max[v(s),bv(t)] &\text{ if }x(s)=bx(t); \end{aligned}\right. \tag{1} \end{multline*} see details on this at the end of this answer; here I am using the (somewhat strange to me) notation $f^0(x;v)$ from the linked paper by Clarke.

So, the generalized gradient is $\{\de_s\}$ if $x(s)<bx(t)$ and $\{b\de_t\}$ if $x(s)>bx(t)$.

Consider now the case $x(s)=bx(t)$:

Consider first the subcase $s\ne t$. We have $$\max(v_s,bv_t)\ge Av_s+Bv_t\quad\text{for all real }v_s,v_t $$ iff $0\le A\le1$ and $B=(1-A)b$. So, if $x(s)=bx(t)$ and $s\ne t$, then the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(s)+(1-A)bv(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[0,1]$.

If $s=t$, $\tau(\{t\})>0$, and $x(t)=bx(t)$ (that is, either $b=1$ or $x(t)=0$), then $f^0(x;v)=\max[v(t),bv(t)]$ and the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[\min(1,b),\max(1,b)]$.

Details on (1): If for a function $x\in L^2(\tau)$ with $\tau(\{s\})>0$ and $\tau(\{t\})>0$ we have $x(s)<bx(t)$, and a function $y\in L^2(\tau)$ is close enough to $x$, and real $h>0$ is small enough, then $y(s)+hv(s)<by(t)+bhv(t)$ and $y(s)<by(t)$, so that $f(y+hv)-f(y)=y(s)+hv(s)-y(s)=hv(s)$, and hence $f^0(x;v)=v(s)$, in the case $x(s)<bx(t)$. Similarly, $f^0(x;v)=bv(t)$ in the case $x(s)>bx(t)$.

To complete the proof of (1), it remains to consider the case $x(s)=bx(t)$. Then the function $x\in L^2(\tau)$ can be however closely approximated by functions $\check x$ and $\hat x$ in $L^2(\tau)$ such that $\check x(s)<b\check x(t)$ and $\hat x(s)>b\hat x(t)$, so that, by what has just been shown, $f^0(\check x;v)=v(s)$ and $f^0(\hat x;v)=bv(t)$. Letting now $\check x\to x$ and $\hat x\to x$ and using the upper semincontinuity of $f^0(\cdot;\cdot)$ (which is item 4 of the list on page 54 in the linked paper by Clarke), we conclude that \begin{equation*} f^0(x;v)\ge\max[v(s),bv(t)]. \tag{2} \end{equation*} On the other hand, for any $y\in L^2(\tau)$ and real $h>0$, \begin{multline*} f(y+hv)-f(y)=\min[y(s)+hv(s),by(t)+bhv(t)]-\min[y(s),by(t)] \\ \le\max[hv(s),bhv(t)], \end{multline*} by the inequality $\min[z_1,w_1]-\min[z_2,w_2]\le\max[z_1-z_2,w_1-w_2]$ for real $z_1,w_1,z_2,w_2$. So, \begin{equation*} f^0(x;v)\le\max[v(s),bv(t)]. \tag{3} \end{equation*} Finally, by (2) and (3), we get $f^0(x;v)=\max[v(s),bv(t)]$, in the case $x(s)=bx(t)$.

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Iosif Pinelis
  • 127.8k
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  • 107
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The$\newcommand{\de}{\delta}$ The notion of the generalized gradient, as defined in Clarke's paper linked in your question, is applicable only to Lipschitz functions. In general, depending on your measure space, your function $f$ will not be Lipschitz, because the functionevaluation functional $L^2(\tau)\ni x\mapsto x(s)$$L^2(\tau)\ni x\mapsto\de_s(x):=x(s)$ for $s\in T$ will not be Lipschitz in general. Therefore, the generalized gradient of your function $f$ will be undefined in general. In particular, it will be undefined if your measure $\tau$ is non-atomic.

 

However, we have $\|x\|_2\ge\tau(\{r\})^{1/2}|x(r)|$ for all $r\in T$. So, if (say)$\tau(\{r\})>0$, then the evaluation functional $s\ne t$$\de_r$ is bounded and hence Lipschitz. So, assuming now that $\tau(\{s\})>0$, and $\tau(\{t\})>0$, we see that your nonlinear function(al) \begin{equation*} f=\min(\de_s,b\de_t) \end{equation*} is Lipschitz. (I am assuming $a=1$, without loss of generality.) Moreover, then the generalized (upper) directional derivative of $f$ at $x$$x\in L^2(\tau)$ in the direction $v$$v\in L^2(\tau)$ is $$\limsup_{y\to x,h\downarrow0}\frac{f(y+hv)-f(y)}h=\max[v(s),bv(t)]; $$\begin{multline*}\tag{1} f^0(x;v):=\limsup_{y\to x,h\downarrow0}\frac{f(y+hv)-f(y)}h \\ =\left\{\begin{aligned} v(s)&\text{ if }x(s)<bx(t), \\ bv(t)&\text{ if }x(s)>bx(t), \\ \max[v(s),bv(t)] &\text{ if }x(s)=bx(t); \end{aligned}\right. \end{multline*} (Isee details on this at the end of this answer; here I am assumingusing the $a=1$, without loss of generality.(somewhat strange to me) Next notation $f^0(x;v)$ from the linked paper by Clarke.

So, wethe generalized gradient is $\{\de_s\}$ if $x(s)<bx(t)$ and $\{b\de_t\}$ if $x(s)>bx(t)$.

Consider now the case $x(s)=bx(t)$:

Consider first the subcase $s\ne t$. We have $$\max(v_s,bv_t)\ge Av_s+Bv_t\quad\text{for all real }v_s,v_t $$ iff $0\le A\le1$ and $B=(1-A)b$. So, hereif $x(s)=bx(t)$ and $s\ne t$, then the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(s)+(1-A)bv(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[0,1]$.

If $s=t$ and, $\tau(\{t\})>0$, then the generalized directional derivative of $f$ atand $x$ in the direction$x(t)=bx(t)$ $v$(that is, either $b=1$ or $x(t)=0$), then $\max[v(t),bv(t)]$$f^0(x;v)=\max[v(t),bv(t)]$ and the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[\min(1,b),\max(1,b)]$.

Details on (1): If for a function $x\in L^2(\tau)$ with $\tau(\{s\})>0$ and $\tau(\{t\})>0$ we have $x(s)<bx(t)$, and a function $y\in L^2(\tau)$ is close enough to $x$, and real $h>0$ is small enough, then $y(s)+hv(s)<by(t)+bhv(t)$ and $y(s)<by(t)$, so that $f(y+hv)-f(y)=y(s)+hv(s)-y(s)=hv(s)$, and hence $f^0(x;v)=v(s)$, in the case $x(s)<bx(t)$. Similarly, $f^0(x;v)=bv(t)$ in the case $x(s)>bx(t)$.

To complete the proof of (1), it remains to consider the case $x(s)=bx(t)$. Then the function $x\in L^2(\tau)$ can be however closely approximated by functions $\check x$ and $\hat x$ in $L^2(\tau)$ such that $\check x(s)<b\check x(t)$ and $\hat x(s)>b\hat x(t)$, so that, by what has just been shown, $f^0(\check x;v)=v(s)$ and $f^0(\hat x;v)=bv(t)$. Letting now $\check x\to x$ and $\hat x\to x$ and using the upper semincontinuity of $f^0(\cdot;\cdot)$ (which is item 4 of the list on page 54 in the linked paper by Clarke), we conclude that \begin{equation*} f^0(x;v)\ge\max[v(s),bv(t)]. \tag{2} \end{equation*} On the other hand, for any $y\in L^2(\tau)$ and real $h>0$, \begin{equation*} f(y+hv)-f(y)=\min[y(s)+hv(s),by(t)+bhv(t)]-\min[y(s),by(t)] \le\max[hv(s),bhv(t)], \end{equation*} by the inequality $\min[z_1,w_1]-\min[z_2,w_2]\le\max[z_1-z_2,w_1-w_2]$ for real $z_1,w_1,z_2,w_2$. So, \begin{equation*} f^0(x;v)\le\max[v(s),bv(t)]. \tag{3} \end{equation*} Finally, by (2) and (3), we get $f^0(x;v)=\max[v(s),bv(t)]$, in the case $x(s)=bx(t)$.

The notion of the generalized gradient, as defined in Clarke's paper linked in your question, is applicable only to Lipschitz functions. In general, depending on your measure space, your function $f$ will not be Lipschitz, because the function $L^2(\tau)\ni x\mapsto x(s)$ for $s\in T$ will not be Lipschitz in general. Therefore, the generalized gradient of your function $f$ will be undefined in general. In particular, it will be undefined if your measure $\tau$ is non-atomic.

However, if (say), $s\ne t$, $\tau(\{s\})>0$, and $\tau(\{t\})>0$, then the generalized (upper) directional derivative of $f$ at $x$ in the direction $v$ is $$\limsup_{y\to x,h\downarrow0}\frac{f(y+hv)-f(y)}h=\max[v(s),bv(t)]; $$ (I am assuming $a=1$, without loss of generality.) Next, we have $$\max(v_s,bv_t)\ge Av_s+Bv_t\quad\text{for all real }v_s,v_t $$ iff $0\le A\le1$ and $B=(1-A)b$. So, here the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(s)+(1-A)bv(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[0,1]$.

If $s=t$ and $\tau(\{t\})>0$, then the generalized directional derivative of $f$ at $x$ in the direction $v$ is $\max[v(t),bv(t)]$ and the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[\min(1,b),\max(1,b)]$.

$\newcommand{\de}{\delta}$ The notion of the generalized gradient, as defined in Clarke's paper linked in your question, is applicable only to Lipschitz functions. In general, depending on your measure space, your function $f$ will not be Lipschitz, because the evaluation functional $L^2(\tau)\ni x\mapsto\de_s(x):=x(s)$ for $s\in T$ will not be Lipschitz in general. Therefore, the generalized gradient of your function $f$ will be undefined in general. In particular, it will be undefined if your measure $\tau$ is non-atomic.

 

However, we have $\|x\|_2\ge\tau(\{r\})^{1/2}|x(r)|$ for all $r\in T$. So, if $\tau(\{r\})>0$, then the evaluation functional $\de_r$ is bounded and hence Lipschitz. So, assuming now that $\tau(\{s\})>0$ and $\tau(\{t\})>0$, we see that your nonlinear function(al) \begin{equation*} f=\min(\de_s,b\de_t) \end{equation*} is Lipschitz. (I am assuming $a=1$, without loss of generality.) Moreover, then the generalized (upper) directional derivative of $f$ at $x\in L^2(\tau)$ in the direction $v\in L^2(\tau)$ is \begin{multline*}\tag{1} f^0(x;v):=\limsup_{y\to x,h\downarrow0}\frac{f(y+hv)-f(y)}h \\ =\left\{\begin{aligned} v(s)&\text{ if }x(s)<bx(t), \\ bv(t)&\text{ if }x(s)>bx(t), \\ \max[v(s),bv(t)] &\text{ if }x(s)=bx(t); \end{aligned}\right. \end{multline*} see details on this at the end of this answer; here I am using the (somewhat strange to me) notation $f^0(x;v)$ from the linked paper by Clarke.

So, the generalized gradient is $\{\de_s\}$ if $x(s)<bx(t)$ and $\{b\de_t\}$ if $x(s)>bx(t)$.

Consider now the case $x(s)=bx(t)$:

Consider first the subcase $s\ne t$. We have $$\max(v_s,bv_t)\ge Av_s+Bv_t\quad\text{for all real }v_s,v_t $$ iff $0\le A\le1$ and $B=(1-A)b$. So, if $x(s)=bx(t)$ and $s\ne t$, then the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(s)+(1-A)bv(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[0,1]$.

If $s=t$, $\tau(\{t\})>0$, and $x(t)=bx(t)$ (that is, either $b=1$ or $x(t)=0$), then $f^0(x;v)=\max[v(t),bv(t)]$ and the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[\min(1,b),\max(1,b)]$.

Details on (1): If for a function $x\in L^2(\tau)$ with $\tau(\{s\})>0$ and $\tau(\{t\})>0$ we have $x(s)<bx(t)$, and a function $y\in L^2(\tau)$ is close enough to $x$, and real $h>0$ is small enough, then $y(s)+hv(s)<by(t)+bhv(t)$ and $y(s)<by(t)$, so that $f(y+hv)-f(y)=y(s)+hv(s)-y(s)=hv(s)$, and hence $f^0(x;v)=v(s)$, in the case $x(s)<bx(t)$. Similarly, $f^0(x;v)=bv(t)$ in the case $x(s)>bx(t)$.

To complete the proof of (1), it remains to consider the case $x(s)=bx(t)$. Then the function $x\in L^2(\tau)$ can be however closely approximated by functions $\check x$ and $\hat x$ in $L^2(\tau)$ such that $\check x(s)<b\check x(t)$ and $\hat x(s)>b\hat x(t)$, so that, by what has just been shown, $f^0(\check x;v)=v(s)$ and $f^0(\hat x;v)=bv(t)$. Letting now $\check x\to x$ and $\hat x\to x$ and using the upper semincontinuity of $f^0(\cdot;\cdot)$ (which is item 4 of the list on page 54 in the linked paper by Clarke), we conclude that \begin{equation*} f^0(x;v)\ge\max[v(s),bv(t)]. \tag{2} \end{equation*} On the other hand, for any $y\in L^2(\tau)$ and real $h>0$, \begin{equation*} f(y+hv)-f(y)=\min[y(s)+hv(s),by(t)+bhv(t)]-\min[y(s),by(t)] \le\max[hv(s),bhv(t)], \end{equation*} by the inequality $\min[z_1,w_1]-\min[z_2,w_2]\le\max[z_1-z_2,w_1-w_2]$ for real $z_1,w_1,z_2,w_2$. So, \begin{equation*} f^0(x;v)\le\max[v(s),bv(t)]. \tag{3} \end{equation*} Finally, by (2) and (3), we get $f^0(x;v)=\max[v(s),bv(t)]$, in the case $x(s)=bx(t)$.

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The notion of the generalized gradient, as defined in Clarke's paper linked in your question, is applicable only to Lipschitz functions. In general, depending on your measure space, your function $f$ will not be Lipschitz, because the function $L^2(\tau)\ni x\mapsto x(s)$ for $s\in T$ will not be Lipschitz in general. Therefore, the generalized gradient of your function $f$ will be undefined in general. In particular, it will be undefined if your measure $\tau$ is non-atomic.

However, if (say), $s\ne t$, $\tau(\{s\})>0$, and $\tau(\{t\})>0$, then the generalized (upper) directional derivative of $f$ at $x$ in the direction $v$ is $$\limsup_{y\to x,h\downarrow0}\frac{f(y+hv)-f(y)}h=\max[v(s),bv(t)]; $$ (I am assuming $a=1$, without loss of generality.) Next, we have $$\max(v_s,bv_t)\ge Av_s+Bv_t\quad\text{for all real }v_s,v_t $$ iff $0\le A\le1$ and $B=(1-A)b$. So, here the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(s)+(1-A)bv(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[0,1]$.

If $s=t$ and $\tau(\{t\})>0$, then the generalized directional derivative of $f$ at $x$ in the direction $v$ is $\max[v(t),bv(t)]$ and the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[\min(1,b),\max(1,b)]$.

The notion of the generalized gradient, as defined in Clarke's paper linked in your question, is applicable only to Lipschitz functions. In general, depending on your measure space, your function $f$ will not be Lipschitz, because the function $L^2(\tau)\ni x\mapsto x(s)$ for $s\in T$ will not be Lipschitz in general. Therefore, the generalized gradient of your function $f$ will be undefined in general. In particular, it will be undefined if your measure $\tau$ is non-atomic.

However, if (say), $s\ne t$, $\tau(\{s\})>0$, and $\tau(\{t\})>0$, then the generalized (upper) directional derivative of $f$ at $x$ in the direction $v$ is $$\limsup_{y\to x,h\downarrow0}\frac{f(y+hv)-f(y)}h=\max[v(s),bv(t)]; $$ (I am assuming $a=1$, without loss of generality.) Next, we have $$\max(v_s,bv_t)\ge Av_s+Bv_t\quad\text{for all real }v_s,v_t $$ iff $0\le A\le1$ and $B=(1-A)b$. So, here the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(s)+(1-A)bv(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[0,1]$.

The notion of the generalized gradient, as defined in Clarke's paper linked in your question, is applicable only to Lipschitz functions. In general, depending on your measure space, your function $f$ will not be Lipschitz, because the function $L^2(\tau)\ni x\mapsto x(s)$ for $s\in T$ will not be Lipschitz in general. Therefore, the generalized gradient of your function $f$ will be undefined in general. In particular, it will be undefined if your measure $\tau$ is non-atomic.

However, if (say), $s\ne t$, $\tau(\{s\})>0$, and $\tau(\{t\})>0$, then the generalized (upper) directional derivative of $f$ at $x$ in the direction $v$ is $$\limsup_{y\to x,h\downarrow0}\frac{f(y+hv)-f(y)}h=\max[v(s),bv(t)]; $$ (I am assuming $a=1$, without loss of generality.) Next, we have $$\max(v_s,bv_t)\ge Av_s+Bv_t\quad\text{for all real }v_s,v_t $$ iff $0\le A\le1$ and $B=(1-A)b$. So, here the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(s)+(1-A)bv(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[0,1]$.

If $s=t$ and $\tau(\{t\})>0$, then the generalized directional derivative of $f$ at $x$ in the direction $v$ is $\max[v(t),bv(t)]$ and the generalized gradient is the set of all linear functionals $\ell$ given by the formula $$\ell(v)=Av(t)\quad\text{for }v\in L^2(\tau) $$ with $A\in[\min(1,b),\max(1,b)]$.

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