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Let $(T,\mathcal T,\tau)$ be a measure space, $a,b\ge0$, $s,t\in T$ and $$f(x):=a\min(x(s),bx(t))\;\;\;\text{for }x\in L^2(\tau).$$

How can we calculate the generalized gradient $\partial_Cf(x)$ of $f$ at $x\in L^2(\tau)$?

We may note that $2\min(u,v)=u+v-|u-v|$ for all $u,v\in\mathbb R$. Now the only point on which $\mathbb R\ni u\mapsto|u|$ is not differentiable (in the classical sense) is $0$, but the generalized gradient at $0$ is easily seen to be $[-1,1]$. Moreover, $\mathbb R\setminus\{0\}\ni u\mapsto|u|$ is continuously differentiable and the generalized gradient at $x\in\mathbb R\setminus\{0\}$ is simply $\{x/|x|\}$. This knowledge should be helpful. However, I'm new to this topic and have no idea how to tackle the problem.

EDIT: As Iosif Pinelis pointed out, we need to assume that the singletons in $(T,\mathcal T,\tau)$ are measurable and admit positive measure to ensure that $f$ is locally Lipschitz continuous.

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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*} 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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  • $\begingroup$ Thank you for your answer. Are there useful sufficient conditions on $\tau$ ensuring that the evaluation maps are locally Lipschitz (or even continuously differentiable)? $\endgroup$
    – 0xbadf00d
    Sep 18, 2019 at 15:29
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    $\begingroup$ @0xbadf00d : Yes, the necessary and sufficient condition for the evaluation map at $s$ to be Lipshcitz is that $\tau(\{s\})>0$. $\endgroup$ Sep 18, 2019 at 15:33
  • $\begingroup$ I see, thanks. It's my fault that I didn't check whether $f$ at least satisfies the necessary conditions. However, I'm still interested in the question under the additional assumption you provided. $\endgroup$
    – 0xbadf00d
    Sep 18, 2019 at 15:36
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    $\begingroup$ @0xbadf00d : I have added the calculation of the generalized gradient in the case when $\tau$ does not have atoms at $s$ or $t$. $\endgroup$ Sep 18, 2019 at 16:00
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    $\begingroup$ @0xbadf00d : I have provided the details you requested, and also a correction: the previous formulas were good only for the more difficult case $x(s)=bx(t)$; the considerations are quite a bit simpler if $x(s)<bx(t)$ or $x(s)>bx(t)$. $\endgroup$ Sep 18, 2019 at 21:04

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