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Iosif Pinelis
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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 $A\in[\min(1,b),\max(1,b)]$.

Iosif Pinelis
  • 127.8k
  • 8
  • 107
  • 229