show that $(y^* \circ F) * \gamma(x) = \int_X {y^* \circ F(x+t)}d\gamma(t)$ is Gateaux differentiable everywhere

Question

Suppose $X$ is a separable Banach space and $Y$ is a Banach space.

If $F:X \rightarrow Y$ is a Lipschitz map and $\gamma$ is a nondegenerate Gaussian probability measure on $X$ with mean $0$, then for every $y^* \in Y^*$, the real-valued map $$(y^* \circ F) * \gamma(x) = \int_X {y^* \circ F(x+t)}d\gamma(t)$$ is Gateaux differentiable everywhere on $X$.

The statement above is taken from here, page $129$.

Can anyone give some hint on how to prove the statement?

UPDATE: Note that $$\left|\int_X{y^*(F(x+hv+t)d\gamma(t))} - \int_X{y^*(F(x+t)d\gamma(t))} \right| $$ $$\leq \int_X \|y^*\| \| F(x+hv+t) - F(x+t) \|d\gamma(t) $$ $$\leq M \| F \|_{Lip} |h| \|v \|_X\int_X d\gamma(t)$$

By the assumption on $\gamma$, we have $\int_X d\gamma(t)=0$.

Hence, by the Squeeze Theorem, we have $$\lim_{h \rightarrow 0}{\dfrac{\int_X{y^*(F(x+hv+t)d\gamma(t))} - \int_X{y^*(F(x+t)d\gamma(t))}}{h}}=0.$$

Is it correct? Why do we need $X$ to be separable?

Answer 1

What is the definition of Gateaux-differentiable? This is the Banach space version of directional derivative, no?

Let us write for a given $y^*\in Y^*$, $F^*\in X^{*}$ is $$F^*(x)=\int_Xy^*(F(x+t))d\gamma(t)$$ Then we are charged with proving that the following exists, for $h\in\mathbb{R}$ and any $v\in X$,

$$\delta(F^*,v)=\lim_{h\rightarrow0}\frac{F^*(x+hv)-F(x)}{h}.$$

You should think about the following, and the proof shouldn't be that far off:

1. Why integration against the Gaussian $\gamma$ makes this well-defined, and
2. How the Lipschitz assumption makes the limit exist, and (this is crucial), how we can use this to show the limit exists.