# Weak Gâteaux Derivative

Suppose $X$ and $Y$ are Banach spaces. Let $F:X \rightarrow Y$ be a function and $U \subset X$ be an open set.

The Gâteaux derivative of $F$ at $u \in U$ in the direction $\phi \in X$ is given by

$$\lim_{t \rightarrow 0}\dfrac{F(u+t \phi) - F(u)}{t}.$$

In the Wikipedia article on the Gâteaux derivative, there is this sentence in the section 'definition'

In some cases, a weak limit is taken instead of a strong limit, which leads to the notion of a weak Gâteaux derivative.

Question: How should one write out the definition of the weak Gâteaux derivative? Since we are talking about weak limit, I suppose we need to talk about the weak topology.

EDIT: Suppose $X$ is a Banach space and $\gamma$ is a Gaussian probability measure on $X$. Define a map $$(F * \gamma) (x) = \int \limits _X{F(x+t) \Bbb d\gamma(t)}$$ What is its weak Gâteaux derivative?

• what is wrong with saying that the weak Gateaux derivative at position $u$ in direction $\phi$ is given by $y \in Y$ if $\lim_{t \rightarrow 0} y'(\dfrac{F(u+t \phi) - F(u)}{t}-y)=0$ for all $y' \in Y'$? – Zimkovic Dec 6 '15 at 15:02
• So I suppose weak Gateaux derivative of $F * \gamma$ is $\lim_{t \rightarrow 0}{y^{\prime}(\dfrac{\int_X{F(u+t \phi + h)-F(u+ h)d\gamma(h)}}{t}}-y)=0$? – Idonknow Dec 6 '15 at 17:07
• yes that seems correct to me – Zimkovic Dec 6 '15 at 17:40
• Why did the answer get the bounty but wasn't accepted? – Viktor Glombik Apr 14 at 11:27

## 1 Answer

In general, we say that a net $(x_i) _{i \in I} \subset X$ converges weakly to $x \in X$ if and only if $\omega (x_i) \to \omega (x) \ \forall \omega \in X^*$.

In your case, the role of $x_i$ is played by the fraction $\dfrac {F(u + t \phi) - F(u)} {t}$ (the role of $i$ is played by $t$ and the role of $X$ is now taken by $Y$). The existence of a weak Gâteaux derivative of $F$ at the point $u \in X$ in the direction $\phi \in X$ (let's call it $WD(F, u, \phi)$ - "$W$" stands for "weak") means that there exist $WD(F, u, \phi) \in Y$ (because $F$ takes its values in $Y$) such that

$$\lim \limits _{t \to 0} \ \omega \left( \frac {F(u + t \phi) - F(u)} {t} \right) = \omega (WD(F, u, \phi)) \ \forall \omega \in Y^* .$$

Applying this to your particular case of $F*\gamma$,

$$\lim \limits _{t \to 0} \ \omega \left( \frac {(F*\gamma) (u + t \phi) - (F*\gamma) (u)} {t} \right) = \lim \limits _{t \to 0} \frac 1 t \omega \left( \int \limits _X F(u + t\phi + y) \Bbb d \gamma(y) - \int \limits _X F(u + y) \Bbb d \gamma(y) \right) = \lim \limits _{t \to 0} \ \omega \left( \int \limits _X \frac {F(u + t\phi + y) - F(u + y)} t \Bbb d \gamma(y) \right) = \omega \left( \lim \limits _{t \to 0} \int \limits _X \frac {F(u + t\phi + y) - F(u + y)} t \Bbb d \gamma(y) \right).$$

(The limit "jumps" inside $\omega$ because $\omega$ is a continuous functional.)

In principle, everything stops here because you do not give any information about $F$. Judging by your other related questions, though, it seems reasonable to assume that $F$ is Lipschitz, in which case we may go further. In order to "slip" the limit inside the integral, we think about applying Lebesgue's dominated convergence theorem; since $\gamma$ is a Gaussian measure (and thus is finite), it will suffice to show that the integrand is bounded (being Lipschitz, therefore continuous, it is already measurable). This is easy to show:

$$\left\| \frac {F(u + t\phi + y) - F(u + y)} t \right\| \le \dfrac {Lip(F) |t| \| \phi \|} {|t|} = Lip(F) \| \phi \| ,$$

from which Lebesgue's theorem transforms the expression where we had stopped into

$$\omega \left( \int \limits _X \lim \limits _{t \to 0} \frac {F(u + t\phi + y) - F(u + y)} t \Bbb d \gamma(y) \right) .$$

Again, this does not seem to be enough, so let us now assume that $F$ is Gâteaux derivable at every point of $X$, in the direction $\phi$ (call this derivative $D$, not to be confused with the weak derivative $WD$ defined above). Then the above expression finally becomes

$$\omega \left( \int \limits _X D(F, u+y, \phi) \Bbb d \gamma(y) \right) = \omega ( D(F, u, \phi) * \gamma ) .$$

We may therefore conclude that if $F$ is Gâteaux derivable at every point in the direction $\phi$, then $F * \gamma$ is weakly Gâteaux derivable at every point in the direction $\phi$ and $WD (F * \gamma, x, \phi) = D(F, u, \phi) * \gamma$, i.e. the $*$ operation "weakens" your Gâteaux derivability.