# Frechet derivative question

I'm trying to show that a map $f$ between Banach spaces $X$ and $Y$ is Frechet differentiable at a point $u$. To do this, it is enough to calculate its Gateaux derivative at $u$ (call it $df(u)$) and show that $df(u)$ is continuous: so for every $\epsilon$, there exists a $\delta$ such that if $$\lVert h_1 - h_2 \rVert_X \leq \delta,$$ then $$\lVert df(u)h_1 - df(u)h_2 \rVert_Y < \epsilon.$$ Is that correct to show Frechet differentiability?

Secondly, my Banach spaces $X$ and $Y$ are Hölder spaces. Specifically, let $X = C^{k, \alpha}(S)$, where $S$ is the closure of a set. Let $g$ be a smooth function in its arguments (so it's continuous). Since $S$ is closed and the norm is over a space of that set, can I say that $g$ is bounded above and hence $$\lVert gh_1 - gh_2 \rVert_X \leq C\rVert h_1 - h_2 \rVert_X$$ for some constant?

Thanks

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Is $g$ a function defined over $S$? And is $gh_1$ pointwise multiplication of functions? – Willie Wong Jul 10 '12 at 15:20
@WillieWong Yes to both questions – soup Jul 10 '12 at 15:53

You may be confusing two kinds of continuity. The Gateaux derivative is a linear functional from $X$ to $Y$, so one can talk about the continuity of this linear functional (which is usually a part of the definition of the Gateaux derivative). But the fact that it is a continuous linear functional does not make it a Frechet derivative. The additional requirement in the definition of the Frechet derivative is that the functional $df(u)h$ approximates $f(u+h)-f(u)$ in a uniform way: specifically, $\|df(u)h-(f(u+h)-f(u))\|/\|h\|\to 0$ as $\|h\|\to 0$.
It is true that a continuous Gateaux derivative is a Frechet derivative. But the continuity here is understood with respect to the point $u$ at which the derivative is taken: that it, $\|df(u_1)-df(u)\|\to 0$ when $u_1\to u$ in the norm.
Re: 2nd question: $g$ being bounded above is not a sufficient reason for your conclusion. You should get your hands dirty with the definition of the Hölder norm, and use the smoothness of $g$ in the process.