# Gateaux and Frechet derivatives and boundedness

Suppose $f:X \to Y$ is map between Banach spaces.

We know that the Frechet derivative of $f$ at $x$ is a bounded map by definition if it exists (satisfying certain properties that I won't write) $f'(x):X \to Y$.

We also know that if one can show that the Gateaux derivative of $f$ at $x$, $DF(x):X \to Y$ is continuous at $x$, then the Frechet derivative at $x$ exists and is given by the Gateaux derivative at $x$.

I don't think that the Gateaux derivative of any function is required to be bounded. I am reading the free book Applied Analysis by Hunter and it doesn't say anything, but I am not sure.

So my question is, if I find the Gateaux derivative of a function at $x$ and show that it is continuous at $x$ without showing that it is bounded (i.e., $\lVert DF(x)h\rVert \leq C\lVert h\rVert$), can I still use the theorem to say that the Frechet derivative exists and hence is bounded? It seems wrong to say so.

By what I wrote above, the answer should be yes. But I have seen a lot of conflicting statements about Gateaux derivatives so I'm not sure.

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Be careful: the Fréchet derivative is a linear map. For many authors, the Gateaux derivative is not necessarily linear. Others say that $\varphi$ is G-differentiable at $u$ if there exists a linear and continuous map $f$ such that $\frac{d}{dt}\varphi(u+tv)\Big|_{t=0}=\langle f,v \rangle$ for every $v$. – Siminore Aug 22 '12 at 13:28
M. Berger, in his Nonlinearity and functional analysis, says that "if the Gateaux derivative is linear and continuous, then the Fréchet derivative exists and coincides with the G-derivative". Also Wikipedia shows the G-derivatives are often defined in different ways, in the literature. However, remember that the space of all linear mappings is hardly turned into a Banach space, if you forget boundedness; it becomes harder to define continuity for G-derivatives. – Siminore Aug 22 '12 at 13:39
@Siminore In the specific case I'm working on, the G-derivative is linear and I can show it's continuous. So I guess I should read the proof of the theorem and see if boundedness is used anywhere.. – Court Aug 22 '12 at 14:05
It doesn't come as a surprise: you need a uniform estimate but you start from an estimate on straight lines. It is reasonable that you need some continuity. – Siminore Aug 22 '12 at 14:24
was discussed here – user31373 Aug 22 '12 at 17:28