Let $X$ and $Y$ be real normed spaces and suppose $F:X\to Y$ has a bounded linear Gateaux derivative $F'$ at $x_0\in X$. (In the following, $F'$ also denotes a Gateaux derivative at other points in $X$, when it exists, and $U$ is an open neighborhood of $x_0$.) What are some conditions on $F$ that imply the existence of a Fréchet derivative at $x_0$? Please comment on (or provide counterexamples to) the following candidates:
- $\require{cancel} \cancel{F\text{ is continuous at }x_0.}$(see edit)
- $F$ is continuous on $U$.
- $F':U\to\mathcal{B}(X,Y)$ is continuous at $x_0$.
- $F':U\to\mathcal{B}(X,Y)$ is continuous on $U$.
In (3) and (4), continuity is with respect to the operator norm on the space of bounded linear maps $\mathcal B(X, Y)$, and $F'$ is assumed to exist on $U$.
Edit: I forgot there are some often-cited counterexamples to (1), e.g., $$f(x, y)=\{x,\text{ if }y=x^2;\;0,\text{ otherwise}\}.$$
$f$ is continuous at $(0,0)$ and has Gateaux derivative $0$, but has no Fréchet derivative there. Meanwhile (3) is claimed to be a sufficient condition in Oliver Diaz's answer, which means the stronger condition (4) is also sufficient. All that remains is condition (2)—is it sufficient?
Edit 2: I just found a counterexample to (2) on Wikipedia, namely $f(x, y)=\frac{x^2y\sqrt{x^2+y^2}}{x^4+y^2}$ for $(x, y)\neq(0,0)$, with $f(0, 0)=0$.
Best Answer
Here is a general result:
Here $\delta_xF:X\rightarrow Y$ is the Gâteaux derivative of $f$ at $\boldsymbol{x}$ which is defined as $$\delta_{\boldsymbol{x}}f(\boldsymbol{v})=\lim_{t\rightarrow0}\frac{F(\boldsymbol{x}+t\boldsymbol{v})-F(\boldsymbol{x})}{t}$$
This can be seen in may books in functional and nonlinear analysis. I think the book of Analysis and Functional Analysis by S. Lang has a detailed proof; also you can take a look at Klaus Diemling's Nonlinear Functional Analysis book.