Let $X$ and $Y$ be Banach spaces, $f:X\to Y$ and $\bar{x}\in X.$
We say that $f$ is strictly differentiable at $\bar{x}$ if there exists $\nabla f(\bar{x})\in L(X,Y)$ such that $$\lim_{x\to \bar{x}, u \to \bar{x},x\ne u}\frac{f(x)-f(u)-\nabla f(\bar{x})(x-u) }{\|x-u\|}=0.$$
We say that $f$ is continuously differentiable around $\bar{x}$ if there exists a neighborhood $U$ of $\bar{x}$ such that $f$ is Frechet differentiable at each point in $U$ and the mapping $\nabla f(\cdot):U \to L(X,Y)$ is norm to norm continuous.
The book I am reading states that if $f$ is continuously differentiable then it is also strictly differentiable at $\bar{x},$ but provides no proof. I have tried to prove this using standard $\epsilon-\delta$ arguments, but have failed. Can you provide this proof or at least point to a reference?
Hint: Given $x$ and $u$, define $\phi:[0,1]\to X$ by $$\phi(t)=f(u+t(x-u)).$$Use the fact that $f$ is continuously differentiable to show that $\phi$ is continuously differentiable. In doing so you'll come up with an expression for $\phi'$ in terms of $\nabla f$.
Now take $$f(x)-f(u)=\int_0^1\phi'(t)\,dt$$and rewrite the integral in terms of $\nabla f$...
Ok, it goes like so: It follows from the definitions that $$\phi'(t)= \nabla f(u+t(x-u))(x-u).$$Hence $$f(x)-f(u)-\nabla f(\overline x)(x-u) =\int_0^1(\nabla f(u+t(x-u))-\nabla f(\overline x))(x-u)\,dt,$$ so$$\begin{aligned}||f(x)-f(u)-\nabla f(\overline x)(x-u)|| &\le\int_0^1||(\nabla f(u+t(x-u))-\nabla f(\overline x))(x-u)||\,dt \\&\le||x-u||\int_0^1||\nabla f(u+t(x-u))-\nabla f(\overline x)||\,dt \\&\le\epsilon||x-u||\end{aligned}$$if $||x-\overline x||<\delta$ and $||u-\overline x||<\delta$ (noting that it follows that $||u+t(x-u)-\overline x||<\delta$).