In what follows, $X, Y$ are Banach and $U \subseteq X$ is open. Consider the map $f: U \to Y$. We say that $f$ is locally uniformly differentiable at $x \in U$ and $h \in X$ if given $\epsilon > 0$ there exists a neighborhood $V$ of $x$ and an $\epsilon_{0} > 0$ such that: $$\bigg{|}\bigg{|}\frac{f(y+th)-f(y)}{t} - \delta_{y}(h)\bigg{|}\bigg{|} < \epsilon $$ holds for every $y \in V$, $||h|| < 1$ and $|t| < \epsilon_{0}$.
I have to prove that if $f$ is locally uniformly continuous on $U$ then it is Fréchet differentiable on $U$ and its Fréchet derivative $Df(x) = \delta_{x}$ is continuous.
I already proved the first part, that is, that the Fréchet derivative exists and satisfies $Df(x) = \delta_{x}$. But I'm stuck at proving this derivative is continuous. I was trying to construct a sequence $\{x_{n}\}_{n\in \mathbb{N}}$ such that $x_{n} \to x$ and proving $||(Df(x_{n})-Df(x))h|| \le \epsilon ||h||$ for every $||h|| < 1$, but I got nowhere because $h$ and $x_{n}$ must be independent so I couldn't use the condition of the definition. Can you help me, please?
Hint: If $\|x - y\|$ and $t$ is small enough, the vectors \begin{align*} f(x) - f(y) &- \delta_y(x - y),\\ f(x + th) - f(y) &- \delta_y(x + th - y),\\ f(x + th) - f(x) &- \delta_x(t h) \end{align*} are small. Now, you can combine these vectors to $(\delta_x - \delta_y)(h)$.