$\newcommand{\vertiii}[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}$Let $(E, |\cdot|_E)$ be a real Banach space and $X$ an open subset of a real Banach space $(Y, |\cdot|_Y)$. Let $\mathcal L(Y, E)$ be the Banach space of bounded linear operators from $Y$ to $E$. We denote by $\| \cdot \|$ the operator norm on $\mathcal L(Y, E)$. For a differentiable function $f: X \rightarrow E$,
- we denote by $\partial f (x) \in \mathcal L(Y, E)$ its Fréchet derivative at $x \in X$.
- we denote by $\partial f$ its Fréchet derivative.
For a map $g:X \to \mathcal L(Y, E)$,
- its supremum norm $\vertiii{g }_\infty$ is defined as $$ \vertiii{g}_\infty := \sup_{x\in X} \|g(x)\|. $$
- we denote by $g(x)[y] \in E$ the value of function $g (x)$ at $y$ for $x\in X$ and $y\in Y$.
To better grasp Fréchet derivative, I'm trying to generalize this result, i.e.,
Let $f_n, f: X \to E$ such that $f_n \in C^1 (X; E)$ for $n \ge 1$ and $g:X \to \mathcal L(Y, E)$. Assume that
- $(f_n)$ converges pointwise to $f$, and
- $(\partial f_n)$ converges locally uniformly to $g$, i.e., for each $x \in X$ there is a neighborhood $U$ of $x$ such that $(\partial f_n)$ converges uniformly (for the norm $\vertiii{\cdot }_\infty$) to $g$ on $U$.
Then $f \in C^1 (X; E)$ and $\partial f=g$. In addition, $(f_n)$ converges locally uniformly to $f$.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
First, we need the following mean value theorem (MVT) for vector-valued functions. In what follows, we use the notation $[\![ x, y ]\!]$ for the straight path $\{x+t(y-x) ; t \in[0,1]\}$ between the points $x, y \in Y$.
Suppose $f: X \rightarrow E$ is differentiable. Then $$ |f(x)-f(y)|_E \leq \sup _{t \in (0, 1)} \| \partial f(x+t(y-x)) \| \, |y-x|_Y $$ for all $x, y \in X$ such that $[\![ x, y ]\!] \subset X$.
Fix $a \in X$. There is $r>0$ such that $B(a, r) \subset X$ and that $(\partial f_n)$ converges uniformly to $g$ on $B(a, r)$. For each $x\in U:= B(a, r/2)$, we consider $$ F : [0, 1] \to E, t \mapsto f_n(a+t(x-a)) - t \partial f_n (a)[x-a]. $$
We have $f_n$ is continuous on $\overline U$ and differentiable on $U$, so $F$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$. Also, $$ F' (t) = \big ( \partial f_n(a+t(x-a)) - \partial f_n (a) \big ) [x-a] \quad \forall t \in (0, 1). $$
By MVT, $|F(1)-F(0)|_E \le \sup_{t \in (0, 1)} |F' (t)|_E$. Then $$ |f_n (x) - f_n (a) - \partial f_n (a) [x-a]|_E \le |x-a|_Y \sup_{t \in (0, 1)} \| \partial f_n(a+t(x-a)) - \partial f_n (a)\|. $$
By triangle inequality, $$ \begin{align*} & \sup_{t \in (0, 1)} \|\partial f_n (a+t(x-a)) - \partial f_n (a)\| \\ \le{} & \sup_{t \in (0, 1)} \|\partial f_n (a+t(x-a)) - g(a+t(x-a))\| + \sup_{t \in (0, 1)} \|g(a) - \partial f_n (a)\| \\ &\quad + \sup_{t \in (0, 1)} \|g(a+t(x-a)) - g (a)\|\\ \le{} & 2\vertiii{(\partial f_n-g)1_{U}}_\infty + \sup_{t \in (0, 1)} \|g(a+t(x-a)) - g (a)\|. \end{align*} $$
Taking the limit $n \to \infty$, we get for all $x \in U \setminus \{a\}$, $$ \begin{align*} \frac{|f (x) - f (a) - g (a) [x-a]|_E}{|x-a|_Y} \le \sup_{t \in (0, 1)} \|g(a+t(x-a)) - g (a)\|. \end{align*} $$
We have $\partial f_n$ is continuous and $(\partial f_n)$ converges uniformly to $g$ on $U$. By uniform limit theorem, $g$ is continuous on $U$. This implies $$ \lim_{x \to a} \sup_{t \in (0, 1)} \|g(a+t(x-a)) - g (a)\|=0. $$
Thus $$ \lim_{x \to a} \frac{|f (x) - f (a) - g (a) [x-a]|_E}{|x-a|_Y} =0. $$
This implies $\partial f(a)=g(a)$. Since this holds for all $a\in X$, we get $\partial f=g$ and thus $f \in C^1 (X; E)$. It remains to prove that $(f_n)$ converges locally uniformly to $f$. We consider $$ G : [0, 1] \to E, t \mapsto (f_n-f)(a+t(x-a)). $$
We have $f_n,f$ are continuous on $\overline U$ and differentiable on $U$, so $G$ is continuous on $[0, 1]$ and differentiable on $(0, 1)$. Also, $$ G' (t) = (\partial f_n - \partial f)(a+t(x-a))[x-a] \quad \forall t \in (0, 1). $$
By MVT, we get $|G(1)-G(0)|_E \le \sup_{t \in (0, 1)} |G' (t)|_E$. Then for $x \in U$, $$ \begin{align*} |f_n (x)-f(x) |_E &\le |(f_n (x)-f(x)) - (f_n (a)-f(a)) |_E + |f_n (a)-f(a)|_E \\ &=|G(1)-G(0)|_E + |f_n (a)-f(a)|_E \\ &\le |x-a|_Y\sup_{t \in (0, 1)} \|(\partial f_n - \partial f)(a+t(x-a))\| + |f_n (a)-f(a)|_E \\ &\le r \vertiii{ (\partial f_n - g)1_{U}}_\infty + |f_n (a)-f(a)|_E. \end{align*} $$
The claim then follows.
