Let $(E, \|\cdot\|)$ be a real Banach space and $X$ an open subset of $\mathbb R$. I'm trying to prove below result which is used in the proof of Hille-Yosida theorem in Brezis' Functional Analysis.
Let $f_n, f, g: X \to E$ such that $f_n \in C^1 (X; E)$ for $n \ge 1$. Assume that
- $(f_n)$ converges pointwise to $f$, and
- $(f'_n)$ converges locally uniformly to $g$, i.e., for each $x \in X$ there is a neighborhood $U$ of $x$ such that $(f'_n)$ converges uniformly (for the supremum norm) to $g$ on $U$.
Then $f \in C^1 (X; E)$ and $f'=g$. 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?
Fix $a \in X$. There is $r>0$ such that $B(a, r) \subset X$ and that $(f'_n)$ converges uniformly to $g$ on $B(a, r)$. For each $x\in U:= B(a, r/2)$, we consider $$ F : [0, 1] \to E, t \mapsto f_n(a+t(x-a)) - t f'_n (a)(x-a). $$
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, $$ F' (t) = \big ( f'_n(a+t(x-a)) - f'_n (a) \big ) (x-a) \quad \forall t \in (0, 1). $$
By mean value theorem, $\|F(1)-F(0)\| \le \sup_{t \in (0, 1)} \|F' (t)\|$. Then $$ \|f_n (x) - f_n (a) - f'_n (a) (x-a)\| \le |x-a| \sup_{t \in (0, 1)} \|f'_n(a+t(x-a)) - f'_n (a)\|. $$
By triangle inequality, $$ \begin{align*} & \sup_{t \in (0, 1)} \|f'_n (a+t(x-a)) - f'_n (a)\| \\ \le{} & \sup_{t \in (0, 1)} \|f'_n (a+t(x-a)) - g(a+t(x-a))\| +\sup_{t \in (0, 1)} \|g(a) - f'_n (a)\| \\ &\quad + \sup_{t \in (0, 1)} \|g(a+t(x-a)) - g (a)\|\\ \le{} & 2\|(f'_n-g)1_{U}\|_\infty + \sup_{t \in (0, 1)} \|g(a+t(x-a)) - g (a)\|. \end{align*} $$
Taking the limit $n \to \infty$, we get for all $x \in U \setminus \{a\}$, $$ \begin{align*} \frac{\|f (x) - f (a) - g (a) (x-a)\|}{|x-a|} \le \sup_{t \in (0, 1)} \|g(a+t(x-a)) - g (a)\|. \end{align*} $$
We have $f'_n$ is continuous and $(f'_n)$ converges uniformly to $g$ on $U$, so $g$ is continuous on $U$. This implies $$ \lim_{x \to a} \sup_{t \in (0, 1)} \|g(a+t(x-a)) - g (a)\|=0. $$
Thus $$ \lim_{x \to a} \frac{\|f (x) - f (a) - g (a) (x-a)\|}{|x-a|} =0. $$
This implies $f'(a)=g(a)$. Since this holds for all $a\in X$, we get $f'=g$ 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, $$ G' (t) = (f'_n-f')(a+t(x-a))(x-a) \quad \forall t \in (0, 1). $$
By mean value theorem, we get $\|G(1)-G(0)\| \le \sup_{t \in (0, 1)} \|G' (t)\|$. Then for $x \in U$, $$ \begin{align*} \|f_n (x)-f(x) \| &\le \|(f_n (x)-f(x)) - (f_n (a)-f(a)) \| + \|f_n (a)-f(a) \| \\ &=\|G(1)-G(0)\|+\|f_n (a)-f(a) \| \\ &\le |x-a|\sup_{t \in (0, 1)} \|(f'_n-f')(a+t(x-a))\| + \|f_n (a)-f(a) \| \\ &\le r\|(f'_n-g)1_{U}\|_\infty + \|f_n (a)-f(a) \|. \end{align*} $$
The claim then follows.
