Good morning, I'm trying to prove that
Suppose $X$ is open in $\mathbb{R}^{n}$ and $F$ is a Banach space. Then $f: X \rightarrow F$ is continuously differentiable if $f$ has continuous partial derivatives.
Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!
My attempt:
For $a \in X$, we define $A \in \mathcal L(\mathbb R^n,F)$ by $$h=\left(h_{1}, \ldots, h_{n}\right) \mapsto A h=\sum_{k=1}^{n} \partial_{k} f(a) h_{k}$$
Our goal is to show that $\partial f(a) = A$ or equivalently $$\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)-A h}{|h|_\infty}=0$$
First, we choose $\varepsilon>0$ such that $\mathbb{B}(a, \varepsilon) \subseteq X$ and let $x_k = a+ (h_1,\ldots,h_k,0,\ldots,0)$ for all $k = \overline{1,n}$. It follows that $$f(a+h)-f(a)=\sum_{k=1}^{n}\left(f\left(x_{k}\right)-f\left(x_{k-1}\right)\right)$$
Let $\{e_1,\ldots,e_n\}$ be the standard basis of $\mathbb R^n$. By definition, we have $$\begin{aligned} \partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) &= \lim_{z \to 0} \frac{f\left(x_{k-1}+t h_{k} e_{k} + ze_k\right) - f\left(x_{k-1}+t h_{k} e_{k}\right)}{z} \\ &= \lim_{z \to 0} \frac{f\left(x_{k-1}+ (th_{k}+z) e_{k}\right) - f\left(x_{k-1}+t h_{k} e_{k}\right)}{z}\\ &= \frac{\partial f\left(x_{k-1}+t h_{k} e_{k}\right)}{\partial (th_k)} \end{aligned}$$
By Fundamental Theorem of Calculus, we have $$\begin{aligned} h_k\int_0^1 \partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) dt &= \int_0^1 \partial_{k} f\left(x_{k-1}+t h^{k} e_{k}\right) d(th_k)\\ &= \partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) \Big|_0^1 \\ &= f\left(x_{k}\right)-f\left(x_{k-1}\right) \end{aligned}$$
As such, $$f(a+h)-f(a)=\sum_{k=1}^{n} h_{k} \int_{0}^{1} \partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) d t$$
Consequently, $$\begin{aligned} \|f(a+h)-f(a) - Ah \| &=\left \|\sum_{k=1}^{n} h_{k} \int_{0}^{1} \left(\partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) -\partial_{k} f(a)\right) d t \right \| \\ &\le \sum_{k=1}^{n} |h_{k}| \int_{0}^{1} \left \| \partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) -\partial_{k} f(a)\right \| d t \\ &\le |h|_\infty \sum_{k=1}^{n} \int_{0}^{1} \left \| \partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) -\partial_{k} f(a)\right \| d t \\ &\le |h|_\infty \sum_{k=1}^{n} \int_{0}^{1} \sup_{t \in [0,1]} \left \| \partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) -\partial_{k} f(a) \right \| d t \\ &\le |h|_\infty \sum_{k=1}^{n} \int_{0}^{1} \sup_{x \in \mathbb{B}(a, \|h\|_\infty)} \left \| \partial_{k} f\left(x\right) -\partial_{k} f(a)\right \| d t \\&= |h|_\infty \sum_{k=1}^{n} \sup_{x \in \mathbb{B}(a, \|h\|_\infty)} \left \| \partial_{k} f\left(x\right) -\partial_{k} f(a)\right \|\end{aligned}$$
We have $h \to 0$ implies $\|h\|_\infty \to 0$, which in turn implies $x \to a$. It follows from the continuity of $\partial_{k} f\left(x\right)$ that $ \sup_{x \in \mathbb{B}(a, \|h\|_\infty)} \left \| \partial_{k} f\left(x\right) -\partial_{k} f(a)\right \| \to 0$ as $x \to a$.
Finally, $$\lim _{h \rightarrow 0} \frac{\| f(a+h)-f(a)-A h \|}{|h|_\infty} \le \lim _{h \rightarrow 0} \sum_{k=1}^{n} \sup_{x \in \mathbb{B}(a, \|h\|_\infty)} \left \| \partial_{k} f\left(x\right) -\partial_{k} f(a)\right \| = 0$$ Consequently, $$\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)-A h}{|h|_\infty}=0$$
Hence $\partial f(a) = A$. Next we prove that $\partial f(\cdot): X \to \mathcal L(\mathbb R^n,F)$ is continuous. We have $$\begin{aligned}\|\partial f(x)h - \partial f(a)h\| &= \left\| \sum_{k=1}^{n} \partial_{k} f\left(x\right) h_{k} - \sum_{k=1}^{n} \partial_{k} f\left(a\right) h_{k} \right\| \\ &= \left\| h_k \sum_{k=1}^{n} ( \partial_{k} f\left(x\right) - \partial_{k} f\left(a\right)) \right\| \\&\le \sum_{k=1}^{n} \left\|\partial_{k} f\left(x\right) - \partial_{k} f\left(a\right) \right\| \cdot |h|_\infty \end{aligned}$$
Consequently, $$\|\partial f(x) - \partial f(y)\| = \sup_{h \in X} \frac{\|\partial f(x)h - \partial f(a)h\|}{|h|_\infty} \le \sum_{k=1}^{n} \left\|\partial_{k} f\left(x\right) - \partial_{k} f\left(a\right) \right\|$$
It follows from the continuity of $\partial_{k} f\left(\cdot\right)$ that $\sum_{k=1}^{n} \left\|\partial_{k} f\left(x\right) - \partial_{k} f\left(a\right) \right\| \to 0$ and thus $\|\partial f(x) - \partial f(y)\| \to 0$ as $x \to a$. Hence $\partial f(x) \to \partial f(y)$.
