# Prove that continuous partial derivatives imply continuous total derivative

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 $$\partial f(x) \to \partial f(y)$$.

I've just figured out a variant that utilized the MVT, so I posted it here. I would be great if someone helps me verify it. Thank you so much for your help!

My attempt:

For $$a \in X$$, we define $$A \in \mathcal L(\mathbb R^n,F)$$ by $$A h=\sum_{k=1}^{n} \partial_{k} f(a) h_{k}$$ for all $$h=\left(h_{1}, \ldots, h_{n}\right) \in X$$. Our goal is to show that $$\partial f(a) = A$$ or equivalently $$\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)-A h}{|h|_\infty}=0$$

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} h_k\partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) &= h_k \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}e_k+(z/h_k)h_ke_{k}\right) - f\left(x_{k-1}+t h_{k} e_{k}\right)}{z/h_k}\\ &= \lim_{z' \to 0} \frac{f\left(x_{k-1}+ th_{k}e_k+z'h_ke_{k}\right) - f\left(x_{k-1}+t h_{k} e_{k}\right)}{z'}\\ &= \lim_{z \to 0} \frac{f\left(x_{k-1}+ (t+z')h_{k}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 t} \end{aligned}

We apply Mean Value Theorem for the map $$\mathbb R \to F, \quad t \mapsto f\left(x_{k-1}+t h_{k} e_{k}\right)$$ and get \begin{aligned}f(x_k) - f(x_{k-1}) &= f(x_{k-1}+1\cdot h_ke_k)-f(x_{k-1}+0\cdot h_ke_k)\\ &= \frac{\partial f\left(x_{k-1}+t h_{k} e_{k}\right)}{\partial t} (t_k) \\ &= h_k \partial_{k} f\left(x_{k-1}+t_k h_{k} e_{k}\right)\end{aligned}

Consequently, \begin{aligned} \|f(a+h)-f(a) - Ah \| &=\left \|\sum_{k=1}^{n} h_{k} \left(\partial_{k} f\left(x_{k-1}+t_k h_{k} e_{k}\right) -\partial_{k} f(a)\right) \right \| \\ &\le \sum_{k=1}^{n} |h_{k}| \left \| \partial_{k} f\left(x_{k-1}+t_k h_{k} e_{k}\right) -\partial_{k} f(a)\right \| \\ &\le |h|_\infty \sum_{k=1}^{n} \left \| \partial_{k} f\left(x_{k-1}+t_k h_{k} e_{k}\right) -\partial_{k} f(a)\right \| \\ &\le |h|_\infty \sum_{k=1}^{n} \sup_{t \in [0,1]} \left \| \partial_{k} f\left(x_{k-1}+t h_{k} e_{k}\right) -\partial_{k} f(a) \right \| \\ &\le |h|_\infty \sum_{k=1}^{n} \sup_{|x-a|_\infty \le |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 $$\partial_{k} f\left(\cdot\right)$$ that $$\sup_{|x-a|_\infty \le |h|_\infty} \left \| \partial_{k} f\left(x\right) -\partial_{k} f(a)\right \| \to 0 \quad (x \to a)$$

Finally, $$\lim _{h \rightarrow 0} \frac{\| f(a+h)-f(a)-A h \|}{|h|_\infty} \le \lim _{h \rightarrow 0} \sum_{k=1}^{n} \sup_{|x-a|_\infty \le |h|_\infty} \left \| \partial_{k} f\left(x\right) -\partial_{k} f(a)\right \| = 0$$ 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 |h|_\infty \sum_{k=1}^{n} \left\|\partial_{k} f\left(x\right) - \partial_{k} f\left(a\right) \right\| \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 $$\partial f(x) \to \partial f(y)$$.