Proof without mean value theorem that continuously partially differentiable implies differentiability

Question

For simplicity we consider the case of $2$ factors. Let $E_1,E_2,F$ be Banach spaces and $f:X\to F$, where $X\subseteq E:=E_1\times E_2$ is open. By first partial derivative we mean the derivaitve of $f(\cdot,e_2)$ when we fix $e_2\in E_2$. Second (or other) partial derivatives are defined analogously.

I'm trying to prove the following theorem from H. Cartan's Differential Calculus:

This is the proof provided in the book:

where the mean value theorem used is this:

My question: Can we avoid the use of the mean value theorem? I think proving $(3.7.2)$ doesn't need the mean value theorem. Here is my proof:

That $f$ is differentiable at $x_0\in E$ with derivative $\partial f(x_0)$ is equivalent to saying that $$f(x)=f(x_0)+\partial f(x_0)(x-x_0)+r(x)\|x-x_0\|,$$ where $r:X\to F$ is continuous at $x_0$ with $r(x_0)=0$. Therefore, we have (again considering the case $n=2$ for simplicity) $$f(x_1,x_2)-f(a_1,x_2)-\frac{\partial f}{\partial x_1}(a_1,a_2)(x_1-a_1)=\Big(\frac{\partial f}{\partial x_1}(a_1,x_2)-\frac{\partial f}{\partial x_1}(a_1,a_2)\Big)(x_1-a_1)+r(x_1,x_2)\|x_1-a_1\|.$$ Since $\frac{\partial f}{\partial x_1}$ and $r$ are continuous at $(a_1,a_2)$, we can choose $\delta>0$ such that, for all $\|x_1-a_1\|+\|x_2-a_2\|<\delta$ (using the $\|\!\cdot\!\|_1$ product norm), we have $\big\|\frac{\partial f}{\partial x_1}(a_1,x_2)-\frac{\partial f}{\partial x_1}(a_1,a_2)\big\|\leq\varepsilon/2$ (operator norm) and $\|r(x_1,x_2)\|\leq\varepsilon/2$, so $$\Big\|f(x_1,x_2)-f(a_1,x_2)-\frac{\partial f}{\partial x_1}(a_1,a_2)(x_1-a_1)\Big\|\leq\varepsilon\|x_1-a_1\|.$$

Is my proof correct? If so, why is the author using the mean value theorem (which does not really simplify the proof in my eyes)?

Answer 1

You error occurs at the point of the proof where you write "we can choose $\epsilon > 0$ such that....".

You are not free to choose $\epsilon$ as you might wish to.

The statement you must prove at this point is:

• For all $\epsilon > 0$ there exists $\eta > 0$ such that if $\|x_1-a_1\| < \eta$ then $$(*) \qquad \|f(x_1,x_2)-f(a_1,x_2)-\frac{\partial f}{\partial x_1}(a_1,a_2)(x_1-a_1) \| < \epsilon \|x_1-a_1\| $$

One never starts a "for all $\epsilon > 0$" proof by saying "choose $\epsilon$ so that...". You do not have the freedom to choose $\epsilon$.

The mechanics of proving this statement are that you are given a value of $\epsilon>0$. Then you must find an appropriate value of $\eta > 0$, and use to prove that if $\|x_1-a_1\| < \eta$ then the inequality $(*)$ is true.

Take a look at what the mean value theorem says (I note that you did not copy the whole theorem in your post; you omitted the conclusion). You'll see how useful it is in carrying out the mechanics of the proof.