# Show that if $f$ is totally differentiable at $x_0$ then some partial derivatives of $f$ exists in a neighborhood of $x_0$

This is the exercise 6 in page 165 of Analysis II of Amann and Escher

Suppose $X$ is open in $\Bbb R^n$, $F$ is a Banach space and $f:X\to F$. Also suppose that $x_0\in X$ and choose $\epsilon>0$ such that $\Bbb B(x_0,\epsilon)\subset X$. Finally let $$x_k(h):=x_0+h_1e_1+\ldots+h_ke_k\quad\text{for }k=1,2,\ldots,n\text{ and }h\in\Bbb B(x_0,\epsilon)$$

Prove that $f$ is differentiable at $x_0$ if and only if for every $h\in\Bbb B(x_0,\epsilon)$ such that $h_k\neq 0$ the limits $$\lim_{\substack{h_k\to 0\\h_k\neq 0}}\frac{f(x_k(h))-f(x_{k-1}(h))}{h_k},\quad k=1,\ldots,n$$ exists in $F$.

First some observations: the vectors $e_k$ are the standard basis of $\Bbb R^n$ and $h:=(h_1,\ldots,h_n)$. It seems a error the definition of $h\in\Bbb B(x_0,\epsilon)$ because its possible that $x_k(h)\notin X$. It seems that it must be $h\in\Bbb B(0,\epsilon)$ instead.

From now, to short things a little bit, I set $x_k:=x_k(h)$. I dont know exactly what to do or what to search for. Searching something for the first implication I search for an useful upper bound to show the existence of the total derivative at $x_0$, by example

$$\frac{\|f(x_0+h)-f(x_0)-\partial f(x_0)h\|}{|h|}=|h|^{-1}\left\|\left(\sum_{k=1}^n(f(x_k)-f(x_{k-1}))\right)-\partial f(x_0)\right\|\\\le\sum_{k=1}^n\frac{\|f(x_k)-f(x_{k-1})-\partial_k f(x_0)h_k\|}{|h_k|}$$

with $|h_k|\neq 0$ for all $k=1,\ldots,n$, and we used the fact that $|h|\ge|h_k|$. But I dont see how to continue from here.

After I thought that if I stablish an equivalence between the existence of the stated limits and the continuity of the partial derivatives of $f$ at $x_0$ Im done. But nothing useful comes from this idea.

Some help will be appreciated.

P.S.: I forget to say that the stated limits are equivalent to state the existence of the partial derivatives $\partial_k f(x_{k-1})$.

• I do not understand what these limits are in "Prove that $f$ is differentiable at $x_0$ if and only if for every $h\in\Bbb B(x_0,\epsilon)$ such that $h_k\neq 0$ the limits $$\lim_{\substack{h_k\to 0\\h_k\neq 0}}\frac{f(x_k(h))-f(x_{k-1}(h))}{h_k},\quad k=1,\ldots,n$$ exists in $F$." – zhw. Sep 3 '17 at 14:57
• @zhw these limits are by definition $\partial_k f(x_{k-1})$ – Masacroso Sep 3 '17 at 16:21
• Are you sure the limit is for $h_k \to 0$ and not for $h \to 0$ (under the constraint that $h_k \neq 0$)? As written, let $g$ the characteristic function of $\mathbb{Q}^n$ and $f(x) = \lVert x\rVert^2\cdot g(x)$. $f$ is differentiable at $0$, but partial derivatives exist only at some points. Take a more complicated $g$ and partial derivatives exist only at $0$. – Daniel Fischer Sep 3 '17 at 16:30
• @Daniel yes, I copied exactly, word by word and $\LaTeX$ by $\LaTeX$. I must add that probably the exercise is wrong, not sure anyway. – Masacroso Sep 3 '17 at 17:01
• @zhw I dont see why they are wrong... they follow exactly the definition of partial derivatives for the points $x_{k-1}$, that is, they are directional derivatives in the directions of the vector basis. – Masacroso Sep 4 '17 at 0:09