A weak definition of multiple differentiability of a function of several variables

Question

Several times I faced the following

Definition 1: A function $f: \mathbb{R^n}\mapsto \mathbb{R}$ is called $k$ times differentiable at $x_0$ iff all the partial derivatives of $f$ of order $k-1$ are differentiable at $x_0$.

Besides this definition, more popular is the other

Definition 2: A function $f: \mathbb{R^n}\mapsto \mathbb{R}$ is called $k$ times differentiable at $x_0$ iff $f$ is $k-1$ times differentiable in a neighborhood of $x_0$ and all the partial derivatives of $f$ of order $k-1$ are differentiable at $x_0$.

It's obvious, that in case $n=1$ these two definitions coincide. Besides that, it can be shown that Def. 2 coincides with general Frechet $k$-times differentiability and obviously it's not weaker than Def 1. On the other hand, it can be shown that Def. 1 is sufficient to prove general Young's theorem. In this regard, the following question arose:

1. Does anyone know an example of a function (say 2 times differentiable of 2 variables) which satisfies the Def. 1, but not the Def. 2? I have some ideas of constructing such an example with the use of Sobolev mollifications, but it's quite complicated and ugly ….

2. If such an example exists, is Taylor theorem with Peano remainder valid for the functions which satisfy the Def. 1?

Answer 1

Here is an example (with $k=2$) of a function satisfying Def. 1 but not Def. 2. With $r:=\sqrt{x^2+y^2}$, let $g(x,y):=xy/r\in[-r,r]$ if $r\ne0$, with $g(0,0):=0$. Everywhere here, $x$ and $y$ are any real numbers. Then $g$ is differentiable everywhere except at $(0,0)$ (and continuous everywhere), but $g$ has both partial derivatives everywhere: $$\text{$g'_x(x,y)=y^3/r^3\in[-1,1]$ and $g'_y(x,y)=x^3/r^3\in[-1,1]$ if $r\ne0$, }$$ and $g'_x(0,0)=g'_y(0,0)=0$.

Let $$f(x,y):=\sum_1^\infty\frac{r^3}{j^2}\,g(x-1/j,y).$$ Then $f$ is not differentiable at any of the points $(1/j,0)$ (and hence is not differentiable in any neighborhood of $(0,0)$), so that $f$ does not satisfy Def. 2.

However, by dominated convergence, $f$ has both partial derivatives everywhere, and (as $r\to0$) $$f'_x(x,y)=\sum_1^\infty\frac{r^3}{j^2}\,g'_x(x-1/j,y)+\sum_1^\infty\frac{3rx}{j^2}\,g(x-1/j,y)\\ =O(r^2)=o(r),$$ $$f'_y(x,y)=\sum_1^\infty\frac{r^3}{j^2}\,g'_y(x-1/j,y)+\sum_1^\infty\frac{3ry}{j^2}\,g(x-1/j,y)\\ =O(r^2)=o(r).$$ So, $f'_x$ and $f'_y$ are both differentiable at $(0,0)$. Thus, $f$ satisfies Def. 1.

Answer 2

Now I got the answer to the second part of my own question: let $f(x,y)$ be twice differentiable at $(0,0)$ by Definition 1. Then $f(x,y)=P(x,y)+o(r^2)$, where $P(x,y)=f(0,0)+f_x(0,0)x+f_y(0,0)y+\frac{1}{2}f_{xx}(0,0)x^2+f_{xy}(0,0)xy+\frac{1}{2}f_{yy}(0,0)y^2$.

Really, denote $g(x,y):=f(x,y)-P(x,y)$. Then $g(x,y)=[g(x,y)-g(0,y)]+[g(0,y)-g(0,0)]$. We have $$g(x,y)-g(0,y) = g_x(\xi_1,y)x=[f_x(\xi_1,y)-f_x(0,0)-f_{xx}(0,0)\xi_1-f_{xy}(0,0)y]x=o(\|(\xi_1,y)\|)x=o(r^2),$$ where $\xi_1=\xi_1(x,y)\in(0,x)$. Similarly, $$g(0,y)-g(0,0)=g_y(0,\xi_2)=o(\|(0,\xi_2)\|)y=o(r^2).$$ So the classical Taylor theorem assumptions can be weakened in finite-dimensional case!?? Anyone saw this somewhere??