Relaxed condition for differentiability - proof check!

Question

I am trying to show that if the two partials of $f:\mathbb{R}^2\to\mathbb{R}$ exist at point, it is enough for only one to be continuous there to imply that $f$ is differentiable there. To show this I have decided to prove that:

$f:\mathbb{R}^2\to \mathbb{R}$ not differentiable at $(a,b)$. If $\partial_x f$ exists and is continuous at $(a,b)$, then $\partial_y f$ cannot exist at $(a,b)$.

My attempt

Assume we have the conditions above.

Take $h,k$ (wlog positive) and a sequence $t_0>t_1>\cdots$ where $t_n\to 0$, and $t_0>0$ sufficiently small.

For each $n$, by the Mean Value theorem, there exists $a<c_n<a+t_nh$ such that:

$$\frac{1}{t_n}\Big[ f(a+t_nh,b+t_nk)-f(a,b+t_nk)\Big]=\partial_x\,f (c_n)$$ As $t_n\to 0$ this converges because $\partial_x f$ exists at $(a,b)$. Write LHS as:

$$\frac{1}{t_n}\Big[ f(a+t_nh,b+t_nk)-f(a,b)\Big]+\frac{1}{t_n}\Big[ f(a,b)-f(a,b+t_nk)\Big]$$ Since the expression on the left does not converge, the one on the right cannot converge either. So $\partial_y f$ does not exist at $(a,b)$.

Question

I would be really happy if someone could tell me if any of this is correct and if not whether I am on track.

Answer 1

Your conjecture is correct but there are problems with your proof. You say at one point that if $Df(a,b)$ fails to exist, then for fixed $h,k,$ the limit

$$\lim_{n\to \infty} \frac{1}{t_n}\Big[ f(a+t_nh,b+t_nk)-f(a,b)\Big]$$

fails to exist. That's not true. The example $xy/(x^2+y^2)$ shows this.

WLOG, $(a,b) = (0,0).$ I'll phrase the result as a positive statement: If $f_x$ is continuous at $(0,0)$ and $f_y(0,0)$ exists, then $Df(0,0)$ exists.

Proof: For $(x,y)$ close to $(0,0),$ we have

$$f(x,y)-f(0,0) = f(x,y)-f(0,y)+f(0,y)- f(0,0)=f_x(c(x,y))x +f_y(0,0)y +o(y).$$

We used the existence of $f_x$ in a neighborhood of $(0,0),$ the mean value theorem, and the existence of $f_y(0,0)$ to get the last expression. This expression equals

$$f_x(0,0)x +f_y(0,0)y +(f_x(c(x,y))-f_x(0,0))x +o(y) = f_x(0,0)x +f_y(0,0)y +o(x) +o(y).$$

The continuity of $f_x$ at $(0,0)$ was used to get the $o(x)$ term. Since $o(x) +o(y) = o(\sqrt {x^2+y^2}),$ we see $Df(0,0)$ is the map $(x,y)\to f_x(0,0)x +f_y(0,0)y.$

Answer 2

No, it is not correct. Define $f(x,y) = \dfrac{xy}{x^2 + y^2}$, and $f(0,0) = 0$. Then $f = 0$ along the axes, so both partial derivatives of $f$ exist and are $0$ at $(0,0)$. On the other hand, $f$ isn't even continuous at $(0,0)$. For instance, the value of on the line $y = mx$ (excluding the origin) is $\dfrac{m}{m^2 + 1}$.