# Does existence of mixed derivative imply continuity?

I would like to does existence of mixed derivative at a point guarantees the function is continuous at that point?

And I would like to ask if $${\frac {\partial ^{2}f}{\partial x^{2}}}$$ exist implies the function is continuous at that point. I believe the answer to this question is no because I can construct a function f(x) such that it equals to $$x^2$$ for $$x=0$$ and equals 1 otherwise.

But how about the first question?Thanks.

Tl;dr: The existence of a mixed partial derivative at some point doesn't imply that the function is continuous at said point. The weaker statement which is related, that is, "the existence of all mixed partial derivatives at some point implies continuity at the point" is also false.

For the following I will be assuming three dimensions (since there isn't much of a difference with higher dimensions).

Let's start with the example you gave, since it seems you already have some intuition built there: $$f(x, y) = \begin{cases}x^2 \quad \mbox{ if x = 0}\\ 0 \quad \mbox{ otherwise.} \end{cases}$$

Since you're looking at $$\frac{\partial^2f}{\partial x^2}$$, I have a feeling that the function you had in mind was to use $$x^2$$ when $$y=0$$ rather than $$x=0$$ like this:

$$f(x, y) = \begin{cases}x^2 \quad \mbox{ if y = 0}\\ 0 \quad \mbox{ otherwise.} \end{cases}$$

The advantage this gives becomes clear when we try to take the partial derivative with respect to $$x$$. In particular, we don't cross the boundaries defining the piecewise function when making small changes to $$x$$. Since this never happens, $$\frac{\partial f}{\partial x} = \begin{cases}2x \quad \mbox{ if y = 0}\\ 0 \quad \mbox{ otherwise.} \end{cases}$$ and $$\frac{\partial^2 f}{\partial x^2} = \begin{cases}2 \quad \mbox{ if y = 0}\\ 0 \quad \mbox{ otherwise.} \end{cases}$$

Of course, $$f$$ is not continuous at $$(3,0)$$, but the second partial derivative with respect to $$x$$ exists at that point. So, the existence of second degree (or any degree, for that matter) partial derivatives in one variable do not imply that the function is continuous.

Now back to your first question. Say we tweak the last example so

$$f(x, y) = \begin{cases}2 \quad \mbox{ if y = 0}\\ 0 \quad \mbox{ otherwise.} \end{cases}$$

Now $$\frac{\partial f}{\partial x} = \begin{cases}0 \quad \mbox{ if y = 0}\\ 0 \quad \mbox{ otherwise} \end{cases}$$, so $$\frac{\partial f}{\partial x} = 0$$ for all $$(x, y)$$. If we now take the partial with respect to $$y$$, the resulting function is still $$0$$ for all $$(x, y)$$. So now we have a function where the mixed derivative $$\frac{\partial^2 f}{\partial x \partial y} = 0$$ exists for all $$(x,y)$$ but is not continuous for $$y = 0$$.

This last result leaves room for other partial derivatives to not exist, namely $$\frac{\partial f}{\partial y}$$. By using a more esoteric piecewise function, we can give a counterexample to this as well.

Consider

$$f(x, y) = \begin{cases}1 \quad \mbox{ if x = y \not = 0}\\ 0 \quad \mbox{ otherwise.} \end{cases}$$

This function is zero everywhere except along the line $$y = x$$, except again at the point $$(0,0)$$. As a result, $$f$$ is not continuous at $$(0,0)$$ since approaching the point through the line $$y=x$$ gives a different result. However, no matter how many partial derivatives you take and in whatever order, the mixed partial derivative is defined (and is equal to $$0$$) at the point $$(0,0)$$.

The last case I can think of (which is well beyond the scope of the question, but still worth thinking about) is whether the existence of every mixed partial derivative at every point implies continuity for all points. My instinct tells me that this may be enough to get around the weird pitfalls that surround continuity for multivariable functions, but I'm far from confident.