To show continuity at a point using boundedness of partial derivatives

Question

Suppose

$$U = (-1,1) \times (-1,1) \subset \mathbb R^2,\ f: U\to \mathbb R\ .$$

Assume that $\partial f/\partial x$ and $\partial f/\partial y$ exist at each point of U and are bounded on U. Show that $f$ is continuous at $(0,0)$.

Show by example that boundedness of the partial derivatives is necessary; mere existence is not enough.

Answer 1

Choose $a,b \in U$. Then we have $f(a)-f(b) = f(a)-f((a_1,b_2)) + f((a_1,b_2)-f(b)$. The one-variable mean value theorem gives $f(a)-f((a_1,b_2)) = \frac{\partial f((a_1,\xi_2))}{\partial y}(b_1-b_2) + \frac{\partial f(\xi_1, b_2)}{\partial x}(a_1-a_2)$ for some $\xi_k \in (a_k,b_k)$. Since the partial derivatives are bounded by some $B$, we have $\|f(a)-f(b)\|_2 \le B(|a_1-a_2|+|b_1-b_2|) \le 2B \|a-b\|_2$, we see that $f$ is Lipschitz continuous on $U$, hence continuous at $0$.

For the other part, let $p(t) = \frac{1}{2}(1+\cos ( 2\pi t)) 1_{[-\pi,\pi]}(t)$. We have $p(0) = 1$, $p'(\pm1) = 0$, and $p(t) = 0$ for $|t| \ge 1$. In particular, $p$ is $C^1$.

Define $f(x,y) = \begin{cases} p(\frac{y-x}{y+x}), & x>0, y>0 \\ 0, &\text{otherwise} \end{cases}$. Note that $f$ is $C^1$ on $\mathbb{R}^2 \setminus \{ (0,0) \}$, and $f$ has partial derivatives at $(0,0)$.

However, for $x>0$, $f(x,x) = 1$, where as $f(0,0) = 0$, hence $f$ is not continuous at $(0,0)$.