# Proving that existence of bounded partial derivatives implies continuity of a function.

Let $E \subset \mathbb{R^n}$ be open and let $f:E \to \mathbb{R}$. Suppose that ${\partial f \over \partial x_1}, ..., {\partial f \over \partial x_n}$ exist and are bounded in E. Prove that $f$ is continuous in $E$

How in the world do I prove this? I thought that the existence of partial derivatives did not imply continuity of the function. How does the condition that each partial is bounded affect that?

Any hints or possibly a sketch of the proof would be welcome. Thanks.

The subscript on the latter partial is an n by the way. I know it's hard to read.

Hint: Apply the mean value theorem coordinate by coordinate to $$f(a+h,b+k)-f(a,b)=\big(f(a+h,b+k)-f(a+h,b)\big)+\big(f(a+h,b)-f(a,b)\big).$$
• (Then generalize to $n$ coordinates) – Pedro Tamaroff Feb 8 '14 at 3:51
• Aren't you only proving that $f$ has all directional derivatives? – Sha Vuklia Apr 9 '17 at 8:41
• @ShaVuklia: Nothing to do with directional derivatives here. We're trying to prove continuity by showing $\lim\limits_{h,k\to 0} |f(a+h,b+k)-f(a,b)| = 0$. – Ted Shifrin Apr 9 '17 at 15:40