# Existence/Boundedness of Partial Derivatives in open set implies continuity

• Sep 15th 2010, 08:14 PM
ajskim
Existence/Boundedness of Partial Derivatives in open set implies continuity
Hi,
If we have $U\subset R^n$, U open, and f:U->R, prove that if all of f's partial derivatives exist and are bounded then f is continuous.
Here is my attempt at a solution:
I have tried approaching this by using the triangle inequality; if the partial exists in the ith direction, then if we restrict the function along that direction then I think it must also be continuous along that direction. From this we can use the triangle inequality, choosing delta_i s.t. |f(x)-f(x')|<E/n for all x' s.t. |x-x'|<delta_i. Doing this over all directions and choosing the smallest delta should let us use the triangle inequality and get continuity.
Is this a valid approach? Also, why is the open set required? To insure existence of the open ball from which to get continuity?
• Sep 17th 2010, 04:32 PM
Iondor
That looks about right. You just have to make sure that the intermediate points you choose for the triangle inequality all lie within the domain U, where f is defined and that might also be the reason why U is required to be an open set, because then it is ensured you can always find such intermediate points, that lie within U.

But you also need the property that the partial derivatives are bounded. From this you can conclude that the function is locally lipschitz continous along the directions of the axis and you have to use this somehow in your inequality.