# Thread: Bounded partial derivatives implies continuity.

1. ## Bounded partial derivatives implies continuity.

Theorem: If $\displaystyle f$ is a function defined on an open set $\displaystyle S \subset \mathbb{R}^{n}$ and the partial derivatives exist and are bounded on $\displaystyle S$, then $\displaystyle f$ is continuous on $\displaystyle S$.

So I sorta understand the proof in the case of $\displaystyle \mathbb{R}^{2}$ given at this site: CHAPTER 2 up until this point:

The two terms in the first bracket on the right hand side differ only in y, those in the second bracket only in x. We can therefore transform both brackets on the right hand side by means of the ordinary mean value theorem of the differential calculus (Volume 1), the first bracket as a function of y alone and the second as a function of x alone. We thus obtain the relation:

where q1 and q2 are two numbers between 0 and 1. In other words, the derivative with respect to y is to be formed for a point of the vertical line joining (x + h, y) to (x + h, y + k) and the derivative with respect to x for a point of the horizontal line joining (x, y) and (x + h, y)

I do not follow how that equality holds nor do I follow the explanation. Can someone explain more clearly how that equality holds and how this proof could be extrapolated to the general case of $\displaystyle \mathbb{R}^{n}$. Thank you.

Okay, I think I understand it, but isn't this proof using an assumption that $\displaystyle f$ is continuous since the mean value theorem is used?

2. The mean value theorem is applicable because, if you fix one variable, the fact that the partial derivatives exist means that the (now one-variable) function is differentiable and they're just applying it in a small interval.

For n variables the argument is the same, although you'll now have to give n segments parallel to the coordinate axis and apply the same argument in each.

,

,

### how can we show partial derivatives are bounded

Click on a term to search for related topics.