# ANALYSIS II: continuity of function in R^n

1. Mar 5, 2008

### jjou

[SOLVED] ANALYSIS II: continuity of function in R^n

Let $$A\subset\mathbb{R}^n$$ and $$f:A\rightarrow\mathbb{R}$$. Show that, if the partial derivatives $$D_jf(x)$$ exist and are bounded on $$B_r(a)\subset A$$, then f is continuous at a.

We know:
$$D_jf(x_1,...,x_n)=\lim_{h\rightarrow0}\frac{f(x_1,...,x_j+h,...,x_n)-f(x_1,...,x_n)}{h} \leq B$$ for all $$1\leq j\leq n$$, |h|<r for some B>0.
f is continuous at a if, for any $$\epsilon>0$$, there exists $$\delta>0$$ such that $$|f(x)-f(a)|<\epsilon$$ whenever $$||x-a||<\delta$$.
Equivalently, f is continuous at a if, for any sequence $$(x_n)$$ converging to a, $$f(x_n)$$ converges to f(a).

I tried showing f to be Lipschitz (and thus continuous) or to estimate |f(x)-f(a)| for $$x\in B_r(a)$$ by the following:
$$|f(x)-f(a)|\leq|f(x)-f(x_1,...,x_{n-1},a_n)|+|f(x_1,...,x_{n-1},a_n)-f(x_1,...,x_{n-2},a_{n-1},a_n)|+...+|f(x_1,a_2,...a_n)-f(a)|$$
since then each term in the right hand sum involves only a change in one coordinate (closer to a partial derivative).

I'm getting stuck though on how to proceed from here since partials involve taking limits, whereas continuity does not. Also, the existence of partials does not guarantee differentiability of f. Am I on the right track or way off? Any hints would be highly appreciated.

Thank you. :)

Last edited: Mar 5, 2008
2. Mar 5, 2008