Let $f$ be a uniformly continuous function defined on $B(0,1)$ above $\mathbb R^n$. Does there exist for every such $f$ a function $g$ which is continuous on $\bar{B}(0,1)$ for which $f=g$ for every $x \in B(0,1)$?
My intuition is yes. Taking $g$ and defining it $=f$ for every $x \in B(0,1)$, and for every $x_0 \in \partial B(0,1)$ ($\{x:||x||=1\})$, define it as $g(x) = \lim_{x \to x_0}f(x)$. My problem is showing that all the limits of $f$ on $\partial B(0,1)$ agree formally. I know I need to use the uniformly continuous trait of $f$ but I'm not sure how to do that. Thanks!
The answer is yes (I consider $\mathbb R^n$ as a normed space so it is a complete topological space). This is my hint. Let $(x_n)_n$ be a sequence in $B(0,1)$ which converges to $x_0 \in \partial B(0,1)$. Then $(x_n)_n$ is Cauchy sequence and since $f:B(0,1) \to \mathbb{R}$ is uniformly continuous in $B(0,1)$, it follows that also $(f(x_n))_n$ is a Cauchy sequence and it converges to some $u_0\in \mathbb{R}$. What happens if we take another sequence $(y_n)_n$ in $B(0,1)$ which converges to $x_0$? By the same argument it converges to some $v_0\in \mathbb{R}$. Now in order to show that $u_0=v_0$, consider the sequence $z_n$ which is equal to $x_n$ if $n$ is even and is equal to $y_n$ if $n$ is odd. Can you take it from here?
To show the continuity of the extension $g$, note that if $(x_n)_n$ is a sequence in $B(0,1)$ which converges to $x_0 \in \partial B(0,1)$ and $(y_n)_n$ is a sequence in $B(0,1)$ which converges to $y_0 \in \partial B(0,1)$ then $$ d(g(x_0),g(y_0)) \le d(g(x_n),g(x_0)) + d(g(x_n),g(y_n)) + d(g(y_n), g(y_0)).$$