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Prove: For every uniformly continuous function $f$ defined on the open ball $B(0,1)=\{ x\in \mathbb{R}^n: \sum_{i=1}^nx_i^2<1\},$ there exists a continuous function $g$ which is defined on the closed ball $\bar B(0,1)$ such that for all $x\in B(0,1): f(x)=g(x).$

My thoughts:

I started by defining $g(x)=f(x), \forall x\in B(0,1),$ so $g$ is obviously continuous on the open ball.

Now, we need to take care of the boundary points.

I thought to take a sequence of points from $B(0,1),$ and use the uniform convergence in order to show in converges to a point on the boundary, but not quite sure if that's the right way, and if so, how to formalize it.

Any help appreciated.

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For any point $x\in \partial B(0, 1)$, let $\{x_n\}_{n=1}^{\infty}\subset B(0, 1)$ such that $x_n\to x$. We will show that $\{f(x_n)\}$ is Cauchy. For $\epsilon > 0$, we consider $\delta_1(\epsilon) > 0$ such that $\|x-y\| < \delta_1(\epsilon)$ implies $\|f(x)-f(y)\| < \epsilon$ for all $x, y\in B(0, 1)$. Now, as $\{x_n\}$ is Cauchy, for any $\delta > 0$ we have some $N(\delta)\in \mathbb{N}$ such that $m, n\geq N$ implies $\|x_n-x_m\| < \delta$. Then, for $m, n\geq N(\delta_1(\epsilon))$, we have $\|x_n-x_m\| < \delta_1(\epsilon)$, and so $$\|f(x_n)-f(x_m)\| < \epsilon$$ Therefore, we let $\tilde{f}(x) = \lim_{n\to \infty} f(x_n)$. We will show that this is well-defined, i.e. for any $\{y_n\}_{n=1}^{\infty}\subset B(0, 1)$, $y_n\to x$ implies $f(y_n)\to \tilde{f}(x)$. As $x_n\to x$ and $y_n\to x$, for any $\epsilon > 0$, we have some $N\in \mathbb{N}$ such that $\|x-x_n\| < \frac{\delta_1(\epsilon)}{2}$ and $\|x-y_n\| < \frac{\delta_1(\epsilon)}{2}$ for $n\geq N$. Then, by the triangle inequality, $$\|x_n-y_n\| = \|(x-x_n)-(x-y_n)\|\leq \|x-x_n\|+\|x-y_n\| < \delta_1(\epsilon)$$ which implies that $\|f(x_n)-f(y_n)\| < \epsilon$ for $n\geq N$. This implies that $f(x_n)-f(y_n)\to 0$, or $\lim_{n\to \infty} f(y_n) = \tilde{f}(x)$. Therefore, we can define $\tilde{f}(x) = f(x)$ on $B(0, 1)$ and $\tilde{f}(x) = \lim_{x'\to x} f(x')$ on $\partial B(0, 1)$ as the continuous extension of $f$ to $\overline{B(0, 1)}$. Note that as $\overline{B(0, 1)}$ is compact, $\tilde{f}$ is actually uniformly continuous.

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  • $\begingroup$ How do we know such sequence $x_n$ exists? $\endgroup$
    – Itay4
    Aug 16 '17 at 6:53
  • $\begingroup$ This is the definition of $\partial B(0, 1)$ as the intersection of $\overline{B(0, 1)}$ and $\overline{B(0, 1)^c}$. Since $\partial B(0, 1)$ is a subset of the closure of $B(0, 1)$, every point of $\partial B(0, 1)$ is a limit point of $B(0, 1)$. $\endgroup$
    – Michael L.
    Aug 16 '17 at 6:55
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Yes, it is the right way. Fix a point on the boundary $ x_0$ Uniiform continuity imllies that for each $\epsilon $ there exists $\delta $ such that $ |f(x)-f (y)|<\epsilon $ for all $ x, y\in B_\delta (x_0) $. Hence you may define $ g (x_0) $ as the limit of $ f $ along any sequence converging to $ x_0$, and the continuity at $x_0$ follows from the triangle inequality plus the above.

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Suppose $0 < r_1 < r_2 < \cdots \to 1.$ For $x\in \overline {B(0,1)},$ define $f_k(x) = f(r_kx).$ Then each $f_k \in C(\overline {B(0,1)}).$ The uniform continuity of $f$ on $B(0,1)$ shows that $f_k$ is uniformly Cauchy on $\overline {B(0,1)}.$ Hence $f_k$ is uniformly convergent on $\overline {B(0,1)}$ to some $g\in C(\overline {B(0,1)}).$ Since $f_k(x) \to f(x)$ for all $x\in B(0,1),$ we have $f= g$ in $B(0,1).$ We're done.

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