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We are given a sequence of real valued functions $\{g_n\}$ that are defined and continuous on the unit sphere $S$ and differentiable inside it (except at the boundary of the sphere $S$ Also, it is given that $\{g_n(0)\}$ is bounded and all partial derivatives of $\{g_n\}$ for all $n=1,2,...$ inside the unit sphere (except boundary) are bounded. Using the Mean Value Theorem, I managed to prove $|g_n(x)-g_n(y)| \leq K|x-y|$ for all $x,y$ in $S$, which shows that the $\{g_n\}$ are Lipschitz continuous and therefore equicontinuous.

I need to show that the sequence of functions $\{g_n\}$ contains a subsequence that converges uniformly on $S$. I am trying to use the Arzela Ascoli Theorem somehow to prove this, but a rigorous proof eludes me. I appreciate any kind of help!

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  • $\begingroup$ Use the (uniform, by the way) equicontinuity and the boundedness of $\{g_n(0)\}$ to establish the uniform boundedness of $\{g_n\}$. Then you have way stronger conditions than you need for Ascoli's theorem. $\endgroup$ – Daniel Fischer Apr 13 '14 at 18:35
  • $\begingroup$ @DanielFischer: Thanks for your help! Can you, please, show me how you prove the uniform boundedness of $\{g_n(x)\}$ for every $x$? Thanks $\endgroup$ – M.Krov Apr 13 '14 at 19:15
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First, let me say that since the family $\{g_n\}$ is not only equicontinuous but equilipschitz (that is, there is a Lipschitz constant that works for all these functions at once), we could prove the uniform boundedness of the family in one short statement. I encourage you to look for that proof.

Let us prove something more general:

Proposition: Let $K$ a compact connected space, and $\mathscr{F}$ an equicontinuous family of real-valued (or $\mathbb{R}^n$-valued, we could even state it in more generality) functions. If there is one point $x_0 \in K$ such that the set $\mathscr{F}(x_0) := \{ f(x_0) : f \in \mathscr{F}\}$ is bounded, then $\mathscr{F}$ is uniformly bounded, i.e. there is a $C < \infty$ such that $\lvert f(x)\rvert \leqslant C$ for all $x\in K$ and all $f\in\mathscr{F}$.

Proof: Let $B = \{ x\in K : \mathscr{F}(x) \text{ is bounded}\}$. Then $B$ is open:
Let $x\in B$ and $C(x) = \sup \{ \lvert f(x)\rvert : f\in\mathscr{F}\}$. By the equicontinuity of $\mathscr{F}$, there is a neighbourhood $U$ of $x$ such that $\lvert f(y) - f(x)\rvert \leqslant 1$ for all $f\in\mathscr{F}$ and all $y\in U$. Hence $\lvert f(y)\rvert \leqslant C(x)+1$ for all $f\in\mathscr{F}$ and all $y\in U$, which implies that $U \subset B$, hence $x\in \overset{\Large\circ}{B}$.

Also, $K\setminus B$ is open, for if $\mathscr{F}(x)$ is unbounded, and $U$ a neighbourhood of $x$ such that $\lvert f(y) - f(x)\rvert \leqslant 1$ for all $f\in\mathscr{F}$ and all $y\in U$, then $U\cap B = \varnothing$. For if we had $\lvert f(y) \rvert \leqslant C$ for some $C\in\mathbb{R}$, $y\in U$ and all $f\in \mathscr{F}$, then we'd have $\lvert f(x)\rvert \leqslant C+1$ for all $f\in\mathscr{F}$, and that would contradict $x\in K\setminus B$.

Since evidently $B\cap (K\setminus B) = \varnothing$, and $K$ is connected, we have $B = \varnothing$ or $K\setminus B = \varnothing$. By assumption, $x_0 \in B$, so it follows that $K = B$.

Now $V_n = \left\{ x \in K : \bigl(\exists U \in \mathscr{V}(x)\bigr)\bigl(\forall y\in U\bigr)\bigl(\forall f\in\mathscr{F}\bigr)\bigl(\lvert f(y)\rvert \leqslant n\bigr) \right\}$ for $n \in \mathbb{N}$ is a nested family of open sets that covers $K$, hence there is an $n\in\mathbb{N}$ with $K = V_n$, i.e. $\lvert f(x)\rvert \leqslant n$ for all $x\in K$ and all $f\in\mathscr{F}$.

Between the general theorem and the very short proof from the equilipschitzness of the family, we have the option to use the uniform equicontinuity of the family to give a shorter proof in the given situation, which we sketch:

By the uniform equicontinuity, there is a $\delta > 0$ such that whenever $\lvert x-y\rvert \leqslant \delta$, then $\lvert g_n(x)-g_n(y)\rvert \leqslant 1$ for all $n$. By assumption, there is a $C_0$ such that $\lvert g_n(0)\rvert \leqslant C_0$ for all $n$. Then $\lvert g_n(x)\rvert \leqslant C_0 + 1$ for all $x$ with $\lvert x\rvert \leqslant \delta$. And hence $\lvert g_n(x)\rvert \leqslant C_0 + 2$ for all $x$ with $\lvert x\rvert \leqslant 2\delta$. Generally, $\lvert x\rvert \leqslant k\delta \implies \lvert g_n(x)\rvert \leqslant C_0 + k$. Choose $k \geqslant \frac{1}{\delta}$.

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