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Suppose that the sequence ${\rm f_{j}}\left(x\right)$ on the interval $\left[0, 1\right]$ satisfies $$ \left\vert\,{\rm f_{j}}\left(t\right) - {\rm f_{j}}\left(s\right)\,\right\vert \leq \left\vert\,s - t\,\right\vert^{\,\alpha} $$ for all $s, t \in \left[0,1\right]$, ${\rm j} = 0, 1, 2,\ldots$, and for some $\alpha \in \left(0,1\right]$. Furthermore, assume that the sequence of functions ${\rm f_{j}}$ converge pointwise to a limit function ${\rm f}$ on $\left[0,1\right]$. Prove that the sequence converges uniformly.

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I wrote an answer, but I would be surprised if this fact this hasn't been proven and given a name already, using the more general assumption that the sequence $(f_n)$ is equicontinuous. –  Stefan Smith Dec 13 '13 at 22:39

2 Answers 2

This requires that property $\lvert f(a) - f(b) \rvert \le M\lvert a - b\rvert$ - I'll find out what that is exactly now...

Additionally convergence on a closed interval => uniform convergence on that interval

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Let $\epsilon > 0$. Let $N \in \mathbb{N}^+$ be big enough so $(1/N)^\alpha < \epsilon/10$. Let $M \in \mathbb{N}^+$ be big enough so

$$n \geq M \Rightarrow \max\{|f_n(0)-f(0)|,|f_n(1/N)-f(1/N)|,|f_n(2/N)-f(2/N)|,\ldots,|f_n(1)-f(1)|\} < \ldots $$

$$\ldots ...<\epsilon/10$$. Applying the triangle inequality (similar to the way it's done in the proof of the Arzela Ascoli Theorem), you should be able to show that if $n \geq M$ and $x \in [0,1]$, then $|f_n(x)-f(x)|<\epsilon$.

This proof is similar to the usual proof of the Arzela-Ascoli Theorem. That theorem doesn't answer the question becuase it only asserts that a subsequence converges uniformly, not the whole sequence. The condition given on $(f_n)$ in this problem could be replaced by the more general condition that the sequence $(f_n)$ is equicontinuous.

I have left some gaps for you to fill in. The "$\epsilon/10$"s are overkill and could probably be replaced by $\epsilon/3$.

ALTERNATE PROOF if you are allowed to use the Arzela-Ascoli Theorem: Let $f$ be the limiting function: that is, $f_n(x) \to f(x)$ as $n \to \infty$ for all $x$. Your assumption on the $f_n$'s implies that $(f_n)$ is equicontinuous. Since $(f_n(0))$ converges, $(f_n(0))$ is bounded. Use this and the equicontinuity to show $(f_n)$ is uniformly bounded. It follows that $(f_n)$ has a subsequence that converges uniformly, and since it converges pointwise to $f$, the subsequence must converge uniformly to $f$. More strongly, any subsequence of $(f_n)$ has a subsequence converging uniformly to $f$.

Now define the sequence $(d_n) \subset [0, \infty)$ by $d_n = \max_{x\in [0,1]}|f_x(x)-f(x)|$. All you have to do now is show $d_n \to 0 $ as $n\to \infty$.

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