Prove sequence converges uniformly 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.
 A: 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 $n\to \infty$.
