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Consider the subset of $C([0, 1])$ given by $$S = \{f \in C([0, 1]) : f(0) = 0, \int_{0}^{1} |f'(x)|^2 dx \leq 1 \}$$ How to prove that any sequence in this set $S$ has a uniformly convergent subsequence using Arzela-Ascoli Theorem (https://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem). I guess one can show that $S$ is uniformly bounded and equicontinuous, but not sure how.

Answers very much appreciated, thanks.

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    $\begingroup$ Hint: key tools will be the fundamental theorem of calculus, $f(z)-f(y) = \int_y^z f'(x)\,dx$, and the Cauchy-Schwarz inequality. $\endgroup$ – Nate Eldredge Dec 6 '19 at 23:18
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@Nate Eldredge already said it all: \begin{align*} |f(z)-f(y)|&=\left|\int_{y}^{z}f'(x)dx\right|\\ &\leq\int_{y}^{z}|f'(x)|dx\\ &=\int_{0}^{1}|f'(x)|\chi_{[y,z]}dx\\ &\leq\left(\int_{0}^{1}|f'(x)| ^{2}dx\right)^{1/2}|z-y|^{1/2}, \end{align*} for $z\geq y$, the rest should be straight forward.

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  • $\begingroup$ I guess this proves uniform boundedness, how would I use the $\epsilon - \delta$ argument to prove equicontinuity. I am kind flustered by the square root term of $|z - y|$ $\endgroup$ – user732491 Dec 7 '19 at 0:18
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    $\begingroup$ Actually you can pick $\delta=\epsilon^{2}$ so if $|z-y|<\delta$, then we have immediately that $|z-y|<\epsilon$. $\endgroup$ – user284331 Dec 7 '19 at 0:20

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