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Let $(g_n)$ be a sequence of twice differentiable functions defined on $[0,1]$, and assume that $g_n(0)=g_n'(0)$ for all $n$. Suppose also that $|g_n'(x)|\leq 1$ for all $n\in\mathbb{N}$ and all $x\in[0,1]$. Prove that there is a subsequence of $(g_n)$ converging uniformly on $[0,1]$.

This problem is screaming at me Arzelà-Ascoli Theorem. However, I am neither sure how to show that $(g_n)$ is uniformly bounded nor how to show that $(g_n)$ is equicontinuous.

Any help would be appreciated.


Edit: $(g_n)$ is uniformly bounded since $$|g_n(x)|=\left|g(0)+\int_0^xg_n'(x)dx\right| \leq |g_n(0)|+\int_0^x|g_n'(x)|dx \leq |g_n(0)|+ \int_0^x dx=1+x\leq2.$$

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You made $g(0)$ disappear in there ; your upper bound should become $2$ instead of $1$. The reason for this is that the sequence $g_n(x) = 1+x$ satisfies the hypothesis, therefore I know your bound is wrong. =) –  Patrick Da Silva Feb 21 '12 at 7:25
    
you are correct i was thinking $g_n(0)=0$ which is not true –  Galois Feb 21 '12 at 7:33
    
Lol even you? XD I mean, you're asking the question! You should've at least read it right. =P This is most often the reason why intelligent people give wrong answers ; they read the question wrong. I am fortunate to have the luck to read things well at this time of the day (2:34 a.m.), so I understand your badluck. ^^ –  Patrick Da Silva Feb 21 '12 at 7:34

3 Answers 3

up vote 3 down vote accepted

For equicontinuity, you have that for some $\zeta \in [0,1]$, $$ |g_n(x) - g_n(y)| = |g'_n(\zeta)||x-y| \le |x-y| $$ by Taylor's theorem, so that $\{ g_n \}$ are all Lipschitz with constant $1$, hence equicontinuous. I don't know why you need twice differentiability ; $C^1$ is enough here.

For uniformly bounded, $$ |g_n(x)| = |g_n(x) - g_n(0) + g_n(0)| = |g_n'(\zeta)x + g_n'(0)| \le |g'_n(\zeta)||x|+|g'_n(0)| \le 1+x. $$ again by Taylor's theorem and since $|g'_n(x)| \le 1$, plus with $x \ge 0$ I can remove the $| \cdot |$.

Hope that helps,

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The fact that the sequence $(g_n)$ is equicontinuous follows immediately from the mean value theorem, and the assumption $|g_n'(x)|\leq 1$. Since $|g_n(x)-g_n(y)|=|g_n'(\xi)|\cdot|x-y|\leq |x-y|$ (where $\xi$ is some element in $(x,y)$).

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Uniformly bounded: For any $n$, we have $$|g_n(x)|=\left|g_n(0)+\int_0^x g_n'(t)dt\right|\leq |g_n(0)|+\int_0^x|g_n'(t)|dt\leq1+\int_0^xdt = 1+x$$ establishing that the sequence is uniformly bounded.

Equicontinuous: Let $\epsilon>0$ and set $\delta=\epsilon$. If $|x-y|<\delta$, then $$|g_n(x)-g_n(y)|=\left|\int_x^yg_n'(t)dt\right|\leq \left|\int_x^y|g'(t)|dt\right|\leq \left|\int_x^ydt\right|=|x-y|<\delta=\epsilon$$ thus the sequence $(g_n)$ is equicontinuous.

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thanks i just realized that and apparently eddited my post as you were answering me –  Galois Feb 21 '12 at 7:19
    
@Galois accidentally pressed enter too soon, which is why my answer was incomplete. I touched it up now and included equicontinuity. –  Alex Becker Feb 21 '12 at 7:22
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@PatrickDaSilva I forgot my apostrophes. –  Alex Becker Feb 21 '12 at 7:25
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But still, your proof is wrong ; the bound is incorrect. You need to compute $$ |g_n(x)| = |g_n(x) - g_n(0) + g_n(0)| = |\int_0^x g_n(t) \, dt + g_n(0)| $$ which leaves you with the correct upper bound $1+x$ when you replace $g_n(0)$ by $g_n'(0)$ which is bounded by $1$. –  Patrick Da Silva Feb 21 '12 at 7:28
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@PatrickDaSilva You're quite right, I misread the OP as saying $g_n(0)=0$. I'll fix that. –  Alex Becker Feb 21 '12 at 7:32

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