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Let $\{f_n\}_n$ be a sequence of real-valued continuous functions defined on $[0,1]$ such that $\int^1_0|f_n(y)|dy\leq3$ for all $n$. Define $g_n:[0,1]\rightarrow\mathbb{R}$ by


I have to prove that $\{g_n\}_n$ contains a subsequence that converges uniformly on $[0,1]$.

I really don't know where to start, could you help me please?

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Sounds typical of problem where the key is to find a compact subset of the space of functions on $[0,1]$ under the uniform norm which contains all $g_n$'s. – Henning Makholm Oct 8 '11 at 23:36
My very first instinct is to grab for Arzelà-Ascoli. – kahen Oct 8 '11 at 23:36
yeah, that was my first instinct too. But I don't know how to apply it – Alex M Oct 8 '11 at 23:51
$|g_n(x)|\leq 3\sqrt{x+1}$ so the sequence is pointwise bounded. $|g_n(x)-g_n(y)|\leq 3 \sup_{z} |\sqrt{x+z}-\sqrt{y+z}| \leq 3\sqrt{|x-y|}$ so it's also equicontinous. – Jose27 Oct 9 '11 at 2:13
yeah the $g_n's$ are equicontinuous and $|g_n(x)|\leq 3\sqrt{2}$ so it suffices to find a compact subspace of the continuous functions with the uniform norm that contains all $g_n's$ and apply Ascoli-Arzelà – Alex M Oct 9 '11 at 2:53

Notice that $$g_n'(x)=\int_0^1\frac{f(y)}{2\sqrt{x+y}}\,\mathrm dy,$$ so that $$|g_n'(x)|\leq \int_1^1\frac{|f(y)|}{2\sqrt{x+y}}\,\mathrm dy\leq\frac1{2\sqrt{x}}\int_0^1|f(y)|\,\mathrm dy\leq \frac{3}{2\sqrt{x}}.$$ This implies that all the derivatives of all the $g_n$ are bounded in $[\varepsilon,1]$ for all $\varepsilon\in(0,1)$. Ascoli-Arzelà implies that there are uniform convergent subsequences there. This does not solve you problem, but may be useful!

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