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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

$g_n(x)=\int^1_0\sqrt{x+y}f_n(y)dy$

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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    A set is precompact if it has compact closure. Arzelà-Ascoli tells you that a subset of $C[0,1]$ is precompact if it is pointwise bounded and equicontinous. Since in metric spaces compactness is equivalent to sequential compactness you have your result.2011-10-09

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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!