2
$\begingroup$

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?

  • 0
    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.2011-10-08
  • 0
    My very first instinct is to grab for [Arzelà-Ascoli](http://en.wikipedia.org/wiki/Arzela-Ascoli).2011-10-08
  • 0
    yeah, that was my first instinct too. But I don't know how to apply it2011-10-08
  • 0
    $|g_n(x)|\leq 3\sqrt{x+1}$ 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.2011-10-09
  • 0
    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-Arzelà2011-10-09
  • 0
    I don't really know why you want to find such a subspace, the sequence is already precompact...(If you must though, the sequence is in the ball centered at 0 and radius $3\sqrt{2}+3$ of $C^{0,1/2}$)2011-10-09
  • 0
    What result are you applying? To say that it's precompact (what you mean precisely with precompact? that there is a subsequence converging?)2011-10-09
  • 0
    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

1 Answers 1

1

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!