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Given a sequence $f_n:[0,1]\to[0,1]$ of continuous functions, define $g_n:[0,1]\to{\Bbb R}$ by setting $$ g_n(x)=\int_0^1\frac{f_n(t)}{(t-x)^{1/3}}dt,\quad x\in[0,1]. $$ Show that $(g_n)_{n\in{\Bbb N}}$ has a uniformly convergent subsequence.


Since $K=[0,1]$ is compact, by Arzela-Ascoli, if one can show

  1. $g_n\in C(K)$;
  2. $\{g_n\}$ is pointwise bounded;
  3. $\{g_n\}$ is equicontinuous on $K$;

then we are done.

I have difficulty in 3. I messed up by only using the definition of equicontinuity. Is there any other theorem which might be useful here?

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1 Answer 1

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It should be useful to estimate this integral: $$\int_0^1\left|\frac{1}{(t-x_2)^{1/3}}-\frac{1}{(t-x_1)^{1/3}}\right|\,dt$$ assuming $x_1<x_2$ for simplicity, you can calculate the integral directly.

There is a theorem you can use instead, if you know it: Translation acts continuosly on $L^1$ functions. Apply it to the function $x\mapsto x^{-1/3}$. The above suggestion is bascially just the corresponding direct calculation.

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  • $\begingroup$ Would you point me to some references of the theorem you said in the answer? $\endgroup$
    – user9464
    Commented Jan 2, 2014 at 1:45
  • $\begingroup$ Every book in existence on (Lebesgue) integration theory will have this theorem. (The book title is often Real Analysis or some variant thereof.) Books by Rudin, Royden, Folland, or McDonald and Weiss come to mind. $\endgroup$ Commented Jan 2, 2014 at 7:58

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