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For $n\in \mathbb Z_{>0}$ let $f_n:[0,1]\to R$ be continuous and differentiable on $(0,1]$ with $$|f_n'(x)|\le \frac{1+|\ln x|}{\sqrt x}$$ on $(0,1]$ and $$\left|\int_0^1f_n(x)\,dx\right|\le 10$$ Prove that $\{f_n\}$ has a uniformly convergent subsequence.

I guess I have to verify the hypotheses of the Arzela-Ascoli theorem: that $f_n$ is pointwise bounded and equicontinuous.

Some ideas for equicontinuity:

$$ |f_n(y)-f_n(x)| = \left|\int_x^yf_n'(t)\,dt\right|\le \int_x^y |f_n'(t)|\,dt\\ \le\int_x^y\frac{1+|\ln t|}{\sqrt t} \, dt \le \int_x^y\frac{1-\ln t}{\sqrt t}\,dt $$

but I'm not sure whether it yields an expression of $|y-x|$ (not sure if it's the correct paths, and it's not obvious how to integrate that). For the boundedness, I guess I need to use the second inequality in question, but I don't know how...

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Your approach for equicontinuity is good. To complete it you need the additional fact that if $g$ is integrable on $[0,1]$ then $\int_a^{b} |f(x)| \, dx \to 0$ as $|x-y | \to 0$. More generally $\int_E |f(x)|\, dx \to 0$ as $m(E) \to 0$ where $m$ is Lebesgue measure. This gives equicontinuity. Now you have to do some more work to show that $\{f_n\}$ is uniformly bounded. By equicontinuity there exists a positive integer $N$ such that $|f_n(x)-f_n(y)| <1$ for all $n$ if $|x-y|<\frac 1 N$. Let $\{x_j:1\leq j\leq m\}$ be the partition obtained by dividing $[0,1]$ into $N$ equal parts. Any $x \in [0,1]$ lies between $x_j$ and $x_{j+1}$ for some $j$. Now $$|f_n(x)-f_n(0)| \leq |f_n(x)-f_n(x_j)|+|f_n(x_j)-f_n(x_{j-1})|+...+|f_n(x_1)-f_n(0)|<j+1\leq N+1$$ ...($\dagger $), so $f_n(x) >f_n(0) -N-1$. This is true for all $x$. Integrating we get $10>\int_0^{1}f_n(x)\, dx > \int_0^{1} [f_n(0)-N-1] \, dx=f_n(0)-N-1$ which shows $f_n(0)$ is bounded above. Similarly the inequality $f_n(x) -f_n(0) <N+1$ for all $x$ shows that $f_n(0) > -10-N-1$ for all $n$. We have proved that $\{f_n(0)\}$ is bounded. Going back to ($\dagger)$ you can now verify that $\{f_n\}$ is uniformly bounded. By Arzela -Ascoli Theorem we conclude now that $\{f_n\}$ has a uniformly convergent subsequence.

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