Proof of a continuity for a function represented by an integral.

Question

Please think this problem easy.

I faced the following problem the other day.

Let $f\in C(0,1]\cap L^{1}(0,1)$. Prove that the function $$ t\mapsto\int_{0}^{t}\frac{f(\tau)}{\sqrt{t-\tau}}d\tau $$ is continuous on $(0,1]$.

It seems not easy to prove. Indeed, since \begin{align*} \left|\int_{0}^{t}\frac{f(\tau)}{\sqrt{t-\tau}}d\tau-\int_{0}^{s}\frac{f(\tau)}{\sqrt{s-\tau}}d\tau\right|&=\left|\int_{0}^{t}\frac{f(\tau)}{\sqrt{t-\tau}}d\tau-\int_{0}^{t}\frac{f((s/t)\tau)}{\sqrt{s-(s/t)\tau}}d\tau\right|\\ &=\left|\int_{0}^{t}\frac{f(\tau)-(s/t)^{1-a}f((s/t)\tau)}{\sqrt{t-\tau}}d\tau\right|\\ &\le\int_{0}^{t}\frac{|f(\tau)-(s/t)^{1-a}f((s/t)\tau)|}{\sqrt{t-\tau}}d\tau, \end{align*} we would like to use the dominated convergence theorem but it is clear that there does not exitst a $L^{1}$-dominate function. Because the integral is convolution type, we might use its properties but I dont't know.

I'm glad if you tell me when you know. Even only a hint is good.

Thank you in advance.

It is $N$-dimensional torus, i.e., $\mathbb{T}^{N}=\mathbb{R}^{N}/\mathbb{Z}^{N}$. — user, Oct 14 '15 at 04:02
My intention was just to try to help your question more readable :) — Yes, Oct 14 '15 at 04:07
Thank you for your advice. It might certainly was unreadable. — user, Oct 14 '15 at 04:09
Perhaps I'm dense but the integral of a measurable function is continuous, is it not? — Cameron Williams, Oct 18 '15 at 00:52
It is just my problem :) I don't know such fact. How is it proved? — user, Oct 18 '15 at 00:56

score 1 · Answer 1 · answered Oct 18 '15 at 01:32

Being fairly naive, I would start like this (my answer is not complete, but I think that it's a good start):

Assume $s < t$.

If $g(t) =\int_{0}^{t}\frac{f(\tau)}{\sqrt{t-\tau}}d\tau $,

$\begin{align*} g(t)-g(s) &\int_{0}^{t}\frac{f(\tau)}{\sqrt{t-\tau}}d\tau-\int_{0}^{s}\frac{f(\tau)}{\sqrt{s-\tau}}d\tau\\ &=\int_{0}^{s}\frac{f(\tau)}{\sqrt{t-\tau}}d\tau +\int_{s}^{t}\frac{f(\tau)}{\sqrt{t-\tau}}d\tau -\int_{0}^{s}\frac{f(\tau)}{\sqrt{s-\tau}}d\tau\\ &=\int_{0}^{s}f(\tau)\left(\frac1{\sqrt{t-\tau}}-\frac1{\sqrt{s-\tau}}\right)d\tau +\int_{s}^{t}\frac{f(\tau)}{\sqrt{t-\tau}}d\tau\\ &=\int_{0}^{s}f(\tau)\frac{\sqrt{s-\tau}-\sqrt{t-\tau}}{\sqrt{t-\tau}\sqrt{s-\tau}}d\tau +\int_{s}^{t}\frac{f(\tau)}{\sqrt{t-\tau}}d\tau\\ &=I+J\\ \end{align*} $

$\begin{align*} J &= \int_{s}^{t}\frac{f(\tau)}{\sqrt{t-\tau}}d\tau\\ &= \int_{0}^{t-s}\frac{f(\tau+s)}{\sqrt{t-s-\tau}}d\tau\\ &= \int_{0}^{u}\frac{f(\tau+s)}{\sqrt{u-\tau}}d\tau \qquad(u = t-s)\\ &\approx f(s)\int_{0}^{u}\frac{1}{\sqrt{u-\tau}}d\tau\\ &\approx 2f(s)\sqrt{u}\\ \end{align*} $

For $I$, we need to work with (remembering that $0 \le \tau \le s < t$)

$\begin{align*} \frac{\sqrt{s-\tau}-\sqrt{t-\tau}}{\sqrt{t-\tau}\sqrt{s-\tau}} &=\frac{1-\frac{\sqrt{t-\tau}}{\sqrt{s-\tau}}}{\sqrt{t-\tau}}\\ &=\frac{1-\sqrt{\frac{t-\tau}{s-\tau}}}{\sqrt{t-\tau}}\\ &=\frac{1-\sqrt{\frac{t-s+s-\tau}{s-\tau}}}{\sqrt{t-\tau}}\\ &=\frac{1-\sqrt{\frac{u+s-\tau}{s-\tau}}}{\sqrt{t-\tau}} \qquad(u = t-s)\\ &=\frac{1-\sqrt{1+\frac{u}{s-\tau}}}{\sqrt{t-\tau}}\\ &\approx\frac{1-(1+\frac{u}{2(s-\tau)})}{\sqrt{t-\tau}} \qquad\text{(this might not be the right approximation)}\\ &=-\frac{u}{2(s-\tau)\sqrt{t-\tau}}\\ \end{align*} $

At this point, I'm not sure what to do, but this seems to me like a good start.

Proof of a continuity for a function represented by an integral.

1 Answers1