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