Existence of solutions for $x(t)=\sin(t)+\int_0^t\frac{x(s)}{\sqrt{s}}ds $ Use the fixed point theorem (or it's generalization) to show there is a continuous function  (unique???) $x:[0,1]\to \mathbb{R}$ such that $$x(t)=\sin(t)+\int_0^t\frac{x(s)}{\sqrt{s}}ds $$
Comment:  We usually define an operator on the space $\mathcal{C}[0,1]$, and using the sup norm $\|\cdot\|_{\infty}$ we prove that for $x,y\in\mathcal{C}[0,1]$, we have  $ |Tx(t)-Ty(t)|\leq L\|x-y\|_{\infty}$, if $L<1$ the theorem guarantees the existence of this unique solution, however the term $\frac{1}{\sqrt{s}}$ near to $0$ makes it difficult. Should I use other theorem rather then this?
If I do this $$|Tx(t)-Ty(t)| = \left|\int_0^t\frac{x(t)-y(t)}{\sqrt{s}}ds \right|\leq \|x-y\|_\infty\int_0^t\frac{1}{\sqrt{s}}ds \leq 2\|x-y\|_{\infty}$$
 A: Let us define the mapping $\Phi:C[0, 1]\rightarrow C[0, 1]$ by
\begin{align}
\Phi(u)(t)= \sin t + \int^t_0 \frac{u(s)}{\sqrt{s}}\ ds. 
\end{align}
Actually, let us first check that $\Phi$ indeed maps $C[0, 1]$ to $C[0, 1]$. By direct computation, we have that
\begin{align}
|\Phi(u)(t)-\Phi(u)(t')| \le&\ |\sin t-\sin t'|+\int^t_{t'} \frac{|u(s)|}{\sqrt{s}}\ ds \\
\le&\ |t-t'| + \|u\|_\infty \int^t_{t'}\frac{ds}{\sqrt{s}}\\
\le&\ |t-t'| +2 \|u \|_\infty |\sqrt{t}-\sqrt{t'}|.
\end{align}
Hence, it is clear that $\Phi(u)(t)$ is continuous if $u$ is continuous (in fact, it suffices to be bounded).
Let us show that $\Phi$ is a contraction on a smaller interval $[0, T]$. Observe that
\begin{align}
\|\Phi(u)-\Phi(v)\|_\infty\le \sup_{t \in [0, T]}\int^t_0 \frac{\|u-v\|_\infty}{\sqrt{s}}\ ds = 2\sqrt{T}\|u-v\|_\infty.
\end{align}
If $2\sqrt{T}<1$, then $\Phi$ is a contraction. Hence, we have a unique solution on $[0, T]$. Repeat this argument a finite number of times, then you will have the desired result.
