How to solve this Integral operator equation? I'm trying to solve this exercise.
Let $ L:C([0,1])\rightarrow C([0,1]) $ such that $$ (Lu)(t) = \int_{0}^{t^2} u(\sqrt{s}) ds. $$ Let $ a\in R $ and consider the following:
$$ Lu = au. $$ Show that necessarily $ u\equiv 0. $
Some ideas to proceed?
 A: We have with the substitution $s=x^2$, formally $ds=2x\; dx$:
$$
(Lu)(t) = 
\int_0^{t^2}u(\sqrt s)\; ds
=
\int_0^{t}u(x)\;2x\; dx\ ,
$$
so in particular $(Lu)(0)=0$. Assume now $u$ is a solution for 
$Lu = au$. 


*

*If $a=0$, then we immediately deduce $u(x)\; 2x\equiv 0$, so $u=0$.

*Else, $u=\frac 1aLu$ is  differentiable, and we compute
$$
u'(t) = \frac 1a(Lu)'(t)
=
\frac 1a u(t)\, 2t\ .
$$
This is a differential equation in $u$, we have also $u(0)=0$, the solution is unique, namely $u=0$. For this we may want to solve the equation or compute the differential of $u(t)\cdot\exp-\frac {t^2}a$. Let us do this explicitly.



It is convenient to use $$c=\frac 2a$$ below, so we can concentrate on the essential part. 


*

*Solving the equation. We have to solve $u'(t) = ct\; u(t)$. We solve locally around points where $u\ne 0$, so that we can rephrase successively equivalently:
$$
\begin{aligned}
u'(t) &= ct\; u(t)\\
\frac{u'(t)}{u(t)} &= ct\qquad\text{Now integrate on both sides:}\\
\log |u(t)| &=ct^2/2 +\text{Constant}\\
u(t) &= \pm Ke^{ct^2/2}\ ,
\end{aligned}
$$
and we can stop here. If $u\ne 0$ on some point of $[0,1]$, then locally we have the above form. Since $u$ is continuous, the above form either has $K=0$, and then $u=0$, or $K\ne 0$, but then the above form has no chance to become $u(0)=0$.

*The last argument may seem unclear, but with its knowledge we can argument as follows. Let $h(t) = u(t)\; e^{-ct^2/2}$. (And we want to show it is constant. So it is constant zero.) Then the derivative of $h$ is
$$
h'(t) = 
u'(t)\; e^{-ct^2/2} + u(t)\; (e^{-ct^2/2})'
=
u(t)\; ct \; e^{-ct^2/2} +
u(t)\; (-ct \; e^{-ct^2/2}) = 0\ .
$$
So $h$ is a constant, and from $h(0)$ we get $h=0$, then $u=0$.
