To find inverse of a linear operator Let $A:C[0,1]\rightarrow C[0,1]$ be a linear operator defined by $Ax(t)=x(t)+t\int_0^t sx(s)\,ds$. It is not difficult to show that $A$ is continuous with respect to the norm $\|\cdot\|$ on $C[0,1]$ defined by $\|f\|=\sup_{t\in [0,1]}|f(t)|$.
Problem:  Prove that the operator $A$ is invertible and find $A^{-1}$.
To show that $A$ is invertable, I begin with $Ax=0$ for some $x\in C[0,1]$. Then, $x(t)+t\int_0^tsx(s)\,ds=0$. Differentiating this twice I got $\ddot{x}(t)+t^2\dot{x}(t)+3tx=0$, where $x(0)=\dot{x}(0)=0$. But I do not know how to solve this differential equation. So, I got stuck here and could not proceed further. Thanks in advance for any help or suggestion.
 A: Choose $b(t)\in C[0,1]$ arbitrarily. We wish to find $x(t)\in C[0,1]$ such that $$(Ax)(t)=x(t)+t\int_0^tsx(s)\mathrm{d}s=b(t)$$ Set $w(t)=\int_0^tsx(s)\mathrm{d}s$ and note $w$ is differentiable. The preceding equation now becomes an IVP $$\frac{dw}{dt}+t^2w=tb(t)$$ $$w(0)=0$$ The particular solution to this IVP is $$w(t)=e^{-t^3/3}\int_0^tse^{s^3/3}b(s)\mathrm{d}s$$ This implies $$x(t)=b(t)-te^{-t^3/3}\int_0^tse^{s^3/3}b(s)\mathrm{d}s$$ which is a nice formula for $(A^{-1}b)(t)$.
A: Let's consider a different operator on $C[0,1]$ for now
$$T=t \int_0^tsx(s)ds$$
and compute the operator norm $\|T\|=\sup_{\|x\|_\infty=1}\|Tx\|_\infty$.
Note: I am using $\|\cdot \|_\infty$ to denote the supremum norm you used in your statement, I want to distinguish it from the operator norm.
$$\|T\|=\sup_{\|x\|_\infty=1}\Big\|t \int_0^t sx(s)ds \Big\|_\infty\leq \sup_{\|x\|_\infty=1} \Big\|t\int_0^t s\|x\|_\infty ds\Big\|_\infty=\big\|\frac{t^3}{2}\big\|_\infty=\frac{1}{2}$$
This tells us that the series $\sum_{k=0}^\infty \|T\|^k$ converges and therefore there series $\sum_{k=0}^\infty T^k$ converges uniformly to $(1-T)^{-1}$ in the operator norm topology. So we have shown that the operator $1-T$ is invertible.
Now we see that we can write $A=1+T$ and so we see that $1-(A-1)=-A$ is invertible and therefore $A$ is invertible with inverse given by $-\sum_{k=0}^\infty T^k$
