find $f \in L^2(0,1)$ such that $f = g + K(f)$ I'm preparing for my qualifying exam and the question bank had the following question under functional analysis section and I'm stuck with this problem.
Let $L^2(0,1)$ be the Hilbert space of square integrable functions on the interval $[0,1]$ with Lebesgue measure. Define $K : L^ 2 (0,1) \rightarrow L^ 2 (0,1)$ by $K(f)(t) =\int_0^t (t − s)f(s)\,ds$. Show that
$||K|| < 1$. Given $g \in L^ 2 (0,1)$ find $f \in L^ 2 (0,1)$ such that $f = g + K(f)$.
My idea for the second part of the problem: I feel like it is the kernel of some map and that the question is asking to find an element in the coset of $g$ (the coset that you get after you quotient out by the kernel).
I really have no idea how to proceed. Any help would be appreciated.
 A: Answer for the second part: $\|K\|<1$ implies that $I-K$ is invertible and its inverse is $\sum\limits_{n=0}^{\infty} K^{n}$. [See my answer here: https://math.stackexchange.com/questions/4021302/prove-existence-of-inverse-of-a-bounded-linear-operator/4021335#4021335 ]
Hence, the unique solution of $f=g+Kf$is $f=(I-K)^{-1} g$.
A: Here's the proof for the first part , I'll try for the second part.
$||Kf||_{L^2(0,1)}=(\int_{0}^1 |Kf(t)|^2 dt)^{\frac{1}{2}}=(\int_{0}^1 |\int_{0}^t (t-s)f(s)ds|^2 dt)^{\frac{1}{2}} \le \int_{0}^t (\int_{0}^1 |(t-s)f(s)|^2dt)^{\frac{1}{2}}ds$ The inequality appears by Minkowski's integral inequality.
Now the last term $=\int_{0}^t (\int_{0}^1 |(t-s)|^2dt)^{\frac{1}{2}}|f(s)|ds$ As translations give isometry on $L^2(0,1)$  and letting $id :(0,1) \to (0,1)$ be $x \mapsto x$
we have, $$\int_{0}^t \Big(\int_{0}^1 |(t-s)|^2dt \Big)^{\frac{1}{2}}|f(s)|ds$$ $$=\int_{0}^t ||id||_{L^2(0,1)}|f(s)|ds$$ $$=\frac{1}{\sqrt{3}}\int_{0}^t|f(s)|ds$$ $$\le \frac{1}{\sqrt{3}} ||f||_{L^1(0,1)}$$ $$\le \frac{1}{\sqrt{3}} ||f||_{L^2(0,1)}$$ Thus $$||Kf||_{L^2(0,1)} \le \frac{1}{\sqrt{3}} ||f||_{L^2(0,1)} \implies ||K|| <1$$
A: There is another way to prove the second statement. Since $K: L^2\to L^2$ for a given $g\in L^2(0, 1)$ you can define a map $T_g: L^2(0, 1)\to L^2(0, 1)$ by
$$T_g(f)=g+K(f).$$
If you show that $T_g$ is a contraction mapping, since $L^2$ is a complete metric space then by the Banach Fixed Point Theorem there exists a unique fixed point $f\in L^2(0, 1)$ of $T_g$, that is
$$f=T_g(f)=g+K(f)$$
which is the solution you are looking for. This is indeed the case, since
$$\|T_g(f_1)-T_g(f_2)\|_{L^2}=\|K(f_1)-K(f_2)\|_{L^2}\leq \|K\|\|f_1-f_2\|_{L^2}$$
and you already proved that $\|K\|<1$.
EDIT: If you want to compute $f$ explicitly, you can go as follows (i'll skip any technical detail). First we write the equation as
\begin{align}f&=g+t\int_0^1 f(s)ds-\int_0^1sf(s)ds\\&=g+C_1t-C_2\end{align}
where
\begin{align}C_1&=\int_0^1f(s)ds,\\ C_2&=\int_0^1 sf(s)ds.\end{align}
Then by replacing $f$ in $C_1$ we get
\begin{align} C_1=\int_0^1 (g(s)+C_1s-C_2)ds=\int_0^1g(s)ds+\dfrac{C_1}{2}-C_2\end{align}
and then
$$\dfrac{C_1}{2}+C_2=\int_0^1 g(s)ds.$$
Similarly by replacing $f$ in $C_2$ we get
$$\dfrac{C_1}{3}-\dfrac{3C_2}{2}=-\int_0^1 sg(s)ds.$$
With these two equations we can obtain $C_1$ and $C_2$, hence by replacing it in the first equation for $f$ you get the full expression only in terms of $g$.
A: There is an easier way prove the first part of this statement using the Cauchy-Swartz inequality. Let $f\in L^2[0,1]$ be arbitrary, and define $g\in L^2[0,1]$ by $g(x)=1-x.$ Notice
\begin{eqnarray*}
|K(f)(t)|^2 & = & \Bigg(\int_0^t(t-x)f(x)dx\Bigg)^2
\\ & \leq & \Bigg(\int_0^1(1-x)f(x)dx\Bigg)^2 \\
 & = & \Big|\big<f,g\big>\Big|^2 \\ & \leq & ||f||^2||g||^2 \\ & = & \frac{1}{3}||f||^2
\end{eqnarray*} Therefore,
$$||K(f)||^2= \int_0^1|K(f)(t)|^2dt \leq \frac{1}{3}||f||^2$$ Hence $||K||\leq \frac{\sqrt{3}}{3}<1$.
