Convolution Integral Linear Operator on $L^2$

Question

Define a linear operator on $L^2[0,1]$ by $K(f)(t) = \int_{0}^{t}(t-s)f(s)ds$. Given $g \in L^2[0,1]$, find $f$ such that $f = g + K(f)$.

I am really lost on how to do this. I already showed that $K$ is a bounded linear operator, so I thought maybe I could use the Riesz representation theorem, but I don't think that's the right direction.

I would really appreciate a hint on where to go. Thanks!

Answer 1

The integral equation may be solved using the Laplace transform. Apply the Laplace transform to both sides, using the fact that the Laplace transform of a convolution of two functions is the product of the Laplace transforms of each function. Doing so, obtain the equation $$F(s) = G(s) + \frac{G(s)}{2(s-1)} - \frac{G(s)}{2(s+1)}$$ where $F$ and $G$ are the Laplace transforms of $f$ and $g$, respectively. Take the inverse Laplace transform term-by-term, using the convolution theorem to find the inverse Laplace of the second and last terms on the right-hand side.

Answer 2

This is the same question as showing that $f\mapsto K(f)-f$ is a surjective linear operator. Because $K$ is a Hilbert-Schmidt operator, it is compact, and $T(f)=K(f)-f$ is therefore Fredholm. In fact, it is Fredholm of index zero, because the Fredholm index doesn't change under addition of compacts. So if we show that the kernel is trivial, then we've shown the operator is surjective.

Assume that $T(f)=0$, or, equivalently, $K(f)=f$. Then for almost every $t$, $f(t)=\int_0^t (t-s)f(s)\,ds$. Because the image of a convolution operator is continuous, we can choose a continuous representative for $f$, and ask for pointwise equality. Now, at any point $t_0$, $\frac{1}{h}[f(t_0+h)-f(t_0)]=\frac{1}{h}\int_{t_0}^{t_0+h}(t-s)f(s)\,ds$, which is less than or equal to $\|f\|_\infty\cdot h$, and therefore the function is differentiable. Now, as we differentiate under the integral, we see that $f'(t_0)=\int_0^{t_0} f(s)\,ds$, and this means that $f''(t_0)=f(t_0)$, so $f(t)=c_1\sin(t)+c_2\cos(t)$. Plugging these in yields immediately that $c_1,c_2=0$, so the kernel is trivial, and the Fredholm index being zero means that the operator $T$ is surjective.