Is there any function from $C[0,1]$ such that $f(x)=x^2+\int_0^x e^{-t}f(t) dt$?

Solution 1:

Since $\|T\| \le k < 1$, the operator $I - T$ is invertible. So the equation $f(x) = x^2 + (Tf)(x)$, which is equivalent to $[(I - T)f](x) = x^2$, has a unique solution.

For a more direct approach, note that if $f\in C[0,1]$ satisfies the above integral equation, then $f(0) = 0$ and $f$ differentiable with $f'(x) = 2x + e^{-x}f(x)$. By method of integrating factors, the solution of this ODE is $$f(x) = e^{-e^{-x}}\int_0^x 2t\, e^{e^{-t}}\, dt$$

Solution 2:

Hint: $f = (I-T)^{-1} g$ where $g(x) = x^2$.

Solution 3:

There isn't really anything to add to Robert Israel's answer, but I am writing this answer because I liked the problem.

As mentioned in Robert's answer, one needs to show that $(I-T)^{-1}$ exists, and then we are done. To see that this is the case, you can define the sequence $S_m:=\sum_{n=0}^{m} T^n$ and show that $S_m$ has a limit and that the limit is the inverse of $(I-T)$.

Before going to the above mention abstract machinery, I often find it fun to do some 'formal' calculations. So, let us see if we can actually answer your question some other way. Taking the derivative of $f$, we find that $f$ must satisfy the following differential equation $$f'(x)=2x+e^{-x}f(x),\;\; f(0)=0.$$ This is a fairly simple ODE and can be solved explicitly to obtain the solution as $$f(x)=e^{-e^{-x}}\int_{0}^{x}(2t)e^{e^{-t}}\;dt.$$

Note that this simple-minded method can be justified and in fact, it says that $f(x)=g(x)+T(f)$ has a solution as long as $g$ is absolutely continuous.

The beauty of the method suggested by Robert is that it says that $f(x)=g(x)+T(f)$ has a solution for every $g\in C[0, 1]$. Even if, we may not be able to explicitly calculate the solution for an arbitrary $g$. This allows you to now connect the dots and see that for $g\in C[0, 1]\cap AC$, we can define $$M(g)=e^{-e^{-x}}\int_0^{x}g'(t)e^{e^{-t}}\;dt.$$ This operator extends as a bounded linear operator, say $\tilde{M}$, to $C[0, 1]$ and $\tilde{M}=(I-T)^{-1}$.