Is this operator on $L^\infty$ injective / surjective?

Question

$$ f \in L^\infty (0,1) \\ Tf(x) = \int_0^x e^{y-x}f(y)dy, x\ge0 $$ I've shown that T is a bounded linear operator from $L^\infty(0,\infty)$ into itself. I've computed its norm (it should be $\|T\| = 1$). Now, I was wondering if it is injective and/or surjective. For injectivity , I have to show that $Tf = Tg \implies f=g \text{ in } L^\infty(0,1)$. This seems to be true $$ Tf(x) = Tg(x) \\ e^{-x}\int_0^x e^y f(y) dy = e^{-x}\int_0^xe^yg(y)dy \\ \int_0^x e^y f(y) dy = \int_0^xe^yg(y)dy $$ Differentiating both sides with respect to $x$ and using the Fundamental Theorem of Calculus: $$ e^x f(x) = e^x g(x) \; a.e.\\ f = g \; a.e. $$ Is this right? However, I do not know how to show surjectivity ( and I do not if it is surjective ) . If it is surjective, then: $$ \forall g \in L^\infty(0,\infty), \exists f \in L^\infty(0,\infty): Tf = g $$ Therefore, I have to solve the following for $f$: $$ e^{-x} \int_0^x e^y f(y) dy = g(x) $$ I try: $$ \int_0^x e^y f(y) dy = g(x)e^x \\ e^x f(x) = g'(x)e^x + g(x)e^x \\ f(x) = g(x) + g'(x) $$ The problem is that I'm writing $g'(x)$ without knowing if $g$ is differentiable (in general, it is not, I think). I do not know how to proceed. Can someone please help? Thank you.

Answer 1

For surjectivity, the image of $T$ consists of continuous functions, so $T$ cannot be surjective.

Your argument for injectivity cannot work as it is. The Fundamental Theorem of Calculus requires that the integrand is continuous (or good enough), which you don't have in this case. The result you need is Lebesgue's Differentiation Theorem.