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

$$f \in L^\infty (0,1) \\ Tf(x) = \int_0^x e^{y-x}f(y)dy, x\ge0$$ 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.

• I was thinking: If $T$ was surjective, then $g$ would be differentiable a.e., since $g(x) = e^{-x} \int_0^x e^y f(y) dy$ . However, it is not true that every function $g \in L^\infty (0,\infty)$ is differentiable a.e. Then, T cannot be surjective. – user3669039 Sep 6 '18 at 15:33
• Or just notice that everything in the range is a continuous function. – zhw. Sep 6 '18 at 17:30

For surjectivity, the image of $T$ consists of continuous functions, so $T$ cannot be surjective.
• $$F: [a,b] \to \mathbb{C}, \text{ the following are equivalent }: \\ a. F \text{ is absolutely continuous on } [a,b]. \\ b. F(x) - F(a) = \int_a^x f(t)dt \text{ for some } f \in L^1([a,b],m) \\ c. \text{F is differentiable a.e. on } [a,b], F' \in L^1([a,b],m) \text{ and } F(x) - F(a) = \int_a^x F'(t)dt \\$$ I thought that, setting $F(x) = \int_0^x e^y f(y) dt$ would make $F$ absolutely continuous and, hence, differentiable a.e. Please tell me where I'm wrong ( I know that I am ). Thank you for your help :D – user3669039 Sep 7 '18 at 10:00