3
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – user3669039 Sep 6 '18 at 15:33
  • 1
    $\begingroup$ Or just notice that everything in the range is a continuous function. $\endgroup$ – zhw. Sep 6 '18 at 17:30
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I meant the Fundamental Theorem of Calculus for Lebesgue Integrals. Is that right? Can I use that? $\endgroup$ – user3669039 Sep 7 '18 at 9:03
  • $\begingroup$ $$ F: [a,b] \to \mathbb{C}, \text{ the following are equivalent }: \\ a. F \text{ is absolutely continuous on } [a,b]. \\ b. F(x) - F(a) = \int_a^x f(t)dt \text{ for some } f \in L^1([a,b],m) \\ c. \text{F is differentiable a.e. on } [a,b], F' \in L^1([a,b],m) \text{ and } F(x) - F(a) = \int_a^x F'(t)dt \\ $$ 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 $\endgroup$ – user3669039 Sep 7 '18 at 10:00
  • $\begingroup$ You are totally right. Sorry about that. $\endgroup$ – Martin Argerami Sep 7 '18 at 18:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.