0
$\begingroup$

I am working with the following problem: Let $C([0, 1])$ be the set of all continous functions $u:[0, 1] \to \mathbb{R}$. For any $u$, define: $$(Lu)(t) = \frac{1}{2}\int_{0}^{1}e^{-u(x)}(x+t)dx$$ Show that $Lu \in C([0, 1])$.

The context for this problem is an introductionary course in real analysis.

I have made some observations: Since all functions $u \in C([0, 1])$ are continous and defined over a closed interval, u is a bounded function. By the fundamental theorem of calculus, the definite integral of a continous function of a bounded interval exists. We know $e^{-u(x)}$ is a continous function.

Intuitively, the problem statement seems to hold according to the vague points above. From the integral above, I get $$\frac{1}{2}\int_{0}^1e^{-u(x)}x dx +\frac{1}{2}\int_{0}^1e^{-u(x)}tdx $$ $$\implies Lu(t) =M +K\cdot t$$ for some constants $ K, M \in \mathbb{R}$, which is a continous function in $C([0,1])$

I believe the argument should be more rigorous, but I am not sure what statements need to be explicitly proven.

Thank you!

$\endgroup$
0
$\begingroup$

Everything is O.K. With $M:=\frac{1}{2}\int_{0}^1e^{-u(x)}x dx$ and $K:= \frac{1}{2}\int_{0}^1e^{-u(x)}dx$ we have

$(Lu)(t) =M +K\cdot t$.

$\endgroup$
  • $\begingroup$ Thank you, Fred! Yes, I meant we define M and K as you describe above. $\endgroup$ – user2005142 Feb 22 '18 at 12:14

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.