# Show $(Lu)(t) = \frac{1}{2}\int_{0}^{1}e^{-u(x)}(x+t)dx$ is in $C([0, 1])$ if u is in $C([0, 1])$

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!

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