The Laplace transform of real measurable functions on $[0,\infty)$ is defined by $L[f] = \int_{0}^\infty f(t)e^{-st}dt $
If $f \in L^{1}[[0, \infty)]$ prove that $L[f]$ exists and is bounded for all $s\geq 0$ . If $f_n ,f \in L^{1}[[0, \infty)]$ for all $n$, and $f_n\to f$ in $ L^{1}[[0, \infty)]$ norm, show that $ L[f_n]\to L[f]$ uniformly on $[0, \infty)$
I can't begin with the question. Tried for 3 days , but couldn't find anything to begin with. Can anybody suggest anything ?
What I have done so far... I don't know if it is correct or not
I have proved existence of $L[f]$ by showing as $f$ is measurable its integrable and hence $\int_{0}^\infty|f|dx \lt \infty$ and using $e^{-sx} \leq 1 \forall x \geq0, s \geq 0$ I have shown $|L[f]| \lt \infty $ which proves it exists I guess
You have already shown that $L[f]$ is bounded: it is bounded by $\int |f|$.
Now $|L[f_n](s)-L[f](s)| =|\int_0^{\infty} [f_n(t)-f(t)] e^{-st} dt| \leq \int |f_n(t)-f(t)| dt$ which shows that $L[f_n](s)\to L[f](s)$ uniformly.