# Consider the operator L given by $Lf(x)=\int_0^\infty e^{-tx}f(t)dt$. Show L is unbounded for 1<p<\infty

Consider the operator $$L$$ given by $$Lf(x)=\int_0^\infty e^{-tx}f(t)dt$$. View $$L$$ as a linear operator on the space $$L^P(0,\infty)$$. Show that $$L$$ is unbounded if $$1 and $$p\neq 2$$. Hint : consider the functions $$f_r(x)=e^{-rx}$$ for $$r>0$$.

We want to show that for $$1 there does NOT exist $$M \in \mathbb N$$ such that $$\|Lf(x)\| < M \|f(x)\|$$ for all $$f \in L^p[0,\infty]$$ .

The hint suggests we consider $$f_r(x)=e^{-rx}$$. Note that $$\lim_{n\to \infty} f_r(x) = 0$$. Consider

$$\|f_r(x)\|_p^p =(\int_0^\infty (e^{-rx})^pdx)=(\frac{1}{rp})\to 0$$

as $$r\to\infty$$ for any $$p$$. Now consider $$\|Lf_r(x)\| = \left\| \int_0^\infty e^{-tx}e^{-rt}dt\right\| = \left\|\int_0^\infty e^{-t(r+x)}\right\| = \left\|\frac{1}{(r+x)}\right\|~$$

Note that $$\frac{1}{r+x}$$ is not in $$L^1[0,\infty]$$ for any $$r$$ but for $$1 < p < \infty \quad \left\|\frac1{r+x}\right\|_p^p = \int_0^\infty \frac1{(r+x)^p}dx=\frac{r}{r^p (p-1)}~.$$
Hence $$\frac{\| Lf(x) \|}{ \| f(x) \|}=\frac{r^(1-p)}{(p-1)}\frac{rp}{1}=r^{2-p}\frac{p}{p-1}$$ which is unbounded for $$1. I am not sure what to say about $$p>2$$ Am I overlooking something obvious?

You got that for all positive $$r$$, $$\sup_{f\in L^p,f\neq 0}\frac{||Lf(x)||_p}{||f(x)||_p}\geqslant\frac{||Lf_r(x)||_p}{||f_r(x)||_p}= r^{2-p}\frac{p}{p-1}.$$ When $$p\neq 2$$, the right hand side can be made arbitrarily big: if $$p\lt 2$$, take $$r$$ big and if $$p>2$$, take $$r$$ close to $$0$$.