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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 $1<p<\infty$ and $p\neq 2$. Hint : consider the functions $f_r(x)=e^{-rx}$ for $r>0$.

We want to show that for $1<p<\infty$ 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 $1<p<2$. I am not sure what to say about $p>2$ Am I overlooking something obvious?

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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 $0$.

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In the case p=2, there is A similaire theoreme than the Plancherel one for the Fourier's Transform on L² : it can be prolonged to an isometry.

The proof use multiplicative convolution on positive real numbers.

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