Show that $f_n(x)=x^ne^{-x^2},n≥0$ convergence in $L^p(\lambda,R)$

Question

Question : Show that

$f_n(x)=x^ne^{-x^2},n≥0$ convergence in $L^p(\lambda,R)$ where $\lambda$ Lebesgue measure

My attempt : $x\to x^ne^{-x^2}$ application measurable as continue

Next : prove : $\int_{R}|x^ne^{-x^2}|^pd\lambda<{+\infty}$

$\exists m$ such that $e^{-x^2}≤\frac{m!}{x^{2m}}$ Then take integral $\int_0^1+\int_1^{+\infty}$ I need solution without gamma function .

Answer 1

This is false. $x^{n}e^{-x^{2}}$ does not converge in $L^{p}$ because no subsequence converges almost everywhere to an $L^{p}$ function. (The limiting function is necessarily $\infty$ for all $x>1$)

To show that $x^{n}e^{-x^{2}}$ is integrable note that this function is bounded by $1$ if $|x| \leq 1$. Now let $|x| >1$. In this case $|x^{n}e^{-x^{2}}| \leq 2^{n} (n!) e^{-x^{2}/2}$ and $e^{-x^{2}/2}$ is integrable. [ To derive above inequality use the fact that $e^{x^{2}/2}\geq \frac {(x^{2}/2)^{n}} {n!}$].