How to give a criterion for strong convergence of in $L_p[0,1]$ for this example

Question

let $x_n=\alpha_n e^{-nt}$ for $n\in \mathbb{N}$ and $1<p<\infty$.

How to give a criterion for strong convergence of in $L_p[0,1]$ for this example:

$x_n\rightarrow 0$ (strong convergence).

strong convergent = convergent in norm.

Attempt:

for fixed n and p: $||x_n||_p=(\int^1_0 |\alpha_n|^p|e^{-ntp}|dt)^{1/p}$=$|\alpha_n|1/(npe^{np})-1/(np))^{1/p}$

Then,

$||x_n||_p$ need to be bounded for strong convergence, so $(\alpha_n)\le M\in \mathbb{R}$

Now, we check:

$||x_n-0||$=$||x_n||_p\rightarrow 0$.

Then, what is the condition for $(\alpha_n)$ s.t.

$|\alpha_n|1/(npe^{np})-1/(np))^{1/p}$.

Answer 1

We have $\Vert x_n \Vert = \vert\alpha_n\vert (\int_0^1 e^{-npt} dt)^{\frac{1}{p}} = \vert\alpha_n\vert \left(\frac{1}{-np e^{np}} + \frac{1}{np} \right)^{\frac{1}{p}} = \left(\frac{\vert\alpha_n\vert^p}{-np e^{np}} + \frac{\vert\alpha_n\vert^p}{np} \right)^{\frac{1}{p}}$. The last expression converges to $0$ iff $\left(\frac{\vert\alpha_n\vert^p}{-np e^{np}} + \frac{\vert\alpha_n\vert^p}{np} \right) \rightarrow 0$ or $\frac{\vert\alpha_n\vert^p}{n}(1-\frac{1}{e^{np}})\frac{1}{p} \rightarrow 0$ which in turn happens iff $\frac{\vert\alpha_n\vert^p}{n} \rightarrow 0$ as the center term in the product above converges to $1$.

Hence, $x_n$ converges to $0$ in norm iff $\frac{\vert\alpha_n\vert^p}{n} \rightarrow 0$.