Convergence of function series $f_n(x) = e^n 1_{[0,\frac{1}{n}]}$ in Lp spaces

Question

We consider the function series $f_n : [0,1] \rightarrow \mathbb{R}, n \geq 1$, such as :

$f_n(x) = e^n 1_{[0,\frac{1}{n}]}(x)$.

I have to study the convergence of the sequence in $L^p([0,1])$, where $p \in ]1,\infty[$.

To begin, I wrote that $\lim_{n \rightarrow +\infty} f_n(x) = \lim_{n \rightarrow +\infty} \begin{cases} e^n & \text{ if } x \in [0,\frac{1}{n}] \\ 0 & \text{ if } x \notin [0,\frac{1}{n}] \end{cases} = \begin{cases} +\infty & \text{ if } x \in \{0\} \\ 0 & \text{ if } x \in ]0,1] \end{cases}$.

But I don't know how to prove if $\lim_{n \rightarrow +\infty} ||f_n-f||_p = 0$.

I'm beginner in spaces $L^p$, someone could help me ? Thank you in advance.

Answer 1

Consider $f\equiv 0$ almost everywhere. So $$ \|f_n-f\|_p^p=\|f_n\|_p^p=\int_0^1|f_n(t)|^p~dt=\int_0^1 e^{pn}1_{[0,1/n]}(t)~dt=\ldots=\frac{e^{pn}}{n}. $$ Hence, $f_n\to f$ in $L^p([0,1])$ if $\lim_{n\to\infty}\frac{e^{pn}}{n}=0$.

Answer 2

Hint: we have $$||f_n-f||_p ^p=\int_{[0,\frac{1}{n}]}|f_n-f|^p + \int_{]\frac{1}{n},1]}|f_n-f|^p $$