Is the vice versa in the corollary below true? If not, could you give me a counterexample?
Thank you so much.
Let $f_{n}\colon\mathbb{R}\to\mathbb{R}$ be given by
$$ \begin{aligned} f_{n}(x):= \begin{cases} 1&\text{ if } x\in[n,n+1]\\ 0& \text{ otherwise} \end{cases} \end{aligned} $$
Then, for each $x\in \mathbb{R}$, we have that $f_{n}(x)$ is eventually $0$. Thus, $f_{n}\to 0 $ pointwise. But, $\|f_{n}\|_{p}=1$ for all $n$. Thus, $f_{n}$ does not converge to $0$ in $L^{p}$.
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Let $f_{n}\colon\mathbb{R}\to\mathbb{R}$ be given by
$$ \begin{aligned} f_{n}(x):= \begin{cases} 1&\text{ if } x\in[n,n+1]\\ 0& \text{ otherwise} \end{cases} \end{aligned} $$
Then, for each $x\in \mathbb{R}$, we have that $f_{n}(x)$ is eventually $0$. Thus, $f_{n}\to 0 $ pointwise. But, $\|f_{n}\|_{p}=1$ for all $n$. Thus, $f_{n}$ does not converge to $0$ in $L^{p}$.