Let $(E,\mathcal E,\mu)$ be a measure space and $p\ge1$. If $(f_n)_{n\in\mathbb N}\subseteq L^p(\mu)$ and $f\in L^p(\mu)$ with $$\left\|f_n-f\right\|_{L^p(\mu)}\xrightarrow{n\to\infty}0\tag1,$$ can we infer that $\left(\frac{f_n}{\left\|f_n\right\|_{L^p(\mu)}}\right)_{n\in\mathbb N}$ is $L^p(\mu)$-convergent as well?
Note that clearly $\left\|f_n\right\|_{L^p(\mu)}\xrightarrow{n\to\infty}\left\|f\right\|_{L^p(\mu)}$. So, the answer should be positive.
If the measure space is finite and $f_n$ is the constant function $(-1)^{n} /n$ then $f_n \to 0$ in any $L^{p}$ but $\frac {f_n} {\|f_n\|_p}$ does not converge.
The conlusion holds if $\|f\|_p \neq 0$ and this is general fact valid in any normed linear space:
Let $(x_n)$ be sequence in a normed linear space $X$ such that $\|x_n-x\| \to 0$. If $x \neq 0$ then $\frac {x_n} {\|x_n\|} \to \frac x {\|x\|}$
Proof: $|\|x_n\|-\|x\|| \leq \|x_n-x\| \to 0$ so $\|x_n\| \to \|x\|$. It is quite easy to see that $\|x_n-x\| \to 0$ together with $a_n \to a, (a_n,a \in [0,\infty)) $ implies $a_nx_n \to ax$. (Use triangle inequality). Now take $a_n=\frac 1 {\|f_n\|_p}, a=\frac 1 {\|f\|_p}$.