Suppose $\mu$ is $\sigma$-finite measure, and $0 \leq f_n$ is sequence of measurable functions that $f_n \rightrightarrows 0$.
Assume that $\int f_n^2 \operatorname d\! \mu = 1$ is constant. Is any of the following necessarily true? $$ \int f_n \operatorname d \! \mu \overset ? \to \infty \quad \quad \quad \int f_n^4 \operatorname d \! \mu \overset ? \to 0 $$
My attempt:
Note that uniform convergence implies that $f_n$ are bounded by $1$ for sufficiently large $n$, and therefore $ \int f_n^4 \operatorname d \! \mu \leq \int f_n^2 \operatorname d \! \mu \leq \int f_n \operatorname d \! \mu$.
It is easy to find examples that satisfy both conditions. For example $f_n = \frac 1 {\sqrt{n}} \chi_{[0, n]}$ and Lebesgue measure. I wasn't able to find counterexample. I tried using standard machine, but this problem seems not to be suited for it: the only thing I came up with is generalization of my example. Take sequence of sets $A_n$ such that $\mu(A_n) < \infty$ and $\mu(A_n) \to \infty$, and put $f_n = \frac 1 {\sqrt{\mu(A_n)}} \chi_{A_n}$. If instead we had disjoint $A_n^{(1)}, A_n^{(2)}, \ldots A_n^{(k)}$, then $\frac {f_n^{(1)} + f_n^{(2)} + \ldots + f_n^{(k)}}{\sqrt k}$ would yield another example. But that won't even be all examples of simple functions that satisfy given conditions, and not even close to all sequences of such. I have no idea what to do in general case.
From context I assume that $f_n \rightrightarrows 0$ denotes uniform convergence.
Both implications are true. For the first implication, suppose for a contradiction that there exists $C \ge 0$ and a sequence $n_k$ such that $\int f_{n_k} d\mu \le C$.
Then $1 = \int f_{n_k}^2 d\mu \le \|f_{n_k}\|_\infty \int f_{n_k} d\mu \le C \|f_{n_k}\|_\infty \to 0$ by the uniform convergence which yields a contradiction.
It is also the case that $\int f_n^4 d\mu \to 0$. For this, just write $$\int f_n^4 d\mu \le \|f_n\|_\infty^2 \int f_n^2 d\mu \to 0.$$