Showing that the following set is closed in $L^p$ Question: Let $1\le p < \infty$ and $1\le q \le \infty$. Prove that the following set is closed in $L^p$.
$$
\{f \in L^p \cap L^q : |f|_q \le 1 \}
$$
My try: Let $f_n$ be a sequence in the above set that converges to $f$ in $L^p$. I can show that $f \in L^p$ but cannot show $|f|_q \le 1$. Any hint?
 A: Hint: there is pointwise a.e. convergence along a subsequence, so apply Fatou's lemma.
A: Let $S=\{f\in L^p \cap L^q(A)\}$ where $1\leq p <\infty$ and $1\leq q<\infty$. Then $S$ is closed in $L^p(A)$. As you say, take $\{f_n\}_{n\geq 0}\subset S$ such that $f_n \to f$ in $L^p$, then $f\in S$.
Indeed, we have
$$\lim_n \int_A |f_n(x)-f(x)|dx =0,$$
this means $f_n$ to $f$ in $L^p$, which in particular (by the inverse triange inequality implies that $f_n$ converges to $f$ in $L^p$-norm as well), i.e.
$$\lim_n \int_A |f_n(x)|^p dx = \int_A |f(x)|^p dx.$$
Then
$$\int_A |f(x)|^p dx = \int_A |f(x)-f_n(x)+f_n(x)|^p dx.$$
Minkowki's inequality ($|a+b|^p \leq 2^{p-1}(|a|^p + |b|^p)$) yields
$$\int_A |f(x)|^p dx \leq 2^{p-1} \int_A |f(x)-f_n(x)|^p dx + 2^{p-1}\int_A |f_n(x)|^p dx.$$
Now, for arbitrarily small $\varepsilon>0$, there is $n_0\in \mathbb{N}$ such that for all $n\geq n_0$ we have $\int_A |f_n(x)-f(x)|^pdx <\varepsilon$. So
$$\int_A |f(x)|^p dx \leq 2^{p-1} \varepsilon +2^{p-1}<\infty,$$
where the latter also follows since the sequence $a_n:=\|f_n\|_{L^p}$ is convergent and hence must be uniformly bounded by some constant $C$.
Observe that $\|f\|_{L^p}\leq 1$ which is immediate! Indeed,
$$1\geq  \int_A |f_n(x)|^p dx \longrightarrow \int_A |f(x)| dx$$
since $\|f_n\|_{L^p}$ is bounded by one and is convergent (converges in norm) so must also $\|f\|_{L^p}$ be bounded by one!
Nevertheless you need $\|f\|_{L^q}\leq 1$ as well to conclude that $f\in S$. This is follows from the fact that $L^p$ convergence implies that there is a subsequence $\{f_{n_k}\}_{k\geq 0}$ that convergences a.e. to $f$, i.e. $\lim_k f_{n_k}(x) = f(x)$ a.e. $x\in A$. Hence, Fatou's lemma allows you to conclude, i.e.
$$\int_A |f(x)|^q dx = \int_A \lim_n |f_n(x)|^q dx \leq \lim_n \int_A |f_n(x)|^q dx \leq \lim_n 1 = 1.$$
You can of course pull the limit out of the power function $|\cdot|^q$ since it is a continuous function and finally observe that the "a.e." convergence is ok here, indeed, the set $N\subset A$ where this convergence may not hold has Lebesgue measure 0 and the integral "does not see" these points.
I hope this helped. Good luck.
