$f \in L^p([0, 1])$, for every $1 \le r < p$, we have $\|f_n - f\|_r \to 0$ as $n \to \infty$.

Question

Let $f_n \in L^p([0, 1])$ with $p > 1$. Assume that $\|f_n\|_p \le 1$ and, as $n \to \infty$, $f_n \to f$ almost everywhere on $[0, 1]$. How do I see the following?

1. We have $f \in L^p([0, 1])$.
2. For every $1 \le r < p$, we have $\|f_n - f\|_r \to 0$ as $n \to \infty$.

Answer 1

It follows directly from Fatou's lemma that $\|f\|_p \leq 1$; in particular, $f \in L^p$. To prove the second claim, we fix $\epsilon>0$ and write

$$\begin{align*} \|f_n-f\|_r^r &= \int_{|f_n-f| \leq \epsilon^{-1}} |f_n-f|^r \,dx + \int_{|f_n-f|>\epsilon^{-1}} |f_n-f|^r \, dx \\ &=: I_1+I_2\end{align*}$$

We estimate the terms separately. Since $r<p$, we can write $r= p+\delta$ for some $\delta>0$. Then

$$\begin{align*} I_2 &\leq \int_{|f_n-f|>\epsilon^{-1}} |f_n-f|^r \frac{|f_n-f|^{\delta}}{(\epsilon^{-1})^\delta} \, dx \\ &\leq \epsilon^{\delta} \left( \int |f_n|^p dx + \int |f|^p \, dx \right) \\ &\leq 2 \epsilon^{\delta}\end{align*}$$

where we used for the last inequality that $\|f_n\|_p \leq 1$ for all $n \in \mathbb{N}$ and $\|f\|_p \leq 1$. Moreover, it follows from the dominated convergence theorem that

$$I_1 \xrightarrow[]{n \to \infty} 0 \tag{1}$$

for each fixed $\epsilon>0$. Consequently, we get

$$\begin{align*} \limsup_{n \to \infty} \|f_n-f\|_r^r &\leq \underbrace{\limsup_{n \to \infty} I_1(n)}_{\stackrel{(1)}{=} 0} + \limsup_{n \to \infty} I_2(n) \leq 2 \epsilon^{\delta}. \end{align*}$$

Since $\epsilon>0$ is arbitrary, this finishes the proof.