Convergence in $L^1$ and Boundedness in $L^p$ Implies Convergence in $L^p$

Question

Let $u_n = u_n(t,x)$ be a sequence of functions, $u_n : (0, \infty) \times \mathbb{R}^N \rightarrow \mathbb{R}$, such that $u_n(t)$ converges to a function $u(t)$ in the $L^1(\mathbb{R}^N)$ norm for all $t \geq 0$.

I.e. $ ||u_n(t,\cdot) - u(t, \cdot)||_{L^1(\mathbb{R}^N)} \rightarrow 0$.

Also, assume the set of $u_n(t) $ is bounded in the $L^p(\mathbb{R}^N)$ norm, for all $p \in [1, \infty)$.

To Prove: $u_n(t) \rightarrow u(t)$ for all $t \geq 0$ in the $L^p(\mathbb{R}^N)$ norm, for all $p \in [1, \infty)$.

My Work So Far

By convergence in $L^1$, we know there exists a subsequence $u_{n_k}$ such that $u_{n_k}(t,x) \rightarrow u(t,x)$ pointwise, for almost all $x \in \mathbb{R}^N$, for all $t \geq 0$.

By boundedness of $u_n(t) $ in $L^p$, we know there exists $\text{lim inf}_{n_k \rightarrow \infty} ||u_{n_k}(t)||_{L^p} < \infty$.

Thus, by Fatou's Lemma, $ ||u(t)||_{L^p} \leq \text{lim inf}_{n_k \rightarrow \infty} ||u_{n_k}(t)||_{L^p} < \infty$ for all $t \geq 0$. In particular, $u(t) \in L^p( \mathbb{R}^N)$ for all $t \geq 0$.

This is the most concrete progress I can make. I would like to proceed as follows, but an unsure if this is a dead end:

Let $f_{n_k} (t,x) := |u_{n_k}(t, x) - u(t, x)|$. Then $f_{n_k}(t)$ is a sequence of functions in $L^p( \mathbb{R}^N)$, for all $p \in [1, \infty)$ and all $t \geq 0$, and we have $f_{n_k}(t) \rightarrow 0$ in $L^1$, for all $t \geq 0$, and $f_{n_k} (t,x) \rightarrow 0$ pointwise for almost all $x \in \mathbb{R}^N$ and all $t \geq 0$.

I would here like to use Lebesgue's Dominated Convergence Theorem somehow to finish the proof, but am having trouble finding a function which dominates $f_{n_k}$. Could someone please tell me how to find such a function, or if a different method is required. Thank you

Answer 1

You can drop the $t$, its not playing any role. Then you can use the following interpolation inequality, which is not hard to prove using Hölder's inequality:

Lebesgue Interpolation. Let $p_0,p_1\in[1,\infty]$, $\alpha\in[0,1]$, and $$\frac1{p_\alpha} = \frac{1-\alpha}{p_0}+\frac{\alpha}{p_1} .$$ Then for any $f\in L^{p_0}\cap L^{p_1}$, $f\in L^{p_\alpha}$ as well, with $$ \|f\|_{L^{p_\alpha}} \le \|f\|^{1-\alpha}_{L^{p_0}} \|f\|^\alpha_{L^{p_1}}.$$ Proof. Using the identity $ \|f\|_{L^p}^p = \| |f|^p \|_{L^1}$, we see that (WLOG $f\ge0$ for ease of notation) \begin{align} \|f\|^{p_\alpha}_{{p_\alpha}} &= \|f^{1-\alpha} f^\alpha\|^{p_\alpha}_{p_\alpha} \\ &=\|f^{(1-\alpha)p_\alpha} f^{\alpha p_\alpha}\|_1 \\ &\le \|f^{(1-\alpha)p_\alpha} \|_{\frac{p_0}{(1-\alpha)p_\alpha}}\|f^{\alpha p_\alpha}\|_\frac{p_1}{\alpha p_\alpha} \\ &= \|f\|^{(1-\alpha)p_\alpha}_{p_0} \|f\|^{\alpha p_\alpha}_{p_0}, \end{align} so the result follows.

Now, for any $p\in(1,\infty)$, there's some some $q$ such that $1<p<q<\infty$, and this defines $\alpha\in(0,1)$ by $\frac 1p = \frac{1-\alpha}1 + \frac\alpha{q}$. We get \begin{align} \|u_n - u\|_{L^p} \le \overbrace{\|u_n - u\|_{L^1}^{1-\alpha }}^{\to 0} \underbrace{\|u_n - u\|_{L^q}^{\alpha}}_{\text {bounded}} \xrightarrow[n\to\infty]{}0.\end{align}

Answer 2

Let $f_n(x)=u_n(t,x)-u(t,x)$ (with $t$ fixed). Then $\|f_n\| \to 0$ and $\|f_n\|^{q} $ is bounded for any $q$. Fix any $ q>p$.

We want to show that $\|f_n\|_p \to 0$. Let $\epsilon >0$ and choose $M$ such that $\mu (|f_n|>M)\leq \frac 1 M \int \|f_n\|_1 <\epsilon$ for all $n$. By holder's inequality $\int_{|f_n| >M} |f_n|^{p} \leq (\int_{|f_n| >M} |f_n|^{q})^{p/q} (\mu (|f_n| >M) ^{(q-p)/q} \leq C (\epsilon)^{(q-p)/p}$ for some finite constant $C$. Next look at $\int_{|f_n| \leq M} |f_n|^{p}$. Bound this by $M^{p-1} \|f_n\|$ to complete the proof.