Convergence to zero in $L^{p}(\mathbb{R}^{N})$

Question

Is true that if $u_{n} \in L^{p}(\mathbb{R}^{N}) \cap L^{q}(\mathbb{R}^{N})$ and $u_{n} $ converges to $0$ in $L^{p}(\mathbb{R}^{N})$ then $u_{n}\rightarrow 0$ in $L^{q}(\mathbb{R}^{N})$.

Edit: I added that $u_{n} \in L^{p} \cap L^{q}$

My guess is that the above statement is true, once that convergence in $L^{p}$ implies in convergence almost everywhere, so $u_{n}(x)$ is bounded. I should find a integrable function $\varphi$ such that $|u_{n}(x)| \leq \varphi(x)$ a.e and conclude the proof using the Dominated Convergence Theorem. Am I right? What function $\varphi$ I can choice?

Answer 1

Actually, for all $p,q\in [1,+\infty)$ such that $p\neq q$, there exists a sequence $\left(f_n\right)_n$ such that $\left\lVert f_n\right\rVert_p\to 0$, $\sup_{n\geqslant 1}\left\lVert f_n\right\rVert_q$ is finite but $\left\lVert f_n\right\rVert_q$ does not converge to zero.

For example, let $f_n(x)=n^{1/q}\mathbf 1_{(0,1/n)}(x)$ if $q>p$ and $f_n(x)=n^{-1/q}\mathbf 1_{(0,n)}(x)$ if $p>q$.

Answer 2

Not true. For one, it is possible for $u_n$ to be in $L^p$ without being in $L^q$ for $q>p$.

Example: Take $f_n(x)= \frac 1 {n\sqrt{x}}$ for $x\in(0,1)$ and $f(x)=0$ anywhere else. That function is in $L^1(\mathbb R)$, and clearly $f_n\rightarrow 0$ in that space. However, $f_n\notin L^2(\mathbb R)$.