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Let $(u_n)_n$ a sequence of $\mathcal C_c^\infty (\mathbb R^n)$ that converge in $L^p$ to $u\in \mathcal C^1_c(\mathbb R)$. If $q>1$, is there a subsequence that converge to $u$ in $L^q$ ?

Attempts

I know that $(u_n)$ is in $L^q$, that $u\in L^q$ also. Now if $q\leq p$, then by Jensen $$\|u_n-u\|_{L^p}\geq \left(\int |u_n-u|^q\right)^{p/q}=\|u_n-u\|_{L^q}^p,$$ and thus $u_n\to u$ in $L^q$. But what happen if $p<q$ ?

user330587
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  • You are using $n$ in two ways: $\mathbb{R}^n$ and $u_n$. –  Oct 18 '17 at 03:06
  • You are also misusing Jensen's inequality; it requires a probability measure space, and $\mathbb{R}^n$ is of infinite measure. –  Oct 18 '17 at 20:56

1 Answers1

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No, you don't get that $u_k\to u$ in $L^q$. Consider the case $p<q$ first. Let $u\equiv 0$, let $u_1$ be any bump function. and define $$ u_k(x) = k^{n/q} u_1(kx) $$ so that $\|u_k\|_q = \|u_1\|_q$ for all $k$ (use the change of variable to see this). Clearly, $u_k$ does not converge to $0$ in $L^q$ norm. But it does converge in $L^p$ for any $p<q$, because $$\|u_k\|_p = k^{n/q-n/p}\|u_1\|_p\to 0$$

The case $p>q$ is similar: let $u$ and $u_1$ be as above, but now $$ u_k (x) = k^{-n/q}u_1(k^{-1}x) $$ Still, $\|u_k\|_q = \|u_1\|_q$ for all $k$. But $$\|u_k\|_p = k^{n/p-n/q}\|u_1\|_p\to 0$$