Suppose that $f_k\in L^{2+\frac{1}{k}}(\Omega)$ with the property that $\|f_k\|_{L^{2+\frac{1}{k}}(\Omega)} = 1$ for all $k\ge 1$. $\Omega$ is a bounded domain in $\mathbb{R}^n$. Can such a sequence of functions strongly converge to ZERO in $L^2(\Omega)$, is it possible?
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I think the answer would be yes. Have you tried taking a characteristic function over a set of size $n^{-1}$ with height $n^{k / (2k + 1)}$. The $(2 + 1/k)$-norm will be $1$ but the (square of the) $L^2$-norm will be $$\frac1{n^{1- \frac{2k}{2k + 1}}}.$$ When fix the $n$ large enough in terms of $k$ to make that expression as small as you want. For instance: $$ n = 2^{k (2k + 1)}. $$ would make it.
Adrián González Pérez
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You computed the square of the $L^2$ there. Nice idea! – daw Feb 22 '19 at 10:48
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This is a nice example, thanks! – Carl Lincoln Mar 04 '19 at 18:58