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Question: If $\|x_n\|_2 \to \infty$ in $L_2$-norm, does $\|x_n\|_{1+\varepsilon} \to \infty$ in $L_{1+\varepsilon}$-norm, for all $\varepsilon > 0$?

Details/Progress: This should follow trivially for $\varepsilon \geq 1$ because $\|x\|_p \leq \|x\|_{q}$ for any $1\leq p \leq q < \infty$, but I am wondering if I can show it also holds for $\varepsilon \in (0,1)$. This clearly does not hold in general, but I have more information for the case at hand: Specifically, I am considering sequences $x_n$ in the space of square-integrable probability densities, i.e.

$$ C = \{x \in L_2(\mathbb{R^n}) | x \geq 0 \text{ and} \int_{\mathbb{R}^n}x(z)dz = 1 \}. $$

Intuitively, I think that because one can only allocate a total area under the curve of size $1$, any sequence $x_n$ in this space that diverges to $\infty$ needs to have some neighbourhood on which $x_n(z)$ grows larger and larger (so that the divergence of $\int_{\mathbb{R}^n}x(z)^{2}dz$ happens purely on that neighbourhood). This makes me think that on that same neighbourhood, $\int_{\mathbb{R}^n}x(z)^{1+\varepsilon}dz$ should also diverge. Is this intuition correct? If so, is there an appropriate/easy way to formalize it? If not, do you know of a counterexample?

Jeremias K
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    https://math.stackexchange.com/questions/76016/is-p-norm-decreasing-in-p –  May 16 '19 at 19:20
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    @Shogun thanks for the link, but the counterexample in the accepted response does not apply to my setting because $\chi_E$ is not in $C$ as I define it (specifically, I need any $x \in C$ to integrate to $1$). – Jeremias K May 16 '19 at 19:52
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    By the way, the thing you consider trivial is not exactly as trivial as you think. The inequality $\Vert x\Vert_p \leq \Vert x\Vert_q$ is not true in general. Our measure space is not finite. What saves you is the assumption that your function have normalized $L^1$ norm. – Severin Schraven May 16 '19 at 22:44
  • @SeverinSchraven You're right, I completely overlooked that -- thanks! – Jeremias K May 17 '19 at 02:54

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Consider $f(z)=\frac{1}{4\sqrt{\vert z \vert}} \chi_{[-1;1]}(z)$. Now define $$ a_n =\int_{[-1/n;1/n]} f(z) dz $$ and define $$ x_n(z) = f(z)\chi_{[-1;1]\setminus{ [-1/n; 1/n]}}(z) + a_n\chi_{[2;3]} $$ Then we have $x_n \in C$. By monotone convergence we get that the $L^2$ norm of $(x_n)$ explodes. However, as $f\in L^{1+\varepsilon}$ for $0<\varepsilon <1$ we get by dominated convergences that the $L^{1+\varepsilon} $ norm stays bounded (in fact it converges to $\Vert f\Vert_{L^{1+\varepsilon}}$).

Severin Schraven
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  • I don't think i understand the argument -- you argue that as $f \in L^{1+\varepsilon}$ prevents $x_n$ from diverging, but from what I can see, $f \in L^{2}$, too. Could you maybe explain in a sentence or two why the $L^2$ norm of your sequence explodes while the $L^{1+\varepsilon}$ norm doesn't? Sorry, it's probably pretty obvious but I don't see it – Jeremias K May 17 '19 at 03:01
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    We know that $(x_n - a_n \chi_{[2;3]})$ converges monotonically increasing to $f$ and $\lim_{n\rightarrow \infty } a_n =0$. Thus, we get for any $1\leq p< \infty$ $$ \lim_{n\rightarrow \infty} \Vert x_n \Vert_p = \Vert f \Vert_p $$ However, for $1\leq p < 2$ we have $\Vert f \Vert_p <\infty$ and $\Vert f \Vert_{q} = \infty$ for $q\geq 2$. – Severin Schraven May 17 '19 at 08:24
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    We have $$ \Vert f \Vert_p^p = \frac{1}{2} \int_{0}^1 z^{-p/2}$$ This integral is finite iff $p<2$. – Severin Schraven May 17 '19 at 08:30
  • Thanks for the explanation, this clarifies it! – Jeremias K May 17 '19 at 12:05