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?
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 $\Vert f\Vert_{L^{1+\varepsilon}}$).