I'm asking myself whether there is a function $f\in L^2(\mathbb{R}_+\to\mathbb{R})$ so that $\lim_{x\to\infty} f(x)^2\log(x)>0$. I think there is no such function. Here is my proof:
Let $f\in L^2$ with $\lim_{x\to\infty} f(x)^2\log(x)>0$. Then $$\infty>\int_{1}^{\infty}f(x)^2\mathrm{d}x=\sum_{n=1}^{\infty}\int_{n}^{n+1}f(x)^2\frac{\log x}{\log x}\mathrm{d}x\geq \sum_{n=1}^{\infty}\frac{1}{\log (n+1)}\int_{n}^{n+1}f(x)^2\log (x)\mathrm{d}x.$$ Since $\sum_{n=1}^{\infty}\frac{1}{\log (n+1)}=\infty$ and $\lim_{x\to\infty} f(x)^2\log(x)>0$, the right hand side of the equation above is $\infty$, which is a contradiction.
Is this proof correct?
Your proof works because $$ \int_{n}^{n+1}f(x)^2\log (x) \, dx > c > 0 $$ for sufficiently large $n$.
Instead of going via infinite series you can also argue that $f(x)^2 \log(x) > a > 0$ for $x \ge x_0$ implies $$ \int_{0}^{x_1} f^2(x) \, dx > \int_{x_0}^{x_1} \frac{a}{\log(x)} \, dx > \int_{x_0}^{x_1} \frac{a}{x} \, dx = a (\log(x_1) - \log(x_0)) $$ for $x_1 > x_0$, and therefore $$ \lim_{x_1 \to \infty}\int_{0}^{x_1} f^2(x) \, dx = \infty \, . $$