Find the values, $p \in [1,\infty)$, such that the sequence $f_n(x) = \frac{n^2}{(n^2+x)\,x^{\frac{1}{3}}}$ is convergent in $L^p((1,\infty),\mu)$

Let, $p \in [1,\infty)$. Observe that, $$ \int_1^\infty \left(\frac{n^2}{(n^2+x)\,x^{1/3}}\right)^p \mathrm d x \le \int_1^\infty \left(\frac{n^2}{n^2\,x^{1/3}}\right)^p \mathrm d x \le \int_1^\infty \frac{\mathrm d x}{x^{p/3}}. $$ We have that, $$ \int_1^\infty \frac{\mathrm d x}{x^{p/3}} = \lim_{k \rightarrow \infty} \int_1^k \frac{\mathrm d x}{x^{p/3}} = \int_1^\infty\frac{\mathrm d x}{x^{p/3}}. $$ Thus, $f_{n} \in L^{p}(1,\infty)$ whenever $p > 3$. Let, $p >3$; thus, $$ \forall n \in \mathbb{N}, ~~\int_1^\infty \left|\frac{n^2}{(n^2+x)\,x^{1/3}} - \frac{1}{x^{1/3}}\right|^p \mathrm d x \le 2 \int_1^\infty \frac{\mathrm d x}{x^{p/3}} < \infty. $$ So, we can use the Lebesgue dominated convergence theorem to conclude that: $$ \lim_{n \rightarrow \infty} \int_1^\infty \left|\frac{n^2}{(n^2+x)x^{1/3}} - \frac{1}{x^{1/3}}\right|^p \mathrm d x = 0, $$ because $\lim_{n \rightarrow \infty} f_n(x) = \frac{1}{x^{1/3}}$, point wise. Now, we wish to show that our sequence does not converge for $p \in [1,3]$. Let $p \in [1,3]$. Suppose, that $f_n$ converges in $L^{p}(1,\infty)$ to some function $g$. Then $f_n$ converges to $g$ in measure, which implies there exists a subsequence $f_{n_k}$ that converges to $g$, point wise. The previous statement implies $g = \frac{1}{x^{1/3}}$, but this is a contradiction since $ g \not \in L^{p}( (1,\infty))$.

Does my proof look good? Thanks.

Solution 1:

You proof looks good, up to a minor thing to modify. The good argument to apply the dominated convergence theorem when $p>3$ is not that $$ \forall n \in \mathbb{N}, ~~\int_1^\infty \left|\frac{n^2}{(n^2+x)\,x^{1/3}} - \frac{1}{x^{1/3}}\right|^p \mathrm d x \le 2 \int_1^\infty \frac{\mathrm d x}{x^{p/3}} < \infty, $$ but rather the pointwise inequality

$$\left|\frac{n^2}{(n^2+x)\,x^{1/3}} - \frac{1}{x^{1/3}}\right|^p \leqslant 2 x^{-p/3}$$ and integrability of $x\mapsto x^{-p/3}$ over $(1,\infty)$.