Hint: Showing $f_{n}(x):=\frac{n\sqrt{x} \sin{(x)}}{1+nx^{2}}$ converges in $L^{p}$ to $x^{-\frac{3}{2}}\sin{(x)}$

Question

Let $f_{n}:]0, \infty[\to \mathbb R$ and $f_{n}(x):=\frac{n\sqrt{x} \sin{(x)}}{1+nx^{2}}$

Show that $f_{n}$ converges in $L^{p}$ to $f$ where $f(x):=x^{-\frac{3}{2}}\sin{(x)}$ and $p \in [1,2[$

My idea:

\begin{align} \lVert f_{n}-f\rVert_{p}^{p}&=\int_{0}^{\infty}\left\lvert\frac{n\sqrt{x} \sin{(x)}}{1+nx^{2}}-x^{-\frac{3}{2}}\sin{(x)}\right\rvert^{p}dx\leq\int_{0}^{\infty}\left\lvert\frac{n\sqrt{x}}{1+nx^{2}}-x^{-\frac{3}{2}}\right\rvert^{p}dx \\ &=\int_{0}^{\infty}\left\lvert\frac{n\sqrt{x}-x^{-\frac{3}{2}}(1+nx^{2})}{1+nx^{2}}\right\rvert^{p}dx=\int_{0}^{\infty}\left\lvert\frac{n\sqrt{x}-x^{-\frac{3}{2}}+nx^{-\frac{1}{2}}}{1+nx^{2}}\right\rvert^{p}dx \end{align}

Am I on the right track? How do I continue from here?

Answer 1

First notice that

$$\frac{n\sqrt{x}\sin x}{1+nx^2} = \frac{\sqrt{x}\sin x}{\frac1n+x^2} \xrightarrow{n\to\infty} x^{-3/2}\sin x$$ pointwise.

Then notice that

$$\left|\frac{\sqrt{x}\sin x}{\frac1n+x^2}\right|^p \le \frac1{x^{p/2}}\chi_{(0,1]}(x) + \frac1{x^{3p/2}}\chi_{[1,\infty)}(x) =: g(x) \in L^1(0,\infty)$$ and also $|x^{-3/2}\sin x|^p$ is dominated by the same integrable function $g$.

Hence

$$\left|\frac{\sqrt{x}\sin x}{\frac1n+x^2} - x^{-3/2}\sin x\right|^p\le 2^p g(x)$$ so Lebesgue dominated convergence theorem yields $$\lim_{n\to\infty}\int_0^\infty \left|\frac{\sqrt{x}\sin x}{\frac1n+x^2} - x^{-3/2}\sin x\right|^p\,dx = \int_0^\infty\lim_{n\to\infty} \left|\frac{\sqrt{x}\sin x}{\frac1n+x^2} - x^{-3/2}\sin x\right|^p\,dx = \int_0^\infty 0\,dx = 0$$

Answer 2

Let $f(x)= x^{-3/2} \sin\, x$. Then $f_n(x) \to f(x)$ for all $x$. $\frac t {1+tx^{2}}$ is an increasing function of $t$ so $\frac n {1+nx^{2}} \sqrt x$ is an increasing sequence of non-negative measurable functions converging at every point to $x^{-3/2}$. By Monotone Convergence Theorem see that $\int_1^{\infty} |f_n(x)-f(x)|dx \to 0$ (because $|\sin\,x | \leq 1$). This proves that the integral from $1$ to $\infty$ tends to $0$ when $p=1$. For the general case not the simple

Lemma If $0\leq g_n$ and $g_n$ increases to $g$ with $\int g^{p} <\infty$ then $\int |g_n-g|^{p} \to 0$.

This lemma follows from Dominated Convergence Theorem

For $x <1$ note that $\frac n {1+nx^{2}} \sqrt x \leq \frac 1 2$, so Dominated Convergence Theorem shows that $\int_0^{1} |f_n(x)-f(x)|^{p}dx \to 0$.

Answer 3

There is a small mistake in your calculation: $$ n x^{1/2} - x^{-3/2} (1+nx^2) = n x^{1/2} - x^{-3/2} \color{red}{-} n x^{\color{red}{+} 1/2} = - x^{-3/2} .$$ Therefore, the integral you actually end up with is $$ \int \limits_0^\infty \frac{\mathrm{d}x}{x^{3p/2} (1+nx^2)^p} , $$ which is divergent at the lower limit. This is because the estimate $\lvert\sin(x)\rvert\leq 1$ is not good enough. Instead we should use $\lvert\sin(x)\rvert\leq x$, which is accurate at the lower limit. It is of course terribly bad for large $x$, but that is irrelevant since convergence at infinity is ensured by the additional $x^{-2p}$ term anyway.

With these changes we find $$ \lVert f_n - f \rVert_p^p \leq \int \limits_0^\infty \frac{\mathrm{d}x}{x^{p/2} (1+nx^2)^p} \stackrel{\sqrt{n} x = t}{=} n^{-\frac{2-p}{4}} \int \limits_0^\infty \frac{\mathrm{d}t}{t^{p/2} (1+t^2)^p}\, .$$ While the remaining integral can be computed explicitly, it is of course sufficient to check that it converges.