Weak and strong convergence in $L^p$

Question

Another practice qual question:

Let $X = [-\pi,\pi]$ and consider the Lebesgue measure. Let $p$ be a real number with $1 \leq p < \infty$. Define for each integer $k \geq 1$ that $f_k(x) = \sin(kx) (x\in X)$. Prove that:

a) The sequence $\left\lbrace f_k \right\rbrace$ converges weakly to $0$ in $L^p(X)$

b) The sequence $\left\lbrace f_k \right\rbrace$ does not converge to $0$ strongly in $L^p(X)$

For part a, I need to show that for all $g \in L^q(X)$, that $\int_{[-\pi,\pi]} f_kg =0$ since the dual space of $L^p$ is $L^q$. Our professor recommended using the Reimann-Lebesgue Lemma, but I don't see how the fact that $\mathbb{F}(L^1(\mathbb{R}^n)) \subset C_0(\mathbb{R}^n)$ where $\mathbb{F}$ is the Fourier transform helps at all.

For part b, I would think I'm done if I can show that the $L^p$ norm of $f_k$ doesn't tend to $0$, I would be done. But how do you integrate $\int_{[-\pi,\pi]} \sin(kx)^p dx$?

Answer 1

For part a): if $g\in C^1[-\pi,\pi]$, $$ \int_{-\pi}^\pi g(t)\sin kt\,dt=\left.-\frac1k\,g(t)\,\cos kt\right|_{-\pi}^\pi+\frac1k\int_{-\pi}^\pi g'(t)\,\cos kt =\frac1k\int_{-\pi}^\pi g'(t)\,\cos kt $$ As $g'$ is continuous in the interval, it is bounded. The integral on the right is thus bounded, and so the whole thing goes to zero as $k\to\infty$.

If now $g\in L^q$ is arbitrary, there exists a sequence $\{g_n\}\subset C^1[-\pi,\pi]$ with $g_n\to g$. Then \begin{align} \left|\int_{-\pi}^\pi g(t)\sin kt\,dt\right| &\leq\left|\int_{-\pi}^\pi g_n(t)\sin kt\,dt\right| +\left|\int_{-\pi}^\pi (g_n(t)-g(t))\sin kt\,dt\right| \\ \ \\ &\leq \left|\int_{-\pi}^\pi g_n(t)\sin kt\,dt\right|+(2\pi)^{1/p}\,\|g_n-g\|_q \end{align} by using Hölder (using that $|\sin x|\leq1$). Now $$ \limsup_{k\to\infty}\left|\int_{-\pi}^\pi g(t)\sin kt\,dt\right| \leq(2\pi)^{1/p}\,\|g_n-g\|_q. $$ As we are free to choose $n$ to make $\|g_n-g\|_q$ as small as we want, $$ \lim_k\left|\int_{-\pi}^\pi g(t)\sin kt\,dt\right|=0. $$

For part b): note that, on $[-\pi/4,\pi/4]$ we have $\sin t\geq1/\sqrt2$. Then $$ \int_{-\pi}^\pi|\sin kt|^p\,dt=\frac1k\,\int_{-k\pi}^{k\pi}|\sin t|^p\,dt\geq\frac1k\,\sum_{j=0}^{k-1}\int_{j\pi-\pi/4}^{j\pi+\pi/4}\sin^pt\,dt \geq\frac1k\,\frac{k-1}{2^{p/2}}\geq\frac1{2^{p/2}+1}. $$

Answer 2

For the first one, it is indeed just the Riemann-Lebesgue Lemma: If $f\in L^q[-\pi,\pi],$ then $f\in L^1[-\pi,\pi].$ Hence

$$\int_{-\pi}^\pi f(x) \sin (kx)\, dx \to 0$$

by RL. So $ \sin (kx) \to 0$ weakly in $L^p$ as desired.

For the second one, we don't need to evaluate the integral exactly. Instead we can just note

$$\int_0^\pi |\sin (kx)|^p\, dx = \frac{1}{k}\int_0^{k\pi} |\sin (y)|^p\, dy = \int_0^{\pi} \vert \sin (y)\vert^p\, dy$$

for all $k.$