Weak and strong convergence in $L^p$ 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 $\int_{[-\pi,\pi]} \sin(kx)^p dx$?
 A: 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 $k.$
A: 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
$$
\int_{-\pi}^\pi|\sin kt|^p\,dt=\frac1k\,\int_{-k\pi}^{k\pi}|\sin t|^p\,dt\geq\frac1k\,\sum_{j=0}^{k-1}\int_{j\pi-\pi/4}^{j\pi+\pi/4}\sin^pt\,dt
\geq\frac1k\,\frac{k-1}{2^{p/2}}\geq\frac1{2^{p/2}+1}.
$$
