what's the relation between functional and norm? Consider the Banach space $L^p(\mu)$,$1\leq p<\infty$.For any $f\in L^p$,we have $$||f||_p=\left(\int_X |f|^pd\mu\right)^{1/p}<\infty.$$ On the other hand,by Riesz Representative Theorem,we have $$\Lambda f=\int_X fd\mu.$$ So what's the relation between them?
For an application,for $1\leq p<\infty$,let $L^p=L^p(-\pi,\pi)$ with respect to Lebesgue measure .And in this $L^p$, $f_n(t)=e^{int},-\pi\leq t\leq\pi$.I want to claim that $f_n$ converges to $0$ weakly but not strongly.How do I start?
 A: If $f \in L^p(\mu)$ and $\frac{1}{p} + \frac{1}{q} = 1$, then $f$ induces a linear functional $F: L^q(\mu) \to \mathbb{C}$ by setting
$$F(g) = \int \! fg \, d\mu.$$
The relation between $f$ and $F$ is that $\|F\| = \|f\|_p$. One of the Riesz Representation Theorems states that every linear functional on $L^q(\mu)$ is of this form. Although this is not the one you quoted in your post, I think you may have meant this one (the one you quote represents the linear functionals on the space of continuous functions as regular Borel measures).
Now to the second part of the question. Let $f_n(x) = e^{inx}$. You want to show that $f_n$ converges weakly, but not strongly, to $f = 0$ in $L^p(-\pi, \pi)$. To see that $f_n$ does not converge strongly to $0$ note that
$$\|f_n\|_p =\left(\int_{-\pi}^\pi \! |f_n|^p \, dx \right)^{1/p} = \left(\int_{-\pi}^\pi \! 1 \, dx \right)^{1/p} = (2\pi)^{1/p}$$
for every $n$. But if $f_n$ converges strongly to $f = 0$, then we would have $\|f_n\|_p \to 0$. Hence $f_n$ does not converge strongly to $f$.
Now we want to show that $f_n$ converges weakly to $f$ in $L^p(-\pi,\pi)$. What exactly does that mean? It means that for every linear functional $F$ on $L^p(\pi, \pi)$ we have
$$\lim_{n \to \infty} F(f_n) = 0.$$
Now we use the Riesz Theorem to see that this actually means that for all $g \in L^q(-\pi, \pi)$ we have
$$\lim_{n\to\infty}\int_{-\pi}^\pi \! g(x) f_n(x) \, dx = \lim_{n\to\infty}\int_{-\pi}^\pi \! g(x) e^{inx} \, dx= 0.$$
Now we instantly recognize that second term is exactly the $-n$-th Fourier coefficient $\hat{g}(-n)$ of $g$ so we want to prove
$$\lim_{n \to \infty} \hat{g}(-n) = 0.$$
But now we simply notice that since $(-\pi, \pi)$ has finite measure we also have $g \in L^1(-\pi, \pi)$ and apply the Riemann-Lebesgue Lemma which asserts exactly the truth of this last statement. I hope this clears up some of your confusion.
A: After considering my arguments, I conclude that with
$$d_f(t) := \lambda\{x : |f(x)| > t \}, \quad \lVert f \rVert_{p,\infty} = \sup_{t>0} t (d_f(t))^{\frac{1}{p}}$$
we have
$$d_{e^{in\cdot}}(t) = \lambda(-\pi,\pi)\chi_{[0,1)}(t) = 2\pi \chi_{[0,1)}(t)$$
and thus
$$\lVert e^{in\cdot} \rVert_{p,\infty} = \sup_{1>t>0}(2\pi)^{\frac{1}{p}} t = (2\pi)^{\frac{1}{p}} \neq 0$$
So your claim is false. $(e^{in\cdot})_n$ does not converge weakly to $0$.
