Let $C([-\pi, \pi])'$ be the set of continuous functions $f$ from $[-\pi, \pi]$ to $\mathbb{C}$ such that $f(\pi) = f(-\pi)$. For $f \in C([-\pi, \pi])', n \in \mathbb{Z}$, define $a_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(\theta) e^{-in\theta} d\theta$. If $f(\theta) = \sum_{n=-\infty}^\infty a_n e^{in\theta}\,\forall \theta\in [-\pi, \pi]$, let $S_N(f)(\theta) = \sum_{n=-N}^N a_n e^{in\theta}$. Define for $f \in C([-\pi, \pi])'. ||f||_2 := \left(\frac{1}{2\pi}\int_{-\pi}^\pi |f(\theta)|^2 d\theta\right)^{1/2},$ $\lVert f\rVert_1 := \frac{1}{2\pi} \int_{-\pi}^\pi |f(\theta)| d\theta$, and $\lVert f(x)\rVert_\infty := \max_{x\in [-\pi, \pi]} |f(x)|$. Consider the following statements for a fixed $f\in C([-\pi , \pi])'$:
- $f(\theta) = \sum_{n=-\infty}^{\infty} a_n e^{in\theta} \,\forall \theta\in [-\pi, \pi]$.
- $\lim\limits_{n\to\infty} ||f-S_n(f)||_2= 0$.
- $\lim\limits_{n\to\infty} ||f - S_n(f)||_1 = 0.$
- $\lim\limits_{n\to\infty} ||f-S_n(f)||_\infty = 0.$
Prove that $(4)\Rightarrow (1), (2)$, and that $(2)\Rightarrow (3)$.
To show that $(4)\Rightarrow (1),$ this follows from the fact that uniform convergence implies pointwise convergence. To show $(4)\Rightarrow (2)$, one can show that $\lVert f\rVert_2 \leq \lVert f\rVert_\infty$ for any $f\in C([-\pi, \pi])'$. Indeed, fix $f\in C([-\pi, \pi])'$. One has $\lVert f\rVert_2^2 = \frac{1}{2\pi} \int_{-\pi}^\pi |f(\theta)|^2 d\theta \leq \frac{1}{2\pi} \int_{-\pi}^\pi \lVert f\rVert_\infty^2 d\theta= \lVert f\rVert_\infty^2.$ However, I'm not sure how to show that $(2)\Rightarrow (3)$. It doesn't seem true that $\lVert f\rVert_\infty \leq \lVert f\rVert_2$ for $f\in C([-\pi, \pi])'$.
Here's my incomplete attempt to prove $(2)\Rightarrow (3)$. If $f$ is identically zero, $\lVert f\rVert_2 = \lVert f \rVert_1$. So suppose f is not identically zero, and let $c = \sup_{x\in [-\pi, \pi], f(x)\neq 0} \frac{1}{|f(x)|}$. The problem is that the set $\{x\in [-\pi, \pi], f(x)\neq 0\}$ doesn't seem to be compact, so the EVT doesn't seem to apply. I might be able to apply the EVT in some other way however.