I have to show that $B=\{x\in \mathcal{H}:||x||\le 1\}$ in the following example is not compact:
Consider the measure space $([0,2\pi],\mathcal{B}([0,2\pi]),\mu)$ where $\mu$ is the probablity measure $\frac{1}{2\pi} \lambda^r_{[0,2\pi]}$ and $\lambda^r_{[0,2\pi]}$ is the lebesgue measure restricted to $[0,2\pi]$.
Furthermore consider the Hilbert space $\mathcal{H}=(L_{\mathbb{C}}^2(\mu),\langle \cdot , \cdot\rangle)$ where $L_{\mathbb{C}}^2(\mu)=\{[f]|f\in \mathcal{L}^2_{\mathbb{C}}(\mu)\}$ and $\langle [f],[g]\rangle=\int_0^{2\pi}f\bar{g} d\mu$ for $f,g\in \mathcal{L}_{\mathbb{C}}^2(\mu)=\{f|Re(f),Im(f)\in \mathcal{L}^2(\mu)\}$.
What I have previously shown is that if H is a Hilbert space with a inner product and norm then B is closed and bounded. (1)
I have also shown that if a hilbert space does not have finite dimension, then there exist a sequence in B that does not have a convergent subsequence. (2)
I have to use result (2) but how do I show that B is not compact in $\mathcal{H}$?
Best Answer
We have a theorem that states:
You already have a norm induced by the inner product.
So it suffices to find an infinite linearly independent sequence of functions in $L^2([0,2\pi])$
Take for instance $\{\frac{1}{\sqrt{2\pi}}e^{imx}:m \in \Bbb{Z}\}\subseteq L^2([0,2\pi])$
Thus the unit ball cannot be compact.