Let $V = \{P(t)h(t):P \text{ is a trigonometric polynomial}\}$. Find $V^\perp$ and show that $V$ is dense in $L^2(\mathbb{T})$.

Question

Let $f \in L^2(\mathbb{T})$ and $f (x) \neq 0$ a.e. Define $$ V = \{P(t)f (x):P \text{ is a trigonometric polynomial}\} $$ Show that

(1) $V$ is a subspace of $L^2(\mathbb{T})$

(2) Find $V^\perp$

(3) $V$ is dense in $L^2(\mathbb{T})$

$\mathbb{T} = \mathbb{R}/(2\pi\mathbb{Z})$, the integral over $\mathbb{T}$ is just an integral taken over any interval of length $2\pi$

The first one is straight forwards, but I am stuck on the second one and not sure how to proceed. For the third one, I think that it has to be something to do with the fact that trigonometric polynomials are dense in $L^2(\mathbb{T})$.

Answer 1

Hints: By Cesaro convergence of Fourier series the closure of $V$ contains all functions of the form $\phi f$ where $\phi$ is continuous on $\mathbb T$. Let $g \in L^{2}$ and consider $\frac g f I_{|f|>\frac 1 n}$. You can approximate this function in $L^{2}$ by a sequence of continuous functions $(\psi_j)$. Use Holder's inequality to show that $f\psi_j$ tends $g I_{|f|>\frac 1 n}$ in $L^{2}$. Conclude that the closure of $V$ is the whole of $L^{2}$ which implies that $V^{\perp} =\{0\}$.