Finding continuous projection from $L^2([0,2\pi),\mu)$ to $M:=\overline{\mathrm{span}\{e^{int}: n=0, 1, 2,\ldots\}}$

Question

Task:

Consider $L^2([0,2\pi),\mu)$ with $d\mu=\frac{1}{2\pi}dx$.

Let $$e_n(t):=e^{int},\ n \in \mathbb Z$$

$$M:=\overline{\mathrm{span}\{e_n: n=0, 1, 2,\ldots\}}$$

Find a continuous projection from $L^2([0,2\pi),\mu)$ to $M$

Maybe this helps:

I don't know how to tackle this problem. Perhaps this helps:

Let $$u_n(t):=\frac{e_{-n}(t)+ne_n(t)}{\sqrt{1+n^2}},\ n \in \mathbb N$$ and

$$N:=\overline{\mathrm{span}\{u_n: n=1, 2,\ldots\}}$$

Show that the projection $P$ from the normed space $X:=M\stackrel.+N$ with $ran\ M$ and $ker\ N$ is not continuous. Hint: consider $Pe_{-n}$.

Because P is linear we are finished if we show that $\|P\|=\infty$

We have $e_{-n}=u_n\sqrt{1+n^2}-ne_n,\ \|e_-{n}\|=\sqrt{(e_{-n},e_{-n})}=1$

Therefore $\frac{\|Pe_{-n}\|}{\|e_{-n}\|}=\frac{\|-ne_n\|}{\|e_{-n}\|}=n$

$\Rightarrow \|P\|=\infty$

Answer 1

$\{ e_n(t) = \frac{1}{\sqrt{2\pi}}e^{int} \}_{n=0}^{\infty}$ is an orthonormal subset of $L^2[0,2\pi]$. So the projection of $f\in L^2[0,2\pi]$ onto the closed linear span of $\{ e_n \}_{n=0}^{\infty}$ is $$ Pf = \sum_{n=0}^{\infty}\langle f,e_n\rangle e_n $$

Answer 2

The $e_n$ form an orthonormal basis to $L_2$, so every $f\in L_2$ is of the form $$f=\sum_{n\in\mathbb{Z}}a_nf_n$$ for some $a_n$.

You need to answer two questions:

1. What is the obvious projection onto the required span?

2. Why is the obvious projection a bounded linear operator (hence continuous)?