Increasing sequence of closed subspaces of $L^2$

Question

Not sure how to start on this question. Writing out a decomposition didnt seem to lead to anything. I think you might have to guess the answer the question first before starting a proof.

Please just hints!

Answer 1

Hints Denote by $P_n$ the projection on $V_n$, i.e. $f_n = P_n(f)$, and by $\langle f,g \rangle = \int f \cdot g \, d\mu$ the canonical scalar product in $L^2(\mu)$.

1. Use that $P_{k-1}^2 = \text{id}$ and $\langle P_{k-1} f,g \rangle = \langle f,P_{k-1} g \rangle$ to deduce that $$\langle f_n-f_{n-1},f_k-f_{k-1} \rangle = 0$$ for any $k < n$.
2. Write $$f_n = f_1 + \sum_{k=2}^n (f_k-f_{k-1}). \tag{1}$$ Conclude from the first step and the fact that $\|f_n\|_{L^2} \leq \|f\|_{L^2}$ that $$\sum_{k=2}^{\infty} \|f_k-f_{k-1}\|_{L^2}^2 < \infty.$$
3. Write $$f_n-f_m = \sum_{k=m+1}^n (f_k-f_{k-1})$$ in order to deduce from step 2 that $(f_n)_n$ is a Cauchy sequence, hence convergent to some element $g \in L^2$.
4. By the definition of the orthogonal projection, $$\|f-P_V(f)\|_{L^2} \leq \|f-g\|_{L^2}$$ for $V:=\bigcup_n V_n$. Conclude that $g=P_V(f)$.