Let $H = L^2(−1, 1)$ and $L \subset H$ be the set of all continuous functions such that $f(0) = 0$. Find the orthogonal projection $P : H → \bar{L}$.
My thoughts on this:
We say $g=P_{\bar{L}}(f)$ iff $f-g \perp \bar{L}$, for $g \in \bar{L}$ and $\forall f \in H$. So we need to find the set describing $g$. In particular, $f-g$ would need to be perpendicular to $g$, that is:
$\int_{-1}^{1} (f-g)\bar{g} \text{d}x = \int_{-1}^{1} f \bar{g} \text{d}x – \int_{-1}^{1} |g(x)|^2 \text{d}x$ $\forall f$ and some $g \in \bar{L}$.
But then I am not sure how to proceed further.
I would appreciate any guidance/hints.
Best Answer
Actually $\overline {L}=H$ so the projection is the identity map!
A well know result in measure theory shows that continuous functions are dense in $H$. Given any continuous function $g$ define $h(x)=xg(\epsilon)$ for $0 \leq x \leq \epsilon$ and $h(x)=g(x)$ for $x \geq \epsilon$. Use a similar construction for $x<0$. Note that $h \in L$. Then $\int |g(x)-h(x)|^{2} dx \to 0$ as $ \epsilon \to 0$. This proves that any $f \in h$ can approximated in the norm of $H$ by an element of $L$.