Can you state the existence of the orthogonal projection on $W$?

Question

Let $V$ be the $ℝ$-space of the continuous functions of $[0,1]$ in $ℝ$, with the product $\langle f, g\rangle = ∫_0^1 f (t) g (t)\, dt$. Let $W$ be the set of constant functions.

Can you state the existence of the orthogonal projection on $W$? If so, describe it, if not, justify it.

Attempt: By contradiction suppose that there exists the projection $E$ of $V$ in $W$. Then as $\dim W< \infty$, it follows that $V = W \bigoplus W^ \perp$. But

$W^ \perp = \{ f \in V; \langle f, c\rangle =0 \,\,\ \forall c \in W \}$.

that is

$0=\langle f, c\rangle = ∫_0^1 f (t) c \,dt \,\,\ \forall c \in W $

that is $0=∫_0^1 f (t) \,dt \,\,\ \forall c \in W $.

But if we take $f (x) = x$, then $f\in V$, $f\notin W$, then it must belong to $W^ \perp$. But $1=∫_0^1 t dt=∫_0^1 f(t) dt$. contradicting $V = W \bigoplus W^ \perp$.

Is this right?? I find it strange because the later exercise says so:

Show a formula for $p_{W^⊥}(f)$ where $p_{W^⊥}$ is the orthogonal projection on $W^⊥$.

And to show this, do I need the above projection of this exercise to be correct?

Answer 1

The problem is to minimise $\|f-c\|$ over constants $c$. This is equivalent to minimising $\|f-c\|^2$ over constants $c$ and is much more tractable.

Expanding gives $\|f-c\|^2 = \|f\|^2 + c^2 - 2 c\langle 1, f \rangle$, and the minimising $c$ is given by $c= \langle 1, f \rangle$.

Hence the projection is given by $Pf = \langle 1, f \rangle$.

It is straighforward to check that $P$ is orthogonal, $\langle c, f - Pf \rangle = c \langle1, f \rangle -c \langle1, f \rangle = 0$.

Answer 2

There is an error, $f(x)=x\in V$ so you have that $f(x)=g(x)+c$ for some function $g\in W^\perp$ and constant function $c$.

The orthogonal space will be

$W^\perp=\{f\in V: \int_0^1 f(x)dx=0\}$

Infact every continuos function $f\in V$ can be written as

$f(x)=(f(x)-\int_0^1f(t)dt)+\int_0^1 f(t)dt$

The orthogonal projection on $W^\perp$ will be simply

$p_{W^\perp}(f)=f(x)-\int_0^1f(t)dt$