Given $X$ Hilbert space, $T\in X^*$, $y$ projection of $x_0$ on $Y=\text{Ker}T$, why does $x-\frac{T(x)}{T(x_0-y)}(x_0-y)\in \text{Ker}T$?

Question

Let

• $X$ Hilbert space
• $T\in X^*$, ie $T:X\to\mathbb R$ linear
• $Y=\text{Ker}T=\{x\in X:T(x)=0\}$ closed subspace of $X$
• $x_0\in X\setminus Y$
• $y\in Y$ orthogonal projection of $x_0$ on $Y$, ie $\langle x_0-y,z \rangle=0$ for every $z\in Y$
• $w=x-\frac{T(x)}{T(x_0-y)}(x_0-y)$, with $x\in X$

Show that

1. $Y$ is closed
2. $w\in Y$ for every $x\in X$

(1) Hints?

(2) In the case $x\in Y$, then $T(x)=0$ and so $w=x-0=x\in Y$.

In the case $x\in X\setminus Y$, by linearity of $T$, it is $T(x_0-y)=T(x_0)-(y)=T(x_0)$, since $y\in Y$.

so $w=x-\frac{T(x)}{T(x_0)}(x_0-y)$.

We can write $X=Y\oplus Y^\perp$, so we can see $x\in X$ as sum of an element of $Y$ with an element of $Y^\perp$. Since $(x_0-y)\in Y^\perp$ and $\frac{T(x)}{T(x_0)}$ is just a scalar, I suppose that the product of them is still an element of $Y^\perp$.

So we can see $w$ as "an element of $Y$" $+$ "an element of $Y^\perp$" $-$ "an element of $Y^\perp$". I suppose that whatever $x$ will be, we can choose the second term in such a way that it will be equal to the third one, so that they will cancel out.

How to prove this in a rigorous way?

Answer 1

$T$ is continous and $Y$ is the inverse image of $\{0\}$ under $T$ so it is closed.

Second part follows from linearity of $T$: $Tw=Tx-\frac {Tx} {T(x_0-y)} {T(x_0-y)}=Tx-Tx=0$.