I don't follow all the steps in a theorem regarding continuous functionals

Question

Hi: I'm reading another document on Hilbert spaces and progressing some but I've come to a theorem in this document, many of the steps of which I don't follow.

I will state the theorem below and then explain which steps are not clear to me.

Theorem 5.1. Let $H$ be a hilbert space and $\phi: H \rightarrow \mathbb{C}$, be a continuous functional. Then,

$\dim N(\phi)^\perp = 1$.

where $N(\phi) = \{h \in H| \phi(h) = 0\}$.

Proof: It is easy to see that $N(\phi)$ is a closed subspace of $H$.

Since, $\phi \neq 0, N(\phi) \neq H$.

Then, $N(\phi) \neq \{0\}$. ( recall that $H = N(\phi) \oplus N(\phi)^\perp$).

Take $x_1 \neq 0, x_2 \neq 0$ in $N(\phi)^{\perp}$.

Then, there exists a complex number $a \neq 0$ such that $\phi(x_1) + a \phi(x_{2}) = 0$.

Then, $\phi(x_1 + a x_{2} ) = 0$ which implies that $x_1 + a x_{2} \in N(\phi) \oplus N(\phi).^{\perp} = \{0\}$.

Thus, $\dim N(\phi)^\perp = 1$.

END OF THEOREM.

There are many steps in above that I don't follow.

1) Then, $N(\phi) \neq \{0\}$. I don't see where that comes from and does that indicate the null set or zero ?

2) $\phi(x_1) + a \phi(x_{2}) = 0$. Is this because the linear combo of 2 objects in a hilbert space is also in that space ?

3) $\phi(x_1 + a x_{2} ) = 0$. This is clearly crucial to the proof. I don't get it.

4) implies that $x_1 + a x_{2} \in N(\phi) \oplus N(\phi).^{\perp} = \{0\}$. Why does that imply it and again, is that the null set or zero ?

5) $\dim N(\phi)^\perp = 1$. Don't follow this either ?

Thanks a lot.

Answer 1

(1) It seems from context that there is a typo and they meant to write $N(\phi)^\perp \ne \{0\}$. That is, $N(\phi)^\perp$ is not equal to the set containing only the zero vector. Since every linear subspace contains the zero vector, this means $N(\phi)^\perp$ must contain some elements other than the zero vector, which justifies the following line.

(2) Note that $\phi(x_1)$ and $\phi(x_2)$ are just (nonzero) complex numbers. Using simple algebra, you can very easily find what $a$ should be in order to make $\phi(x_1) + a \phi(x_2) = 0$.

(3) Recall that $\phi$ is a linear functional.

(4) I think they meant to write $x_1 + a x_2 \in N(\phi) \cap N(\phi)^\perp$. We have just shown that $\phi(x_1 + a x_2) = 0$, which is the very definition of $x_1 + a x_2 \in N(\phi)$. And we started out assuming that $x_1, x_2 \in N(\phi)^\perp$, and $N(\phi)^\perp$ is a linear subspace... Finally, recall that a set and its orthogonal complement always have only the zero vector in common. (Any vector common to both must be orthogonal to itself.)

(5) We have just shown that $x_1 + a x_2 = 0$ for arbitrary $x_1, x_2$, which is to say that any two vectors in $N(\phi)^\perp$ are linearly dependent. If the dimension of $N(\phi)^\perp$ was at least 2, we would be able to find two linearly independent vectors, and we've just shown that can't be done.

Answer 2

The other answer covers it, but here is another approach, too long for a comment:

$H=N(\phi)\oplus N(\phi)^\perp$ (because $H$ is a Hilbert space) and $H/N(\phi)\cong \mathbb C$ (By first isomorphism theorem), so the dimension of the quotient is $1$.

There is an $h_1\in H$ such that $h_1+N(\phi)$ generates $H/N(\phi)$. Now, $h_1=h+k$ for some $ h\in N(\phi),\ k\in N(\phi)^{\perp},$ so $h_1+N(\phi)=k+N(\phi).$

Therefore, if $x\in N(\phi)^{\perp},$ then $[x]=a(k+N(\phi))$ for some $a\in \mathbb C$, from which it follows that $x-ak\in N(\phi)$, but since each of these terms is contained in $N(\phi)^{\perp},$ we must have $x-ak=0$, and so $x=ak$. The result follows.