# Riesz's Theorem - Null Space

I'm studying Riesz's Theorem using Kreyszig's Functional Analysis textbook.

I have problem with part of the proof. Wonder if someone could help explain it to me.

Every bounded linear functional $f$ on a Hilbert space $\textit{H}$ can be represented in terms of the inner product, namely, $f(x) = \langle x, z \rangle$

...

$\mathcal{N}(f)$ is a vector space and is closed. Furthermore, $f \neq 0$ implies $\mathcal{N}(f) \neq \textit{H}$ so that $\mathcal{N}(f) \neq \{0\}$ by Direct Sum Theorem ; Hence $\mathcal{N}(f)^{\perp}$ contains a $z_0 \neq 0$.

Question:

1) How does $f \neq 0$ implies $\mathcal{N}(f) \neq \textit{H}$?

2) Which in turn implies $\mathcal{N}(f) \neq \{0\}$ by Direct Sum Theorem?

Hope I provide enough information for my question.

Thank you!

• If $f \neq 0$, there must be a vector not sent to zero, and this vector is therefore not in the null space. – Chappers Jul 19 '17 at 23:28
• It should probably be $\mathcal{N}(f)^{\perp} \neq \{0\}$. – Chappers Jul 19 '17 at 23:29

1) If $f \neq 0$ there exists $x \in H$ such that $f(x) \neq 0,$ but $\mathcal{N}(f)=H$ means that $f(x)=0$ for all $x \in H.$
2) Since $\mathcal{N}(f)$ is closed in $H$ and $H$ is a Hilbert space, the Direct Sum Theorem states that $\mathcal{N}(f) \oplus \mathcal{N}(f)^{\perp}=H.$ If $\mathcal{N}(f) \neq H,$ then $\mathcal{N}(f)^{\perp}\neq 0.$