# If $\dim (\operatorname{Ker}f)^{\perp} = 1$ then $f$ is continuous for $f$ linear functional on Hilbert space

I need to prove the following result:

If $\dim (\operatorname{Ker}f)^{\perp} = 1$ then $f$ is continuous for $f$ linear functional on Hilbert space.

I have the following ingredients in mind:

• Use sequential characterization of continuity knowing that since $\operatorname{Ker}(f)$ is closed, projection theorem gives that $x_n = m_n + u_n \to x = m + u$ is equivalent to $m_n \to m \land u_n \to u$.

• Someone suggested me to further prove that $H = \operatorname{Ker}(f) \oplus \operatorname{Ker}(f)^{\perp} = \operatorname{Ker}(f) \oplus\operatorname{Img}(f)$ but I don't see how this is of any help.

Any ideas?

• Let $0\neq y\in(Ker(f))^{\bot}$. Then $f(y)\neq0$. If $x\in ((Ker(f))^{\bot})^{\bot}$, then $f(x)=0$. In fact, if $f(x)\neq0$ then $y-\frac{f(y)}{f(x)}x$ would be in the kernel and in particular in $((Ker(f))^{\bot})^{\bot}$ which is impossible, because it would imply that $y\in((Ker(f))^{\bot})^{\bot}$. Therefore $Ker(f)=((Ker(f))^{\bot})^{\bot}$, which is closed. Feb 8, 2018 at 0:01

## 3 Answers

Suppose $\dim(Ker(f)^{\perp})=1$, where $f$ is a linear functional. Then there exists a unit vector $x$ such that $x\perp Ker(f)$. In particular $f(x)\ne 0$. So, $$f\left(y-\frac{f(y)}{f(x)}x\right) = 0,$$ which forces the following for all $y$: $$\left\langle y-\frac{f(y)}{f(x)}x,x\right\rangle = 0 \\ \langle y,x\rangle = \frac{f(y)}{f(x)} \\ f(y) = \langle y,\overline{f(x)}x\rangle,\;\; y\in H.$$ So $f$ is a continuous linear functional because it has a Riesz representation.

Perhaps this is obvious, but it does not leap out at me that $\ker f$ is closed.

This is essentially @orole's argument in the comment above.

For example, if we take $A \subset l_2$ to the the subspace generated by elements of the form $(0,x_2,x_3,...,x_n,0,0,...)$, then $A^\bot = \operatorname{sp} \{ e_1\}$, but while $A \subset A^{\bot \bot}$, they are clearly not equal.

Suppose $\phi:\mathbb{H} \to \mathbb{R}^n$ is linear and $\dim (\ker \phi)^\bot = n$, then $\ker \phi$ is closed.

Let $y_1,...,y_n$ be a basis for $(\ker \phi)^\bot$. Then $\phi (y_1),...,\phi (y_n)$ are linearly independent and hence span ${\cal R} \phi$.

Let $z \in (\ker \phi)^{\bot \bot}$. Choose $\alpha_k$ such that $\phi(z) = \phi (\sum_k \alpha_k y_k)$, then $z-\sum_k \alpha_k y_k \in \ker \phi \subset (\ker \phi)^{\bot \bot}$. Since $z \in (\ker \phi)^{\bot \bot}$ this gives $\sum_k \alpha_k y_k \in (\ker \phi)^{\bot \bot}$ and hence $\sum_k \alpha_k y_k = 0$ and so $\phi(z) = 0$, or $z \in \ker \phi$. Hence $\ker \phi$ is closed (and so $\phi$ is continuous).

(Looking back, it would have sufficed to do this for $n=1$.)

Let $x$ non zero in the orhogonal of $Ker(f)$, the linear function $\langle x, .\rangle$ is continuous, this implies that its kernel $Ker(f)$ is closed and $f$ is continuous.

Closed kernel implies continuous linear functional : Zorn's Lemma

• Why is $\ker f$ closed? Feb 8, 2018 at 0:00
• because it is the kernel of $f(y)=\langle x,y\rangle$ Feb 8, 2018 at 0:01
• You haven't proved that. Feb 8, 2018 at 0:02
• That $f(y)=(x,y)$ ... which is even false for the arbitrary $x\in (Ker(f))^{\bot}$ that you chose. Feb 8, 2018 at 0:06
• It is not an argument, it is a request for clarification. Feb 8, 2018 at 0:14