# equivalnce of linear functions, which one's kernel includes the other's

The following is from my homework. PLEASE don't reveal all the solution, but leave at least something for my imagination.

Let $X$ be a normed space. Let $\phi,\psi : X → \mathbb C$ be linear functionals such that $\ker \phi ⊂ \kerψ$. Show that there exists $λ \in \mathbb C$ such that $ψ = \lambda\phi$.

The reason that I'm asking for help is that this proposition don't even sound right to me. The result is VERY strong, while $\ker \phi \subset \ker\psi$ isn't (at least the way I see it).

What I've tried so far was assuming that $\not \exists \lambda \in C$, therefore there are $x_1,x_2\in X$ with $\psi(x_1)=\lambda_1\phi(x_1)$, $\psi(x_2)=\lambda_1\phi(x_2)$, and $\lambda_1 \neq \lambda_2$. I tried to find $z\in X$ that would be in $\ker\phi - \ker\psi$, using the linearity.

My main problem was that I don't feel I have enough "tools" - no strong theorems and nothing much to work with. This is the kind of help I would like the most - if you could give such lead.

Thanks so much!

• – Hagen von Eitzen Nov 16 '14 at 21:53
• @HagenvonEitzen: Thanks. two (significant) differences: 1. I'm possibly dealing with infinite dimentions so it doesn't quite work. I guess I could say that the kernel is of co-dimension 1? 2. I don't understand (in the question you provided) why "the possible values for dimkerT are dimV, dimV−1". could you assist with that? I guess that would make my understanding better (the case dimKerT=dimV is obvious if T=0, but what if T=!0? why it can't be higher co-dimension? – user188400 Nov 16 '14 at 21:57
• Yes, I am aware of that - but working with codimension instead of dimension you can still do the same. Take a basis of $\ker\phi$. To extend it to a basis of $X$ takes only one additinal vector (or none). – Hagen von Eitzen Nov 16 '14 at 22:00

## 1 Answer

The candidate $\lambda$ can be computed: suppose $\ker(\phi) \neq X$ (if so, both are $0$-maps, and we take $\lambda = 0$). Then for some $x_0 \in X$, $\phi(x_0) \neq 0$. Now both $\psi(x_0)$ and $\phi(x_0)$ are reals, the latter non-zero. If $\lambda$ exists, we can now compute it: we need $\psi(x_0) = \lambda \phi(x_0)$, so define $\lambda = \frac{\psi(x_0)}{\phi(x_0)}$.

Also, $\ker(\phi)$ is a maximal linear subspace, being the kernel of a functional, so either $\ker(\psi) = X$ (and we are done), or $\ker(\psi) = \ker(\phi)$. (Proof sketch: suppose that $f$ is a functional on $X$ and there exists a linear subspace $V \subset X$, with $\ker(f) \subset V \subset X$, all inclusions being proper. Then pick $x \in V \setminus \ker(f), y \in X \setminus V$, and note that $\{x,y\}$ spans a two-dimensional space on which $f$ is injective (as neither $x$ nor $y$ is in the kernel of $f$), which cannot be, as the image is $\mathbb{R}$ which is one-dimensional).

Now consider the functional $\psi - \lambda\phi$, which vanishes on this common kernel, and also on $x_0$, by construction, so vanishes on $X$ and we are done.

• mm I understood what you did here, but I can't understand where it goes. – user188400 Nov 16 '14 at 22:07
• Saw the edit. That's not what bothered me, I just couldn't show that this $\lambda$ actually works with what we want to prove – user188400 Nov 16 '14 at 22:19