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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!

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  • $\begingroup$ Cf. math.stackexchange.com/questions/1024930/… $\endgroup$ – Hagen von Eitzen Nov 16 '14 at 21:53
  • $\begingroup$ @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? $\endgroup$ – user188400 Nov 16 '14 at 21:57
  • $\begingroup$ 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). $\endgroup$ – Hagen von Eitzen Nov 16 '14 at 22:00
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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.

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  • $\begingroup$ mm I understood what you did here, but I can't understand where it goes. $\endgroup$ – user188400 Nov 16 '14 at 22:07
  • $\begingroup$ 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 $\endgroup$ – user188400 Nov 16 '14 at 22:19

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