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!
 A: 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.
