0

Let $X$ be a normed space, and $f, g$ be linear functionals on $X$ such that they have the same kernels. I have to prove that there exists some scalar $c$ such that $f=cg$.

If $f$ and $g$ are zero functionals, there is nothing to prove. So I assume that f is non-zero, therefore there exists some $v_0$ such that f($v_0$)$\neq$0, and consequently g($v_0$) is also non-zero. I chose $c$ to be $\frac{f(v_0)}{g(v_0)}$, if I can prove that kernel of $f-cg$ is the whole of $X$, I'd be done. Can someone please give a hint for that?

K.defaoite
  • 12,536
Aryan Saxena
  • 33
  • I guess you forgot a hypothesis. It is true if $f$ and $g$ are linear forms. – Zag Mar 26 '23 at 11:50
  • I've written that f and g are linear functionals. – Aryan Saxena Mar 26 '23 at 11:51
  • I think a natural way to go from here is to note that your choice of $v_0$ gives you a direct product decomposition $X = \ker f \oplus \langle v_0 \rangle$. (ie, can you show that given an $x \in X$, you can write $x = x' + \lambda v_0$ for some $x' \in \ker f$ and some scalar $\lambda$?). – Izaak van Dongen Mar 26 '23 at 12:09

0 Answers0