$\bigcap_{S \in L(E,F)} ker(S) = \{0\}$

Question

Let $E$ a Banach space, $F$ a normed space and $L(E,F)$ a set of bounded linear operator from E to $F$. Is true that $$\bigcap_{S \in L(E,F)} ker(S) = \bigcap_{S \in L(E,F)} S^{-1}(0) = \{0\}.$$ If not, is there a simple counter-example?

Answer 1

The answer is no if $F$ is trivial.

Now, if $F$ is non-trivial, I think that the answer is yes and can be proven the following way:

Pick a one dimensional subspace $\mathbb C y \leq F$.

For $x \in E, x \neq 1$ by Hanh -Banach you can extend $$\phi : \mathbb C x \to \mathbb C \,;\, \phi(kx)=k$$ to a linear functional $\phi: E \to \mathbb C$.

Then $$T(z)=\phi(z) y$$ is a linear operator from $E$ to $F$ and $x \notin \ker(T)$.

P.S. The proof basically boils down to:

• By HB the statement is true if $F$ is one dimensional.
• If $F$ is any non-trivial space it has a one dimensional subspace and we can use 1.

Answer 2

I assume that $F\neq \{0\}$.

Pick $x\in E$. By the Hahn-Banach theorem, there is $e^*\in E^*$ be such that $\langle e^*, x\rangle =\|x\|$. Choose non-zero $y\in F$ and define $S_xz = \langle e^*, z\rangle y$. In particular, $x\notin \ker S_x$. Then 0 is the only element of $E$ which belongs to kernels of all possible operators from $E$ to $F$.