What's the name of this condition so I can google it? Let $x$ be an element of a Banach space $X$, and let $L$ be the space of bounded linear transforms from $X$ to $X$ with $\ell \in \mathcal{L}(X,X)$.
What is it called when, for a non-zero element $x\in X$, $\ell x=0$ implies $\ell =0$?
 A: As I said in the comments, this is a property that applies precisely to non-zero points $x$ in one-dimensional Banach spaces $X$. I will assume you can see how to prove this property in one-dimension; I'll disprove it in more dimensions.
Suppose that $X$ is at least two dimensions (possibly infinite) and $x \in X \setminus \{0\}$. Then $\{x\}$ is a linearly independent subset of $X$, and since $X$ has more than one-dimension, it is not spanning. This makes $\operatorname{span}\{x\}$ a proper subspace of $X$, and since it is finite-dimensional, it is closed.
Choose any $y \in X \setminus \operatorname{span}\{x\}$. Using Hahn-Banach, there exists a functional $f \in X^*$, with $\|f\| = 1$, such that $\operatorname{span}\{x\} \subseteq \operatorname{ker} f$ and $f(y) = \|y\|$. Most of this information is not relevant; the important bit is that $f(x) = 0$ and $f(y) \neq 0$. Fix any $v \neq 0$ (indeed $v = x$ is allowed), and define:
$$l(w) = f(w)v$$
for all $w \in X$. Then $l$ is linear, by the linearity of $f$. We also have $l(x) = f(x)v = 0$, but $l(y) = f(y)v \neq 0$, as both $f(y) \neq 0$ and $v \neq 0$. So, $l \in L(X, X)$ maps $x$ to $0$, but is not the $0$ operator.
So, this property does not apply to any non-zero point in any Banach space, other than one-dimensional Banach spaces.
