Prove that $L$ is the zero fucntion

Question

Let $L$ be a linear transformation $L:\mathbb{R^n} \to \mathbb{R^m}$. If $\lim_{\vec{x} \to {\vec{0}}} ||L(\vec {x})||/||\vec{x}||=0$ then $L$ is the zero constant

I'm having trouble proving this result: I know that given $\epsilon > 0$ there exists $\delta_{\epsilon}> 0$ such that $||L(\vec {x})||<\epsilon||\vec{x}||$

But I don´t know how to follow from this point. I would really appreciate any hint or suggestion.

Answer 1

Assume that there is $x \in \mathbb R^n$ such that $L(x) \ne 0$. Then we have that $x \ne 0$ and we can assume that $||x||=1$. Put $x_n = \frac{1}{n}x \quad (n \in \mathbb N).$ Then $x_n \to 0$, hence

$\frac{||L(x_n)||}{||x_n||} \to 0$.

But $\frac{||L(x_n)||}{||x_n||}=||L(x)|| \ne 0$ for all $n$, a contradiction.

Answer 2

I will drop the arrows for simplicity. Fix and $x \neq 0$ and replace $x$ in the hypothesis by $tx$ ($t$ real) conclude that $\frac {\|L(tx)\|} {\|tx\|} \to 0$ as $t \to 0$. By linearity of $L$ and homogeinity of the norm this says $\frac {\|L(x)\|} {\|x\|} \to 0$. But the left side does not depend on $t$ so this means $\frac {\|L(x)\|} {\|x\|} =0$ which gives $Lx=0$..