open ball definition and bounded

Question

Let $A: V \rightarrow W$ be a linear map between normed space. Then the following conditions are equivalent:

$\left ( a \right ):A$ is continuous at$ \vec{0}$

$\left ( b \right ) A$ is continuous

$\left ( c \right )A$ is bounded

Proof:

$\left ( a \right )IFF\left ( c \right ):$ Assuming A is continuous at $\vec{0}$,then using the open ball definition of continuity, for all $\epsilon >0$ there exists $\delta >0$ such that $v \in B_{\delta}\left ( \vec{0} \right )\Rightarrow Av \in B_{\epsilon}\left ( \vec{0} \right )$.

I don't understand how using the open ball definition for continuity leads to the above implication.

Can someone shed some light?

Answer 1

Hint: Walk the proof $(b)\rightarrow(a)\rightarrow(c)\rightarrow(b)$

$(b)\rightarrow(a)$ is clear

$(a)\rightarrow(c)$:From continuity follows, that $\delta > 0$ exists with $A B_V(0,\delta) \subseteq B_W (0,1)$, therefore $||Ax||_W \leq 1$ for every $x \in B_V(0,\delta)$. Then by homogenity $||A||_W \leq \frac{1}{\delta} ||x||_V$ for every $x \in V$.

$(c)\rightarrow(b)$: Let $x \in X$ and you know $||Ax-Ay||_W\leq M||x-y||_V$ ; so from $||x-y||_V\leq \frac{\epsilon}{M}$ follows $||Ax-Ay||_W\leq \epsilon$ and A is uniformly continuous and therefore continous.

q.e.d.

Answer 2

Hint: Given continuity at $\vec{0}$, to prove continuity at any other point $\vec{v}$, consider a sequence $\{\vec{v_{n}}\}_{n\geq 1}$ such that $\vec{v_{n}}\rightarrow \vec{v}$. What can you say about the sequence $\{\vec{v}-\vec{v_{n}}\}_{n\geq 1}$?