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how do I go about proving these: Lets F be a family of linear operator and the following are equivalent:

1) F is equicontinuous at some point $v_{0} \in V$

2) F is equicontinuous at all points of $V$.

3) F is equibounded whereby there exists $M>0$ such that $||T(v)|| \leq M||v||$ for all $v \in V$ and $T \in F$

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up vote 1 down vote accepted

1) implies 2):

Let $F$ be equicontinuous at $v_0$. Let $\varepsilon > 0$. By assumption we have that for $f \in F$ there is a $\delta_{v_0}$ such that $|v - v_0| < \delta_{v_0}$ implies $|f(v) - f(v_0)| < \varepsilon$. Now let $v_1$ be an arbitrary point.

We claim that if $|v - v_1| < \delta_{v_0}$ then $|f(v) - f(v_1)| < \varepsilon$.

Proof: Let $|v - v_1| < \delta_{v_0}$.

Then $|f(v) - f(v_1)| = |f(v - v_1)|$. Now translate $v_1$ to $v_0$ by subtracting $v_1 - v_0$:

$$\begin{align} |f(v) - f(v_1)| = |f(v - v_1)| & = |f(v - (v_1 - v_0) + (v_1 - v_0) - v_1)| = \\ & = |f(v_0 - (v_1 - v)) -f(v_0)| \end{align}$$

Now let $v\prime := v_0 - (v_1 -v)$. Then $|v^\prime - v_0| < \delta_{v_0}$ by assumption. Hence $|f(v) - f(v_1)| = |f(v^\prime) - f(v_0)|< \varepsilon $.

Hope this helps.

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Thanks Matt. Your explanation is very clear. Thank You. How about Part 3 on equibounded? Any thoughts? Surprisingly no books that I came across talk about equibounded. – Sandra Apr 2 '12 at 16:06
@sandra You're welcome. Glad I could help : ) I'm not sure what you mean by "any thoughts on 3", doesn't Xabier do all the remaining parts in his answer? – Rudy the Reindeer Apr 2 '12 at 16:57
Hi Matt, oh sorry, yes Xabier had answered my questions. He post is right after(2 mins) I post my question. So now the question is actually solve. – Sandra Apr 2 '12 at 17:06
Once again thanks for both of your help. ;p – Sandra Apr 2 '12 at 17:07

(3)$\Rightarrow$(2): Fix $v\in V.$ For every $v'\in V$ and $T\in F$ you have $$||T(v)-T(v')||=||T(v-v')||\le M||v-v'|| $$ which implies that $F$ is equicontinuous at $v.$

(1)$\Rightarrow$(2) is in Matt's answer.

(2)$\Rightarrow$(1) is trivial.

(2)$\Rightarrow$(3): Since $F$ is equicontinuous at zero, there exists $\delta>0$ such that $||v||\le \delta, T\in F\Rightarrow ||Tv||\le 1.$ For any $x\not=0$ and $T\in F,$ $||Tx||=\frac{||x||}{\delta} ||T(\frac{\delta}{||x||}x)||\le \frac{1}{\delta}||x||.$

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Oh thanks Xabier. I never thought of $(3) \Rightarrow (2)$. You had opened up my mind and thoughts. Thank you. – Sandra Apr 2 '12 at 16:21

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