# Continuous at some and at all points

Can I have someone to show me an insight on how to prove these? I had referred to a number of books but most authors merely state them as definitions or theorems without proof.

Let $T:X \to Y$ be a linear operator. Then the following are equivalent:

1. $T$ is continuous at some point $v_{0} \in V$

2. $T$ is continuous at all points of $V$

3. $T$ is bounded

4. $T$ is uniformly continuous.

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Usefull intuition is that the linearity allows us to transport properties at origo to other points and vice versa (injectivity, continuity etc.). – KPK Apr 5 '12 at 21:46

Those results are standard and appear in (almost) any functional analysis textbooks. The equivalence between (1), (2) and (3) appears in the text "Analysis Now" by Petersen (full proof with all the details).

$\bf Edit:$ $(1)\rightarrow (2)$ Suppose $T$ is continuous at $v_0$ and let $x\in X$ be given. Let $\epsilon>0$ be given. Find $\delta>0$ so that $\| T(v_0)-T(v)\|<\epsilon$ whenever $\| v_0-v\|<\delta$. Note that if $\|x-y\|<\delta$ then $\|x-y\|=\| v_0-(x-y+v_0)\|<\delta$. It follows from linearity that $T(x)-T(y)=T(v_0)-T(x-y+v_0)$, hence $\|T(x)-T(y)\|<\epsilon$. Thus $T$ is continuous at $x$.

$(2)\rightarrow (3)$: If $T$ is continuous at zero then there exists $\delta>0$ so that $\|T(x)\|\leq 1$ for all $\|x\|\leq \delta$. This implies that, for all $x\ne 0$, $\|T(\frac{\delta\cdot x}{\| x\|})\|\leq 1$, hence $\| T(x)\|\leq \frac{\| x\|}{\delta}$. Therefore $T$ is bounded.

$(3)\rightarrow (4)$: Follows easily from Beni's observation and $(4)\rightarrow (1)$ is obvious.

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Thanks Azarel. Guess you can't find (4) because I added in myself. Do you know how to prove (4)? I am thinking of the relation (1) \rightarrow (4) \rightarrow (2) \rightarrow (3) \rightarrow (1) but still thinking on how to get them. – Sandra Apr 5 '12 at 4:29
@sandra (4) comes from $\|Tx-Ty\|\leq \|T\| \|x-y\|$ – Beni Bogosel Apr 5 '12 at 6:20
Dear Beni. Ok Thanks. Sadly Peterson's book is not in our library. Have to look for others book now. – Sandra Apr 5 '12 at 16:43
@sandra I'll add a full proof then. – azarel Apr 5 '12 at 16:52
Thank You Azarel. – Sandra Apr 5 '12 at 23:59