If $S \in L(X,Y)$ and lim$_{r \to 0}\frac{\|Sr\|}{\|r\|}=0$,then $S=0$.

Question

Here is a lemma whose proof is as under:

If $S \in L(X,Y)$ and lim$_{r \to 0}\frac{\|Sr\|}{\|r\|}=0$,then $S=0$.

Proof:

The condition lim$_{r \to 0}\Big(\frac{\|Sr\|}{\|r\|}\Big)=0$means that for each $\epsilon \gt 0$ there is a $\delta \gt 0$ such that $\Big(\frac{\|Sr\|}{\|r\|}\Big)\leq \epsilon $ whenever $0\lt \|r\| \lt \delta$

Let $\text{u}\in X$ be a non-zero vector.Choose a non-zero $t\in \mathbb R$ so that $\|t\text{u}\|\lt \delta $ .Then $\Big(\frac{\|S(t\text{u})\|}{\|t\text{u}\|}\Big)= \Big(\frac{\|S\text{u}\|}{\|\text{u}\|}\Big)\leq \epsilon $ and therefore $\|S\text{u}\|\leq\epsilon \|\text{u}\|.$

This is true for any $\epsilon \gt 0.$Hence $S\text{u}=0$ for all $u \in X$ This means that $S=0$

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I can't understand the step why did we take a vector $\text{u}\in X$ and then introduce $t$ in proof? Please help....

If we can prove for any non-zeros vector $u \in X$, $Su = 0$, then we can conclude $S =0$ — Petite Etincelle, Sep 25 '14 at 06:09
Because Su=0 is obvious if u=0 and because ||Su||/||u|| does not exist if u=0. — Did, Sep 25 '14 at 06:10
@Did why did we take a non-zero t∈R so that $∥tu∥$<$\delta $.Won't $u$ alone would be enough ..... — coool, Sep 25 '14 at 06:34
Same as above: Because S(tu)=0 is obvious if t=0 and because ||S(tu)||/||tu|| does not exist if t=0. — Did, Sep 25 '14 at 06:35

Did · Accepted Answer · 2014-09-25T07:24:42.537

The setting of the proof is that one fixes some vector $u$ and one wants to show that $Su=0$. If $u=0$, this is easy but, if $u\ne0$, how to do that? The only information one has about the function $S$ is its behaviour near $0$. Hence the idea to consider $Sw$ for some vectors $w$ such that (i) $w$ is close to $0$, and (ii) $Sw$ is somehow related to $Su$. These two constraints are met simultaneously if one chooses $w=tu$ when the real number $t$ is close to $0$. Ergo.

score 1 · Answer 2 · answered Sep 25 '14 at 06:56

We know that $\|Su\|\leq \epsilon\|u\|$ for all $u$ with $\|u\|<\delta$. Now we want to show that this extends to any $u\in X$. So pick any $u$. For this $u$, we can conclude that $$ \|S \delta u/(2\|u\|)\|\leq \epsilon \|\delta u/(2\|u\|)\|, $$ because $\|\delta u/(2\|u\|)\|=\delta/2<\delta$. Now we use linearity to conclude $$ \|S u\|\leq \epsilon \|u\|, $$ which was what we wanted. The variable $t$ just plays the role of $\delta/2\|u\|$, because it is easier to write that way.

If $S \in L(X,Y)$ and lim$_{r \to 0}\frac{\|Sr\|}{\|r\|}=0$,then $S=0$.

2 Answers2