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let $E, F$ be two Banach spaces and $T : E \to F$ be a continuous linear operator.

show that :

$$T \text{ is injective and has closed range } \iff \exists c > 0, \, \forall x \in E, \, \, \|T(x)\| \geq c\|x\|$$

my Attempt :

"$\implies$"

case 1 : $x = 0$ is trivial

case 2 : $x \neq 0$

assume that $\forall c > 0, \, \exists z \neq 0 \in E, \frac{\| T(z)\|}{\|z\|} < c$

this means that $\|T(z)\| = 0 \iff T(z) = 0 \iff z \in \ker T \iff z =0$, since $T$ is injective.

$z$ can't be zero so it's a contradiction.

"$\impliedby$"

let $x, y \in E$ such that $T(x) = T(y)$

then :

there exists $c > 0$ such that : $T(x -y ) = \|T(x-y)\|0 \geq c \| x - y \|$

therefore $x = y$ so $T$ is injective.

let $(y_n)_n \subset T(E)$, for one we know that $y_n \to y \in F$ and that there exists $(x_n)_n \subset E$ such that $Tx_n = y_n$ and $x_n \to x \in E$

both the norm and $T$ are continuous :

$$ 0\leq \|y_n - Tx\|=\|T(x_n - x)\| \leq \|T\| \| x_n - x \|$$

by the squeeze theorem then $y_n \to y = Tx \in T(E)$ so $T(E)$ is closed

could someone please check if my work is correct, I have some doubts because I haven't used all assumptions.

the_firehawk
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    There is some leap above at the line beginning "assume...". How do you suddenly conclude that $|Tz|= 0$? – copper.hat Jan 26 '20 at 22:04
  • @copper.hat well generally speaking if a nonnegative number is strictly less then any possible positive constant then it is equal to 0,

    it can be shown by contraposition, it's equivalent to saying that every nonzero number is either negative or positive

     – the_firehawk Jan 26 '20 at 22:09
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    I am familiar with such reasoning, but there is a big difference between $\exists z $ $\forall c$ and $\forall c$ $\exists z$. – copper.hat Jan 26 '20 at 22:11
  • @copper.hat you're right, in this case $z$ depends on $c$, thanks for noticing, I'll review my work. – the_firehawk Jan 26 '20 at 22:13
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    One standard proof of $\implies$ uses the open mapping theorem on the range of $T$. – copper.hat Jan 26 '20 at 22:15

1 Answers1

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In case 2 of "$\implies$" you have mis-interpreted $$\forall c>0\,\exists z \ne 0\,\in E\,\left(\frac {\|T(z)\|}{\|z\|}<c\right)$$ as $$\exists z\ne 0\,\in E\,\,\forall c>0\,\left(\frac {\|T(z)\|}{\|z\|}<c\right)$$ which renders the argument invalid. By way of comparison let $f(z)=z$ for all $z\in \Bbb R.$ Then $$\forall c>0\,\exists z\ne 0\,(|f(z)|<c)$$ but it is NOT true that $$\exists z\ne 0\,\forall c>0\,(|f(z)|<c).$$ You MUST use the "Banach" (completeness) property of $E$ and $F$. For example let $E$ be the set of every real sequence $(r_n)_{n\in \Bbb N}=(r_n)_n$ such that $\{n\in \Bbb N:r_n\ne 0\}$ is finite. With norm $\|(r_n)_n\|=\max \{|r_n|: n\in \Bbb N\}.$ Let $F=E$ and let $T((r_n)_n)=(r_n/n)_n.$ Then $T:E\to F$ is a continuous linear bijection but $\inf \{\|T(e)\|: \|e\|=1\}=0.$ However this normed linear space $E$ is NOT a Banach space.

Addendum. Another example, to show that the closed range of $T$ must be used: Let $E=F=l_{\infty}.$ Let $T((x_n)_n)=(x_n/n)_n.$ Then $(1/\sqrt n\,)_n\in \overline {T[E]}\setminus T[E]$ and $\inf \{\|T(e)\|:\|e\|=1\}=0.$

DanielWainfleet
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  • I d.k. how to prove "$\implies$". But I have seen the assertion that if $E, G$ are Banach spaces and $T:E\to G$ is a continuous linear bijection then $T^{-1}$ is continuous. Proving that would prove your "$\implies$" with $G=T[E]={T(e):e\in E}.$ – DanielWainfleet Jan 27 '20 at 09:16
  • I added an afterthought (Addendum) after Acceptance. – DanielWainfleet Jan 29 '20 at 01:06