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A linear operator $T:X\rightarrow X$ is bounded below if for all $x\in X$, $||Tx||\geq c||x||$ for some $c>0$. So $T$ is not bounded below if there exists some $x\in X$, $||Tx||<c||x||$ for all $c>0$. This implies $\frac{||Tx||}{||x||}<c$ for all $c>0$. Does this imply $Tx=0$? If $T$ is injective, how can they are not bounded below?

glS
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Ken.Wong
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    You got it wrong. The negative is not that there exists $x\in X$ such that $||Tx||<c||x||$ for all $c>0$, but that for each $c>0$ there is some $x\in X$ (which depends on $c$) such that $||Tx||<c||x||$. – Mark Jul 02 '22 at 12:27
  • Every injective compact operator $T,:X \to X,$ where $X$ is an infinite dimensional Banach space, is not bounded below. Both answers give examples of that sort. – Ryszard Szwarc Jul 02 '22 at 15:30
  • related: https://math.stackexchange.com/q/2618055/173147 – glS Aug 27 '24 at 22:37

2 Answers2

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Define $T:C[0,1] \to C[0,1]$ by $Tf(x)=\int_0^{x} f(t)dt$. ($C[0,1]$ is given the usual sup norm). Then $T$ is a bounded injective operator but there is no positive number $c$ such that $\|Tf\| \geq c\|f\|$ for all $f$. To see this note that if $f_n(x)=x^{n}$ the $\|Tf_n|| \to 0$ but $\|f_n\|=1$ for each $n$.

Kavi Rama Murthy
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Take $T: l_1 \to l_1$ as $T(a_1,a_2,a_3,...) = (a_1/1,a_2/2,a_3/3,...)$ Clearly, if $Tx = 0$ then $x=0$. A the same time $||T(0,0,...1,0,0...)||= 1/n$

Salcio
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