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Could you please explain why the operator $T:X \to Y$ (from a Banach space $X$ to a Banach space $Y$) is injective if there is a constant $c>0$ such that for all $x \in X$: $$ c \|x\| \leq \|Tx\| $$

Thank you for your help.

glS
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Konstantin
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1 Answers1

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If $Tx=0$, then $\|Tx\|=0$, so $c\|x\|\leq 0$, so $\|x\|\leq 0$ since $c>0$, so $\|x\|=0$, and hence $x=0$, so $\ker T=\{0\}$.

user284331
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  • So $Tx=0 \implies x=0 \implies \exists T^{-1} \implies T ; \text{is injective}$. Thank you very much for the great reply! – Konstantin Jan 23 '18 at 19:29