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Let $X$ and $Y$ be two Hilbert spaces and $A\in\mathcal{L}(X,Y)$. Suppose that there's $\beta > 0$ such that $$\inf_{z\ \in\ \text{Ker}(A)}\|x-z\|\ \leq\ \beta\|A(x)\|,\quad \forall\ x\in X.$$ Show that, $\text{R}(A) = \text{Im}(A)$ is closed.


Please, somebody can help with this problem? I tried to prove that $\text{R}(A) = \text{Ker}(A^*)^{\perp}$ in order to show that $\text{R}(A)$ is closed, but I really don't know how I can use the fact of $\text{dist}(x, \text{Ker}(A)) \leq \beta\|A(x)\|$. Please, I need some hints.

Thanks in advance.

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  • $\begingroup$ What about factoring through $X/\ker A$? $\endgroup$ – Berci May 20 '13 at 1:59
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Consider well defined linear operator $$ \hat{A}: X/\operatorname{Ker}(A)\to Y:\hat{x}\mapsto A(x) $$ It is bounded and have the same image as $A$. It is enough to prove that $\hat{A}$ have closed image.

  1. Since $X$ is complete and $\operatorname{Ker}(A)$ is closed then $X/\operatorname{Ker}(A)$ is complete. Note that $Y$ is also complete.

  2. Inequality in your question preciesly means that $\hat{A}$ is bounded below. Just recall the definition of the quotient norm.

  3. Any bounded below operator between Banach spaces have closed range.

From $1$, $2$, and $3$ it follows that $\hat{A}$ have closed range, and so does $A$. Note that this proof valid for any Banach spaces. Hilbertability is redundant here.

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