Statement: Given a Hilbert space $\mathscr{H}$ and $\mathscr{K}$ and a bounded operator $A \in \mathscr{B}(\mathscr{H}, \mathscr{K})$. Show that $A$ is invertible if and only if $A$ is bounded below and has dense range:

Attempted proof:

$\Rightarrow$ Since we are invertible then the range of $A$ will be equal to the $\mathscr{K}$. Furthermore, if $A$ is invertitble then $A^{-1}A = \mathbb{I}$ hence $\forall h \in \mathscr{H}\, , \|h\|=\|A^{-1}Ah\| \leq \|A^{-1}\|\|A(h)\| \rightarrow \frac{\|h\|}{\|A^{-1}\|}\leq \|A(h)\|$ so we are bounded below.

$\Leftarrow$ (this is the dirrection I get stuck in) We now assume that $A$ is bounded below and has dense range then the opeartor is invertible. Since we are bounded below then we are injective. To see this we assume by contradiction, then $\exists h\in \mathscr{H}, \, s.t. \, h \neq 0 \rightarrow A(h) = 0$, and since we are bounded below then $ \exists \delta >0 \, s.t \, \delta\|h\| \leq \|A(h)\| = 0$ so we arrive at a contradiction since the L.H.S > 0 but the R.H.S = 0.

now all that remains to be shown is surjectivity. Let $\{h_n\}_{1}^{\infty}$ be a Cauchy sequence, since we are bounded from below then $\delta\|h_n-h_m\| \leq \|A(h_n-h_m)\| = \|A(h_n)-A(h_m)\|$

which implies that $\|h_n-h_m\| \leq \frac{1}{\delta}\|A(h_n)-A(h_m)\|=\frac{1}{\delta}\|A(h_n-h_m)\|\leq\frac{\|A\|}{\delta}\|h_n-h_m\|< \varepsilon$ since the last term is Cauchy we can let $\|h_n-h_m\| <\frac{\delta^2}{\|A\|} = \varepsilon$

So we know that if we are Cauchy is $\mathscr{H}$ then we are Cauchy is $\mathscr{K}$.

And I don't know how to proceed from this point.

  • 2
    $\begingroup$ You need it the other way round, if you're Cauchy in $\mathscr{K}$, then you're Cauchy in $\mathscr{H}$. More direct: bounded below means it's a homeomorphism to its image. $\endgroup$ Commented Apr 18, 2014 at 20:06
  • $\begingroup$ If we are Cauchy in $\mathscr{K}$ then we arrive at $\delta\|h_n-h_m\| \leq \|A(h_n) - A(h_m)\| < \varepsilon$. Would we use the denseness of $A$ to show that we converge to an element in $\mathscr{K}$ implying closedness? $\endgroup$ Commented Apr 18, 2014 at 20:35
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    $\begingroup$ Since $\mathscr{K}$ is Hilbert (Banach is plenty sufficient, by the way), if $(A(h_n))$ is Cauchy, it converges to something. Since $(h_n)$ is also Cauchy, and $\mathscr{H}$ also complete, that too converges to something. What follows? $\endgroup$ Commented Apr 18, 2014 at 20:38
  • $\begingroup$ I think I got it. If we are Cauchy in $\mathscr{K}$ then we are Cauchy in $\mathscr{H}$ (by the previous comment). Since we are in a Hilbert space then we are complete. Thus if $\lim_{n\rightarrow \infty} h_n = h$ then we have $\lim_{n\rightarrow \infty} A(h_n) = A(h)\in \mathscr{K}$ so that $\mathscr{K}$ is closed. Now this would imply that $R(A) = \mathscr{K}$ hence surjective (and injective by the first half in the proof above, going in $\Leftarrow$ direction). So we are bijective, therefore invertible. $\endgroup$ Commented Apr 18, 2014 at 21:15
  • 1
    $\begingroup$ Where you wrote "$\mathscr{K}$ is closed", that should have been $R(A)$. Yes, bounded below implies the range is complete, hence closed. Since a closed dense subspace hasn't much choice what to be, bijective. $\endgroup$ Commented Apr 18, 2014 at 21:19

2 Answers 2


A solution was worked out in comments, so I'll add a more general version of the difficult direction, to show that linearity isn't necessary.

Suppose $X$ and $Y$ are complete metric spaces, $f:X\to Y$ is a continuous map with dense range, and there is $\delta>0$ such that $d_Y(f(a),f(b))\ge \delta d_X(a,b)$ for all $a,b\in X$. Then $f$ is a homeomorphism between $X$ and $Y$.

Proof. It suffices to show that $f(X)$ is closed, because being closed and dense, it must coincide with $Y$. Take a sequence $(y_n)$ in $f(X)$ that converges to $y\in Y$. Write $y_n=f(x_n)$. Since $d_X(x_n,x_m)\le \delta^{-1} d_Y(y_n,y_m)$, the sequence $(x_n)$ is Cauchy in $X$. By completeness of $X$, there is a limit $x=\lim x_n$. Since $f$ is continuous, $y=f(x)\in f(X)$. $\quad\Box$


Here's a direct proof of surjectivity that doesn't assume completeness of the co-domain.

Let $T: X \to Y$ be a bounded linear operator between normed vector spaces, with $X$ complete. Suppose that $T(X)$ is dense in $Y$ and that there exists $C \in (0,\infty)$ such that $\|Tx\| \geq C\|x\|$ for all $x \in X$. Let $y \in Y$. Since $T(X)$ is dense in $Y$ there exists a sequence $(y_n)_{n=1}^{\infty}$ in $T(X)$ such that $\lim y_n = y$. Let $(x_n)_{n=1}^{\infty}$ be a sequence in $X$ such that $Tx_n = y_n$. Now note that $$ \|x_n - x_m\| \leq \frac{1}{C} \|T(x_n - x_m)\| = \frac{1}{C} \|Tx_n - Tx_m\| $$ for all $n,m \in \mathbb{N}$. The sequence $(Tx_n)_{n=1}^{\infty}$ is convergent and hence Cauchy, and so $\|Tx_n - Tx_m\| \to 0$ as $n,m \to \infty$. Therefore, $\|x_n - x_m\| \to 0$ as $n,m \to \infty$, i.e., $(x_n)_{n=1}^{\infty}$ is Cauchy. Since $X$ is complete there exists some $x \in X$ such that $x_n \to x$. Then since $T$ is continuous (because it's bounded) it follows that \begin{align*} Tx = \lim_{n \to \infty} Tx_n = \lim_{n \to \infty} y_n = y. \end{align*}

Hence, for each $y \in Y$ there is some $x \in X$ such that $Tx = y. \qquad \square$


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