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I'm struggling with a problem from Young's Introduction to Hilbert Space (7.30 to be more specific)

Let $\mathbb{H}$ be a Hilbert space, $A\in B(\mathbb{H})$ (ie a bounded operator) and $(x_n)_{n=1}^{\infty}$ a sequence in $\mathbb{H}$. Prove that if $A^*x_n\rightarrow y$, there exists a sequence $(y_n)_{n=1}^{\infty}$ such that $A^*Ay_n \rightarrow y$.

I'm 99% sure this has something to do with the relationship between the kernel and the image of $A$ and $A^*$ ($\ker A^*$ being the orthogonal of the image of $A$, etc), but I haven't been able to make much progress. Could I get a hint on how to proceed?

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As you say: use that ${\Bbb H}=\ker A^*\oplus \overline{\operatorname{range}A}$. Then we can split accordingly $$ x_n=z_n+z_n^\bot,\quad z_n\in\ker A^*. $$ Apply $A^*$: $A^*x_n=A^*z_n^\bot\to y$. But $z_n^\bot\in\overline{\operatorname{range}A}$, i.e. there exists $v_{n(k)}$ such that $Av_{n(k)}\to z_n^\bot$ as $k\to\infty$. For every $n$, pick $k_n$ that makes $\|Av_{n(k)}-z_n^\bot\|\le\frac1n$ for the tail $k\ge k_n$. Can you prove that $y_n=v_{n(k_n)}$ works?

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Recall that $\ker A = \ker A^*A$. Namely, clearly $\ker A \subseteq \ker A^*A$. Conversely

$$x \in \ker A^*A \implies A^*Ax = 0 \implies 0 = \langle A^*Ax, x\rangle = \langle Ax, Ax\rangle \implies Ax = 0\implies x \in \ker A$$

Since $(\ker T)^\perp = \overline{\operatorname{Im} T^*}$, taking orthogonal complements in the above relation gives

$$\overline{\operatorname{Im}A^*} = \overline{\operatorname{Im}A^*A}$$

Therefore, $A^*x_n \to y$ implies that $y \in \overline{\operatorname{Im}A^*}$, so $y \in \overline{\operatorname{Im}A^*A}$ implying that there exists a sequence $(y_n)$ such that $A^*Ay_n \to y$.

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The image of a linear operator is closed, so $y$ is a limit of elements of an image iff it is an element of that image. So this is equivalent to showing that given any $v$, there is some $u$ such that $A^* Au = A^*v$.

$v$ can be written as $h+p$, where $h$ is in $ker A^*$ and $p$ is orthogonal to $ker A^*$, and $A^*v =A^*p$. So we need, given arbitrary $p$, to find $u$ such that $Au = p$.

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    $\begingroup$ It's not true that the image of every linear operator is closed. This is not even true in general if the operator is bounded. $\endgroup$ – PhoemueX May 14 '18 at 21:13
  • $\begingroup$ @PhoemueX can you give a counterexample? $\endgroup$ – Acccumulation May 14 '18 at 21:18
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    $\begingroup$ $A : \ell^2 \to \ell^2$ given by $A(x_n)_n = \left(\frac1n x_n\right)_n$. Notice that $\operatorname{Im} A$ contains the canonical vectors $(e_n)_n$ so $\operatorname{Im} A$ is dense in $\ell^2$. If it were closed, we would have $\operatorname{Im} A = \ell^2$, but clearly $\left(\frac1n\right)_n \in \ell^2 \setminus \operatorname{Im} A$. $\endgroup$ – mechanodroid May 14 '18 at 21:22

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