1
$\begingroup$

I'm studying a proof of the spectral theorem for compact operators. The first part of it reads as follows:

Let $X$ be an infinite dimensional inner product space and let $A: X \to X$ be a compact and self-adjoint linear operator with infinite-dimensional range.

Now as $A$ is self-adjoint we have $\sup_{\Vert x \Vert=1} |(Ax,x)| = \Vert A \Vert > 0$. So there exists a sequence $(x_n)_{n=1}^\infty$ such that $|(Ax_n,x_n)| \to \Vert A \Vert$ as $n \to \infty$.

Consequently, there exists a subsequence $(x_m)_{m=1}^\infty$ and $\lambda_1 \in \mathbb{R}$ such that:

  • $\Vert x_m \Vert = 1$ for all $m \ge 1$
  • $|(Ax_m,x_m)| \to \lambda_1$ as $n \to \infty$
  • $|\lambda_1| = \sup_{\Vert x \Vert = 1}|(Ax, x)| = \Vert A \Vert > 0.$

Questions

  1. Why does there exist the sequence $(x_n)_{n=1}^\infty$ such that $|(Ax_n,x_n)| \to \Vert A \Vert$ as $n \to \infty$?
  2. Why can we say $\Vert x_m \Vert = 1$?
$\endgroup$
  • $\begingroup$ The first assertion follows more or less from the definition of supremum. As for the second assertion, notice that you take the supremum over vectors of norm one. $\endgroup$ – Mathematician 42 Jun 17 '16 at 11:44
1
$\begingroup$

It follows from the definition of $\sup$. If $$ M=\sup\left( f(x)\ : x \in B\right), $$ then there exists a sequence $x_n\in B$ such that $$ M=\lim_{n\to \infty} f(x_n).$$ Apply this observation with $B=\text{unit sphere}$, $f(x)=|(Ax_n, x_n)|$ and $M=\|A\|$.

$\endgroup$
  • $\begingroup$ Ok I see it now thanks. For the $\Vert x_m \Vert = 1$ part..how can we say for sure that the elements $x_m$ have a norm of 1? Why are they not just some $c \in \mathbb{R}$? $\endgroup$ – csss Jun 17 '16 at 11:59
  • 1
    $\begingroup$ You have written that $\|A\|=\sup \left( |(Ax, x)|\ :\ \|x\|=1\right)$. So when you take the sequence $x_n$, all its elements satisfy the condition $\|x_n\|=1$. (Written $x_n\in B$ in the main text of this answer) $\endgroup$ – Giuseppe Negro Jun 17 '16 at 12:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.