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 $n \to \infty$
  • $|\lambda_1| = \sup_{\Vert x \Vert = 1}|(Ax, x)| = \Vert A \Vert > 0.$


  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 $\Vert x_m \Vert = 1$?
  • $\begingroup$ 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

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 $M=\|A\|$.

  • $\begingroup$ 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

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