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Let $(H, \langle \cdot, \cdot\rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $T:H\to H$ be a self-adjoint bounded linear operator. Let $B_H$ be the closed unit ball of $H$. Let $S$ be the unit sphere of $H$. Let $$ \begin{align} m_1 &:= \inf_{u\in S} \langle Tu, u\rangle, \\ m_2 &:= \sup_{u\in S} \langle Tu, u\rangle. \end{align} $$

I would like to prove a result in my lecture notes, i.e.,

$\|T\| = \max \{ |m_1|, |m_2| \}$.

Could you elaborate on how to prove the challenging direction $\|T\| \le \max \{ |m_1|, |m_2| \}$?

Clearly, $m_1 \le m_2$. By Cauchy–Schwarz inequality, $$ \begin{align} m_1 &\ge \inf_{u\in S} (-|Tu|) = -\|T\|,\\ m_2 &\le \sup_{u\in S} |Tu| = \|T\|. \end{align} $$

Then $-\|T\| \le m_1 \le m_2 \le \|T\|$. Then $\max \{ |m_1|, |m_2| \} \le \|T\|$.

Analyst
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    Use this equivalence https://math.stackexchange.com/q/859328/27978 and note that it a real space. – copper.hat May 29 '23 at 22:19
  • @copper.hat I have formulated your idea as an answer below. Could you have a check on it? Thank you so much! – Analyst May 30 '23 at 00:07

1 Answers1

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Let $m := \sup_{u\in S} |\langle Tu, u\rangle|$. Then $m=\|T\|$ because $T$ is self-adjoint. Let $$ \begin{align} S_+ &:= \{u \in S : \langle Tu, u\rangle \ge 0\}, \\ S_- &:= \{u \in S : \langle Tu, u\rangle < 0\}. \end{align} $$

WLOG, we assume $S_+ \neq \emptyset$ and $S_- \neq \emptyset$. Then $$ \begin{align} m_1 &= \inf_{u\in S_-} \langle Tu, u\rangle, \\ m_2 &= \sup_{u\in S_+} \langle Tu, u\rangle, \\ m &= \max \{ m_2, - m_1 \}. \end{align} $$

Because $m_2 \le |m_2|$ and $-m_1 \le |m_1|$, we get $m \le \max \{ |m_1|, |m_2| \}$. The claim then follows.

Analyst
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