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Given $V$ a pre-Hilbert space and T a self-ajoint linear Operator $V \to V$ show that$$ \sup \|Tx\| =\sup |\langle Tx,x \rangle | \in \mathbb{R} \cup \{ \infty \}$$ for all $ \|x\| =1$.

I know that $\| x \| = \sup |\langle x,y \rangle | $ over all $y \in V$. from cauchy-schwarz $|\langle Tx,x \rangle | \le \|Tx\| \| x \| = \|Tx\| $ But I cannot derive the wanted + I dont know how to bring the self-ajointness in.

Greetings.

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This is a very standard result in Hilbert space theory. A proof can be seen from Hunter and Nachtergaele's Applied Analysis Lemma 8.26. The proof there does not use the completeness and thus it is applicable to the pre-Hilbert space case. To understand the proof, you should note that $$ \|T\|=\sup\{\|Tx\|:\|x\|=1\}. $$ Also, note that in Hunter-nachtergaele, linearity is in the second argument of the inner product.

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For every nonzero vector $x$, we have $$|\langle Tx, x\rangle| =\|x\|^2|\langle T\|x\|^{-1}x,\|x\|^{-1}x\rangle| \le \sup_{\|u\|=1}\langle Tu,u\rangle\|x\|^2 $$

Now from $$4Re\langle Tx,y\rangle =\langle T(x+y),x+y\rangle-\langle T(x-y),x-y\rangle$$we have $$4Re\langle Tx,y\rangle \le \sup_{\|u\|=1}\langle Tu,u\rangle(\|x+y\|^2+\|x-y\|^2)=2(\|x\|^2+\|y\|^2)\sup_{\|u\|=1}\langle Tu,u\rangle$$

Let $\langle Tx,y \rangle=e^{i\theta}|\langle Tx,y \rangle|$. For $x'=e^{-i\theta}x$, we have $|\langle Tx,y \rangle|=e^{-i\theta}\langle Tx,y \rangle=\langle Tx',y\rangle=Re\langle Tx',y\rangle$. Thus for $\|x\|=\|y\|=1$(when $\|x\|=1$,observe that $\|x'\|=1$) we have $$|\langle Tx,y \rangle|=|Re\langle Tx',y\rangle| \le \sup_{\|u\|=1}\langle Tu,u\rangle$$

Thus, $$\|T\|=\sup_{\|x\|=1,\|y\|=1}|\langle Tx,y \rangle| \le \sup_{\|u\|=1}\langle Tu,u\rangle$$

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