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Suppose that $T$ is any bounded operator. I want to show that $\| T^{\ast} T \| = \| T \|^2$. Can you please check my proof.

Consider that $\| T \|^2 = \sup_{\| x \| \leq 1} \| Tx \|^2 = \sup_{\| x \| \leq 1} \left| \langle x, T^{\ast}Tx \rangle \right| \leq \| T^{\ast}T \|$ and $\| T^{\ast} T \| = \sup_{\| x \| \leq 1} \| T^{\ast}Tx\| = \sup_{\| x \| \leq 1} \left| \langle x, T^2 x \rangle \right| \leq \| T \|^2$.

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    Seems correct $\,$ – 2017-02-25
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    How do you get $\sup_{\|x\|\leq1}\|T^*Tx\|=\sup_{\|x\|\leq1}|\langle x,T^2x\rangle|$? – 2017-02-25
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    @Aweygan How would I justify this? I used the reasoning from the other direction. – 2017-02-25
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    @user3359 I don't know how, that's what I'm asking. – 2017-02-25
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    @Aweygan How would I then show that $\| T^{\ast} T \| \leq \| T \|^2$? – 2017-02-25
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    Show that $\|T\|^2\leq\|T^*T\|\leq\|T^*\|\|T\|$, then apply the same reasoning for $T^*$ in place of $T$. – 2017-02-25

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You make a similar gap in both inequalities you are trying to show. Specifically, you claim that $\sup_{\|x\|\leq 1}|\left< x, Ax\right>| \leq \|A\|$. This claim requires justification.

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    What would the intermediary step be? – 2017-02-25
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    I don't know why this is a big deal, $\sup_{\|x\|\leq 1}|\left< x, Ax\right>| \leq\sup_{\|x\|\leq1}\|A\|\|x\|^2=\|A\|$ – 2017-02-25