Yes, if the operator is self-adjoint. Here is a proof from Conway's book, A Course in Functional Analysis:
Let $M = \{\sup |\langle Ax, x\rangle| : \|x\| = 1\}$. If $\|x\| = 1$, then $|\langle Ax,x\rangle | \leq \|A\|$ (since this holds for all pairs of vectors, so certainly works if we use the same in each slot), so $M \leq \|A\|$.
Now let $h,g$ be unit vectors. Self-adjointness gives $$\langle A(h \pm g) , h \pm g\rangle = \langle Ah,h\rangle \pm 2\mbox{Re}\langle Ah,g\rangle + \langle Ag,g \rangle.$$ Subtracting these equations gives $$\langle A(h + g), h + g \rangle - \langle A(h-g), h-g \rangle = 4 \mbox{Re} \langle Ah,g \rangle.$$ Since $|\langle Af, f \rangle| \leq M\|f\|^2$ for any $f$, by Cauchy-Schwarz, we have using the parallelogram law that \begin{align*}&4\mbox{Re}\langle Ah,g\rangle \\\leq &M(\|h + g\|^2 + \|h-g\|^2) \\ =&2M(\|h\|^2 + \|g\|^2) \\ = &4M. \end{align*}
To finish up, choose $e^{i\theta}$ so that $\mbox{Re}(e^{i\theta}\langle Ah, g\rangle) = |\langle Ah,g \rangle|$. Then $$|\langle Ah, g \rangle| = \mbox{Re}(\langle Ae^{i\theta}h, g\rangle ) \leq M.$$ Take the supreumum over all $h,g$ to finish.
Hint: Denote by
$$D:(C^{1}[0,1],\Vert\cdot\Vert_{\infty})\rightarrow(C[0,1],\Vert\cdot\Vert_{\infty})$$
the differential operator. Let $f_{n}(x)=\sin(nx)$. Note that $\Vert f_{n}\Vert_{\infty}=1$ (for $n\geq 2$). However, $(Df_{n})(x)=n\cos(nx)$, and hence $\Vert Df_{n}\Vert_\infty=n$. What does this mean?
Best Answer
$ \newcommand{\norm}[1]{\left\|{#1}\right\|} \newcommand{\ip}[1]{\left\langle{#1}\right\rangle} $You might find it easier to use an alternative but equivalent definition of the operator norm: if $S \in B(H)$, then $$ \norm{S} = \sup_{\norm{x}=1} \norm{Sx}. $$
Now:
On the one hand, $$ \norm{T} = \sup_{\norm{x}=1} \norm{Tx} = \sup_{\norm{x}=1} \sqrt{\ip{Tx,Tx}}. $$ Since $\ip{Tx,Tx} = \ip{x, T^\ast T x} \leq \norm{T^\ast T}\norm{x}$, what can you conclude?
On the other hand, $$ \norm{T^\ast T} = \sup_{\norm{x}=1} \norm{T^\ast Tx}. $$ Since $\norm{T^\ast T x} \leq \norm{T^\ast} \norm{Tx}$, where $\norm{T^\ast}=\norm{T}$ (why?), what can you conclude?
In terms of wider context, the identity $\norm{T^\ast T} = \norm{T}^2$ is called the $C^\ast$-identity, and is the key fact that makes the theory of $C^\ast$-algebras (e.g., $B(H)$ for $H$ a Hilbert space) so much easier (for lack of a better word) than the theory of more general Banach ($\ast$-)algebras, just as the theory of Hilbert spaces is so much easier than the theory of more general Banach spaces.