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Let $A$ be a $C^*$-algebra, and denote by $A_{sa}$ its self-adjoint (or 'hermitian') elements, i.e. the fixed-point set of the involution. Any $x\in A$ may be written as sum $$x=h+ik\qquad\text{where both }\; h={x+x^*\over 2}\;\text{ and }\;k={x-x^*\over 2i}\in A_{sa}$$ are uniquely determined. Note that $\|h\|, \|k\|\le\|x\|$.

If $t\in\mathbb{R}$ then \begin{align} \|h+ik\|\: & =\; \left\|e^{it}(h+ik)\right\| \\[.7ex] &=\; \|\cos t\cdot h-\sin t\cdot k\,+\, i(\sin t\cdot h +\cos t\cdot k)\| \tag{0}\\[.7ex] &\ge\; \|\cos t\cdot h-\sin t\cdot k\| \end{align} hence $$T(x):=\,\sup_{t\in [0,2\pi]}\|\cos t\cdot h-\sin t\cdot k\|\;\le\;\|x\|\tag{1}$$

My main question alias $\mathbf{Q_0}$: Is $\,T\,$ the C*-norm in disguise?
Equivalently, do we have equality $\,T(x)=\|x\|\,$ for all $x\in A$ in (1) ?

If the answer is negative, here's $\mathbf{Q_{0.1}}$:
Is it true at least if $A=M_n(\mathbb{C})$, that is the $C^*$-algebra of $n\times n$ complex matrices?

I couldn't find any $2\times 2$ matrix displaying inequality ... neither at my desk nor after literature search, so finally $\mathbf{Q_1}$:
Do you know / Can you provide references regarding this subject?

Addendum after acceptance of Jonas Meyer's response.

Look again at equation $(0)$, it further implies $$\|x\|\:\le\;\|\cos t\cdot h-\sin t\cdot k\| + \|\sin t\cdot h + \cos t\cdot k\| \; \le\; 2T(x)$$ and Jonas' concrete example, in which the factor $2$ appears, shows that the preceding estimate is optimal.

Summary

  • $T$ is equivalent to the $C^*$-norm, in any C*-algebra $A$: $$T(x)\:\le\:\|x\|\:\le\:2T(x)\quad\forall x\in A\tag{2}$$
  • The bounds in (2) are optimal.
  • When evaluated for normal elements then $T$ is equal to the $C^*$-norm.
    In particular, $\,T=\|\!\cdot\!\|\,$ holds in commutative $C^*$-algebras.
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    Any thoughts on the answers? – 2017-02-08
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    @Aweygan Afraid $\exists$ a fairly obvious counter-example I've overlooked, but frankly written I believe in $T = \|\cdot\|$. – 2017-02-08

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Try $x=\begin{bmatrix}0&1\\0&0\end{bmatrix}$, which has $\|x\|=1$, and $\|\cos(t)h-\sin(t)k\|=\frac12$ for all $t$.

Looking at the easiest example of a nonnormal element of a $C^*$-algebra is natural because the equality would hold for normal elements. Equivalently, this holds in $C(X)$ when $X$ is a compact Hausdorff space. In that case, for $f\in C(X)$ there exists $x_0\in X$ such that $$\|f\|=\max\limits_{x\in X}|f(x)|=|f(x_0)|=|h(x_0)+ik(x_0)|=\sqrt{h(x_0)^2+k(x_0)^2},$$ and $$\cos(t)h(x_0)+\sin(t)k(x_0)=\sqrt{h(x_0)^2+k(x_0)^2}\sin(t+\theta)$$

for some $\theta\in[0,2\pi)$, so by taking $t=\frac\pi2-\theta$ you get $\cos(t)h(x_0)+\sin(t)k(x_0)=\|x\|$.

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    So it's noncommutativity which inhibits $T$ to coincide in general with the $C^*$-norm. _Thank you for settling the question, especially for pointing at & elaborating on normal elements._ My insights gained from your answer lead to an appendix in the OP, and I'm kind of satisfied that I missed $x=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ ... – 2017-02-10