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Let $M_n(\mathbb{C})$ denote the algebra of $n \times n$ complex matrices, and $H_n(\mathbb{C})$ denote the linear space of $n \times n$ Hermitians. Both spaces are endowed with the usual operator norm. Assume that $$ \Phi \colon H_n(\mathbb{C}) \rightarrow H_n(\mathbb{C})$$ is a (real) linear map of norm $1.$

Then $\Phi$ has a natural linear extension to $M_n(\mathbb{C})$ by

$$\Phi(A) := \Phi\left({A+A^* \over 2}\right) + i\Phi\left({A-A^* \over 2i}\right).$$

Could you give me an example with $\|\Phi\| > 1 ?$ Or an explanation why this might happen?

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    The $\mathbb{C}$-linear extension of $\Phi$ as of above, let's denote it by $\tilde{\Phi}$. In the context of '$\mathbb{C}$omplexification' there's another look at it as a tensor product: $H_n(\mathbb{C})$ is an $\mathbb{R}$-linear subspace of $M_n(\mathbb{C})$ and every element in $M$ equals $h+ik$ with unique $h,k\in H$, hence $$M_n(\mathbb{C}) = \mathbb{C}\otimes_\mathbb{R}H_n(\mathbb{C}) \quad\text{and}\quad \tilde{\Phi} = \operatorname{id}\otimes_\mathbb{R}\Phi $$ If now $\|\text{id}\otimes_\mathbb{R}\Phi\|=\|\text{id}\|\cdot\|\Phi\|$ we were done. – 2017-02-05
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    This question is unclear. Operator norms do not exist independently. They are always induced by some norm. What is the norm that induces the so-called "usual" operator norm in question? If the inducing norm is the maximum absolute row sum norm, examples of $\|\Phi\|>1$ clearly exist (just let $\Phi(A)$ be the plain transpose of $A$); if the inducing norm is the Frobenius norm (hence the induced norm is the maximum singular value of an $n^2\times n^2$ matrix), the answer is clearly no; if the inducing norm is the maximum singular value, the answer is rather inobvious. – 2017-03-07
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    @user1551: Good point. While I'm sure zoli will clarify, I hadn't thought of the potential ambiguity perhaps in part because of the tag "operator algebras," where the norm on $M_n(\mathbb C)$ considered as an algebra is usually by default the $C^*$-algebra one, or equivalently the maximum singular value, or equivalently the norm induced from acting on $\mathbb C^n$ with Euclidean norm. – 2017-03-07
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    Yes, the norm is the maximum singular value, and we have the Euclidean norm on $\mathbb{C}^n.$ For me, this is the usual operator norm on the matrix algebra $M_n(\mathbb{C}).$ Please, notice that in the commutative case, if we consider the $L^\infty[0,1]$ algebra ($[0,1]$ is equipped with the Lebesgue measure), the linear extension of $\Phi$ from real functions to complex functions does not increase the norm of $\Phi$ - this is probably a well-known result and goes back to M. Riesz, and A. E. Taylor. And the point is whether the same phenomena appears in the non-commutative setting, or not. – 2017-03-07
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    I asked a question extending this one: http://math.stackexchange.com/questions/2185942 – 2017-03-14

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I think I have an example showing the norm can be greater than $1$. For Hermitian $H=\begin{bmatrix}a&b\\\overline{b}&d\end{bmatrix}\in H_2(\mathbb C)$, define $\Phi(H)=\begin{bmatrix}0&\dfrac{a+di}{\sqrt2}\\\dfrac{a-di}{\sqrt2}&0\end{bmatrix}.$ We have $\|\Phi(H)\|=\frac1{\sqrt2}\sqrt{a^2+d^2}\leq\max\{|a|,|d|\}\leq\|H\|,$ with equality holding in case $|a|=|d|$ and $b=0$.

Consider the extension applied to $A=\begin{bmatrix}1&0\\0&i\end{bmatrix}$. We have $\|A\|=1$ while $\tilde{\Phi}(A)=\begin{bmatrix}0&0\\\sqrt2&0\end{bmatrix},$ so $\|\tilde{\Phi}(A)\|=\sqrt2$.

Motivation: To have the norm of the extension increase, it makes sense to look for cases where the real and imaginary parts of $A$ are "nonoverlapping," so that $A+A^*$ and $A-A^*$ can be as large as possible without increasing the norm of $A$. Then if we can have $\Phi$ map each of those parts in such a way that they do "overlap," an increase in norm can happen. This was done by taking distinct diagonal entries and placing them in the same off-diagonal positions. The off-diagonal was needed to allow the diagonal entries from the Hermitian matrices to be sent to real and imaginary parts, keeping the norm of $\Phi$ from being too large on $H_2(\mathbb C)$.

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    That's a nice example. – 2017-03-09
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    I don't know if $\sqrt2$ is worst possible. – 2017-03-10
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    Yep, you are right. My bad. – 2017-03-13
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    @Jonas: You mean $\sqrt{2}$ is the worst in your example, or in general? – 2017-03-13
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    @zoli: It is worst in my example, because for general $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, $\|\tilde{\Phi}(A)\|=\frac{1}{\sqrt2}\max\{|a+id|,|a-id|\}\leq \frac{1}{\sqrt2}(|a|+|d|)\leq \sqrt2\max\{|a|,|d|\}\leq \sqrt2 \|A\|$. So in this case $\|\tilde{\Phi}\|=\sqrt 2$, but can it be that $\|\tilde{\Phi}\|>\sqrt2$ for other examples? – 2017-03-13
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    @Jonas: I would like to think that $\sqrt{2}$ is the worst factor in any case... – 2017-03-14
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    @zoli: I know no reason to believe that isn't true, or that it is true, as of yet. Seems plausible. – 2017-03-14
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    A bit similar question is: Does it follow that $\|(\Phi(A)^2 + \Phi(B)^2)^{1/2}\| \leq \sqrt{2} \|\Phi\| \|(A^2+B^2)^{1/2}\|,$ where $A,B$ are Hermitians? – 2017-03-16