Dual norm of a product space Suppose we have two spaces $X, Y$, each is a linear space equipped with norm $\|\cdot\|_X, \|\cdot\|_Y$, then we define a norm on the product space $Z = X\times Y: \|(x,y)\| = \sqrt{a\|x\|_X^2+b\|y\|_Y^2}$, where $a,b > 0$. 
I am wondering how to compute the dual norm of this norm. By the definition of dual norm, we need to find for $z^* \in Z^*$, $\|z^*\|_* = \underset{\|z\|=1}{\mathrm{sup}} \;z^*(z)$, but what is $z^*(z)$ here? What are the elements in $Z^*$ like? 
 A: Your dual space $(X \times Y)^*$ is isometrically isomorphic to the product of dual spaces $X^* \times Y^*$ equipped with the norm $\|(f,g)\| = \sqrt{a\|f\|^2 + b\|g\|^2}$.
Proof:
Define $J : X^* \times Y^* \to (X \times Y)^*$ as
$$[J(f,g)](x,y) = af(x) + bg(y)$$
for $(f,g) \in X^* \times Y^*$.
$J$ is clearly well-defined and linear. Check that its inverse is given by
$$J^{-1}(F) = \left(\frac1a F(\cdot, 0), \frac1b F(0, \cdot)\right)$$
for $F \in (X \times Y)^*$ so $J$ is bijective.
We have
\begin{align}
[J(f,g)](x,y) &= |af(x) + bg(y)| \\
&\le a|f(x)| + b|g(y)| \\
&\le a\|f\|\|x\| + b\|g\|\|y\| \\
&\le \sqrt{a\|f\|^2 + b\|g\|^2}\sqrt{a\|x\|^2 + b\|y\|^2} \\
&= \|(f,g)\|\|(x,y)\|
\end{align}
so $\|J(f,g)\| \le \|(f,g)\|$.
To prove the converse inequality, let $\varepsilon > 0$ and pick unit vectors $x \in X$, $y \in Y$ such that $f(x) \ge (1-\varepsilon)\|f\|$ and $g(y) \ge (1-\varepsilon)\|g\|$.
Consider $\left(\|f\|x, \|g\|y\right) \in X \times Y$. Its norm is $$\left\|\left(\|f\|x, \|g\|y\right)\right\| = \sqrt{a\|f\|^2\|x\|^2 + b\|g\|^2\|y\|^2} = \sqrt{a\|f\|^2 + b\|g\|^2} = \|(f,g)\|$$
We have
\begin{align}
J(f,g)(\|f\|x, \|g\|y) &= af(\|f\|x) + bg(\|g\|y) \\
&= a\|f\|f(x) + b\|g\|g(y) \\
&\ge a(1-\varepsilon)\|f\|^2 + b(1-\varepsilon)\|g\|^2 \\
&= (1-\varepsilon)(a\|f\|^2 + b\|g\|^2) \\
&= (1-\varepsilon)\|(f,g)\|^2 \\
&= (1-\varepsilon)\|(f,g)\|\|(\|f\|x, \|g\|y)\|
\end{align}
so $\|J(f,g)\| \ge (1-\varepsilon)\|(f,g)\|$. Since $\varepsilon$ was arbitrary, we conclude $\|J(f,g)\| \ge \|(f,g)\|$.
Therefore $\|J(f,g)\| =\|(f,g)\|$ so $J$ is an isometric isomorphism.
This gives you an explicit formula for the dual norm of $F \in (X \times Y)^*$:
$$\|F\| = \sqrt{\frac1a\|F(\cdot, 0)\|^2 + \frac1b\|F(0, \cdot)\|^2}$$
A: Perhaps I'm wrong but it seems to me that an element in $Z^*$ is of the form:
$\phi(x)+\rho(y)$ where $\phi\in X^*$, $\rho \in Y^*$.
It is first easy to see that elements of this form are indeed functionals on $X\times Y$. Furthermore, all $(x,y)\in X\times Y$ can be written as $(x,y)=(x,0)+(0,y)$. Hence by linearity of the functionals, we know that for all $\varphi \in Z$ we have:
$\varphi(x,y)= \varphi(x,0)+\varphi(0,y)$. Where $\phi(x):=\varphi(x,0)$ and $\rho(y):=\varphi(0,y)$ are functionals on $X$ and $Y$ accordingly.
