Isometric isomorphism between $(X\oplus Y)'$ and $X' \oplus Y'$ Let $X$ and $Y$ be normed spaces and consider $X\oplus Y$ with norm $\Vert (x,y)\Vert_1 = \Vert x \Vert + \Vert y \Vert$. I have shown that $(X\oplus Y)'$ is isomorphic to $X'\oplus Y'$ with norm $\Vert (x',y')\Vert_\infty=\max\{\Vert x'\Vert,\Vert y'\Vert\}$ using the map $T : X'\oplus Y' \to (X\oplus Y)'$ given by $$[T(x',y')](x,y)=x'(x)+y'(y).$$ Linearity and boundedness can be easily verified and I prove bijectivity using the restriction of a given $f \in (X\oplus Y)'$ to the subspaces $\{(x,0):x\in X\}$ and $\{(0,y):y \in Y\}$ of $X\oplus Y$. The bounded inverse theorem implies the boundedness of $T^{-1}$. However, I would like to know if it is possible to make this same map an isometric isomorphism by taking a different norm in $X'\oplus Y'$ (or even with the same norm). Does anyone know a similar construction?
 A: Your map $T$ already is an isometric isomorphism.
Let $(x',y') \in X'\oplus Y'$ be arbitrary.
For any  $(x,y)\in X \oplus Y$ we have
\begin{align}
|T(x',y')(x,y)| &= |x'(x)+y'(y)|  \le |x'(x)|+|y'(y)| \le \|x'\|\|x\| + \|y'\|\|y\|\\
&\le \max\{\|x'\|,\|y'\|\}(\|x\|+\|y\|) = \|(x',y')\|_\infty\|(x,y)\|_1
\end{align}
so $\|T(x',y')\| \le \|(x',y')\|_\infty$.
Conversely, let $\varepsilon > 0$ be arbitrary. WLOG assume that $\|x'\| \ge \|y'\|$. By definition of the operator norm for $x'$, there exists $x \in X$ such that $\|x\|=1$ and $|x'(x)| > \|x'\|-\varepsilon$. Moreover, by multiplying this $x$ with a suitable scalar on the unit sphere, we can further assume that $x'(x)$ is a nonnegative real number.
We then have
$$T(x',y')(x,0) = x'(x)+y'(0) \ge  \|x'\| - \varepsilon = \max\{\|x'\|,\|y'\|\} - \varepsilon = \|(x',y')\|_\infty - \varepsilon$$
and therefore since $\|(x,0)\|_1 = \|x\|+\|0\| =\|x\|$, we have
$$\|T(x',y')\| \ge \frac{|T(x',y')(x,0)|}{\|(x,0)\|_1} \ge \|(x',y')\|_\infty - \varepsilon.$$
Letting $\varepsilon\to 0^+$ shows $\|T(x',y')\| \ge \|(x',y')\|_\infty$.
Therefore, for all $(x',y') \in X'\oplus Y'$ we have  $\|T(x',y')\| = \|(x',y')\|_\infty$ so $T$ is an isometry.
