Folland Real Analysis Exercise 5.21 - An isometry between $\mathcal X^* \times \mathcal Y^*$ and $(\mathcal X \times \mathcal Y)^*$ The following is problem 5.21 in Folland's Real Analysis 2nd Edition:

If $\mathcal X$ and $\mathcal Y$ are normed vector spaces, define
$\alpha:\mathcal X^*\times\mathcal Y^*\to(\mathcal X\times\mathcal Y)^*$ by
$$\alpha(f,g)(x,y)=f(x)+g(y).$$
Then $\alpha$ is an isomorphism which is isometric if we use the norm
$\|(x,y)\|=\max(\|x\|,\|y\|)$ on $\mathcal X\times\mathcal Y$, the
corresponding operator norm on $(\mathcal X\times\mathcal Y)^*$, and
the norm $\|(f,g)\|=\|f\|+\|g\|$ on $\mathcal X^*\times\mathcal Y^*$.

This question is already posted, with an answer, here, but I did not understand the answer and would like to know if my partial solution is correct.
Here is what I did:
First we show that $\alpha$ is a linear bijection. Let $\lambda \in \mathbb C$ $f_1, f_2 \in X^*$ and $g_1, g_2 \in Y^*$. Then
\begin{align*}
\alpha((f_1, g_1) + \lambda(f_2, g_2))(x, y) & = (f_1 + \lambda f_2)(x) + (g_1 + \lambda g_2)(y) \\
& = f_1(x) + g_1(y) + \lambda(f_2(x) + g_2(y) \\
& = \alpha((f_1, g_1))(x, y) + \lambda \alpha((f_2, g_2))(x, y)
\end{align*}
and $\alpha$ is seen to be linear. Suppose that $\alpha(f, g) = 0$. Then $f(x) = g(y) = 0$ for all $x \in X$ and $y \in Y$, and it follows that $f = g = 0$. Thus $\alpha$ is injective. Now let $\varphi \in (X \times Y)^*$. Define $f(x) = \varphi(x, 0)$ and $g(y) = \varphi(0, y)$. Note that $f$ and $g$ are linear and bounded, and that we can write $\varphi(x, y) = f(x) + g(y)$. Thus $\alpha$ is surjective and hence an isomorphism.
We now show that $\alpha$ is an isometry with the given norms. We want to show that $||\alpha(f, g)|| = ||(f, g)|| = ||f|| + ||g||$. First note that
\begin{align*}
||\alpha(f, g)|| & = \sup_{||(x, y)|| = 1}|\alpha(f, g)(x, y)| \\
& = \sup_{\max(||x||, ||y||) = 1}|\alpha(f, g)(x, y)| \\
& = \sup_{\max(||x||, ||y||) = 1} |f(x) + g(y)| \\
& \leq ||f|| + ||g|| \\
& = ||(f, g)||.
\end{align*}
My question is:

How to prove the converse inequality? Namely, that $\| \alpha(f, g)\| \geq \|(f, g)\|$.

 A: Given $\varepsilon>0$, you can find $x\in X$, $y\in Y$ with $\|x\|=\|y\|=1$, such that
$$\|f\|<f(x)+\varepsilon/2,\qquad \|g\|<g(y)+\varepsilon/2.$$
Then we have
$$\|(f,g)\|=\|f\|+\|g\|<f(x)+g(y)+\varepsilon\leq\|\alpha(f,g)\|+\varepsilon.$$
Since $\varepsilon>0$ was arbitrary, the reverse inequality holds.
A: Let's prove you have equality where you've written inequality. I claim that
$$\sup_{\max\{\|x\|, \|y\|\} = 1} |f(x) + g(y)| = \sup_{\max\{\|x\|, \|y\|\} = 1} |f(x)| + |g(y)|.$$
To prove this, I propose that we show the following:

Given any $(x, y) \in X \times Y$ such that $\max\{\|x\|, \|y\|\} = 1$, there exists some $(x, y') \in X \times Y$ such that $\max \{\|x\|, \|y'\|\} = 1$ and
$$|f(x) + g(y')| = |f(x)| + |g(y')| = |f(x)| + |g(y)|.$$

How does this help? It means that $\sup_{\max\{\|x\|, \|y'\|\} = 1} |f(x) + g(y')|$ is greater than or equal to $|f(x)| + |g(y)|$ for any $x \in X$ and $y \in Y$ such that $\max\{\|x\|, \|y\|\} = 1$. In particular, this proves that
$$\sup_{\max\{\|x\|, \|y'\|\} = 1} |f(x) + g(y')| \ge \sup_{\max\{\|x\|, \|y\|\} = 1} |f(x)| + |g(y)|.$$
The reverse inequality follows easily from the triangle inequality on $\Bbb{C}$.
To prove this claim, take an arbitrary $(x, y) \in X \times Y$ such that $\max\{\|x\|, \|y\|\} = 1$. Note that if $g(y) = 0$ or $f(x) = 0$, then $|f(x) + g(y)| = |f(x)| + |g(y)|$ trivially, so we can simply take $y' = y$. Otherwise, assume $f(x), g(y) \neq 0$.
Let $\lambda = \frac{|g(y)| \cdot g(x)}{|f(x)| \cdot g(y)}$, and $y' = \lambda y$. Note that $|\lambda| = 1$, so $\|y'\| = \|y\|$ and $|g(y')| = |g(y)|$. We also get
\begin{align*}
|f(x) + g(y')| &= |f(x) + \lambda g(y)| \\
&= \left|f(x) + \frac{f(x)}{|f(x)|} |g(y)|\right| \\
&= |f(x)| \cdot \frac{|f(x)| + |g(y)|}{|f(x)|} \\
&= |f(x)| + |g(y)|.
\end{align*}
Thus, our choice of $y'$ has all the properties we want, proving
$$\sup_{\max\{\|x\|, \|y\|\} = 1} |f(x) + g(y)| = \sup_{\max\{\|x\|, \|y\|\} = 1} |f(x)| + |g(y)|,$$
as desired.
We can then prove that
$$\sup_{\max\{\|x\|, \|y\|\} = 1} |f(x)| + |g(y)| = \sup_{\|x\| = \|y\| = 1} |f(x)| + |g(y)|.$$
Note that $\ge$ is clear, simply because $\|x\| = \|y\| = 1 \implies \max\{\|x\|, \|y\|\} = 1$. To get $\le$, replace $x$ and $y$ with $x' = x / \|x\|$ (when $x \neq 0$) and $y$ with $y' = y / \|y\|$ (when $y \neq 0$). If $x = 0$, then let $x'$ be an arbitrary norm $1$ vector, and similarly for $y$. Then $\|x'\| = \|y'\| = 1$, $|f(x')| \ge |f(x)|$ and $|g(y')| \ge |g(y)|$. Thus,
$$|f(x')| + |g(y')|\ge |f(x)| + |f(y)|,$$
proving $\le$ as required.
Finally, note that
$$\sup_{\|x\| = \|y\| = 1} |f(x)| + |g(y)| = \sup_{\|x\| = 1}\sup_{\|y\| = 1} |f(x)| + |g(y)| = \left(\sup_{\|x\| = 1} |f(x)|\right) + \left(\sup_{\|y\| = 1} |g(y)|\right),$$
completing the proof.
