$(X \oplus_p Y)^*$ isometric to $(X^*\oplus_q Y^*)$ Let $X,Y$ be Banach spaces.
For $1 < p < \infty$, define a norm on $X \oplus Y$ by
$\|(x,y)\|_p=(\|x\|_X^p+\|y\|_Y^p)^{1/p}$.
Homework asks to prove that:
$(X \oplus_p Y)^*$ is isometric to $(X^*\oplus_q Y^*)$ ($^*$ denotes dual).
Ofcourse, my candidate for the isometry $T:(X^*\oplus_q Y^*) \rightarrow (X \oplus_p Y)^*$ is defined by $T(\lambda_1,\lambda_2)=((x,y)\mapsto \lambda_1(x)+\lambda_2(y))$.
I used the definitions, but couldn't find a reason my $T$ preserves the norm.
 A: For $x\in X$, $y\in Y$, $\lambda_1\in X^*$, $\lambda_2\in Y^*$, we have the inequality
\begin{align}
|\lambda_1(x)+\lambda_2(y)|&\leq|\lambda_1(x)|+|\lambda_2(y)|\leq \|\lambda_1\|\,\|x\|+\|\lambda_2\|\,\|y\|\\ \ \\ &\leq (\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/q}\,(\|x\|^p+\|y\|^p)^{1/p}
\end{align}
(where we used Hölder in the last $\leq$). This shows that
$$
\|T(\lambda_1,\lambda_2)\| \leq \|\lambda_1,\lambda_2\|_q.
$$
Now fix $\varepsilon>0$ and let $x'\in X$, $y'\in Y$ with $\|x'\|=\|y'\|=1$, and $\|\lambda_1\|\leq|\lambda_1(x')|-\varepsilon$, $\|\lambda_2\|\leq|\lambda_2(y')|-\varepsilon$. By multiplying each by an appropriate complex number with absolute value 1, we may assume that $\lambda_1(x')\geq0$, $\lambda_2(y')\geq0$. Let
$$
x=\frac{\|\lambda_1\|^{q-1}}{\|(\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\,x', \ \ \ 
y=\frac{\|\lambda_2\|^{q-1}}{\|(\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\,y'.
$$
Then
$$
\|(x,y)\|_p^p=\frac{\|\lambda_1\|^{p(q-1)}}{\|\lambda_1\|^q+\|\lambda_2\|^q}+\frac{\|\lambda_2\|^{p(q-1)}}{\|\lambda_1\|^q+\|\lambda_2\|^q}=\frac{\|\lambda_1\|^q+\|\lambda_2\|^q}{\|\lambda_1\|^q+\|\lambda_2\|^q}=1,
$$
so $\|(x,y)\|_p=1$. And
\begin{align}
T(\lambda_1,\lambda_2)(x,y)&=\lambda_1(x)+\lambda_2(y)=\frac{\|\lambda_1\|^{q-1}}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\,\lambda_1(x')+\frac{\|\lambda_2\|^{q-1}}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\,\lambda_2(y')\\ \ \\
&\geq\frac{\|\lambda_1\|^{q-1}}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\,(\|\lambda_1\|-\varepsilon)+\frac{\|\lambda_2\|^{q-1}}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\,(\|\lambda_2\|-\varepsilon)\\ \ \\
&=\frac{\|\lambda_1\|^q+\|\lambda_2\|^q}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}-\varepsilon\,\frac{\|\lambda_1\|^{q-1}+\|\lambda_2\|^{q-1}}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\\ \ \\
&=(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/q}-\varepsilon\,\frac{\|\lambda_1\|^{q-1}+\|\lambda_2\|^{q-1}}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\\ \ \\
&=\|(\lambda_1,\lambda_2)\|_q-\varepsilon\,\frac{\|\lambda_1\|^{q-1}+\|\lambda_2\|^{q-1}}{(\|\lambda_1\|^q+\|\lambda_2\|^q)^{1/p}}\\ \ \\.
\end{align}
As $\varepsilon$ was arbitrary, we get that $\|T(\lambda_1,\lambda_2)\|\geq\|(\lambda_1,\lambda_2)\|_q$.
So $\|T(\lambda_1,\lambda_2)\|=\|(\lambda_1,\lambda_2)\|_q$ for all $\lambda_1,\lambda_2$, i.e. $T$ is an isometry.
It remains to show that $T$ is onto, but I assume you can do that.
