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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.

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    Do you remember how to prove that $(\mathbb{R} \mathbin{\oplus_p} \mathbb{R})^\ast = \mathbb{R}^\ast \mathbin{\oplus_q} \mathbb{R}^\ast$? If not, you should probably revisit the proof of the duality between $L^p$ and $L^q$. As soon as you have the two-dimensional case then the rest should not be too hard (use Hahn-Banach).2012-11-14

1 Answers 1

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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 $\|\lambda_1\|\leq|\lambda_1(x')|-\varepsilon$, $\|\lambda_2\|\leq|\lambda_2(y)|-\varepsilon$. 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.

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    It doesn't. I'll edit soon.2018-01-28