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Let $X=L^2(\mu)\times L^2(\mu)=\{(f,g)|f,g\in L^2(\mu)\}$ be the linear space normed by $\|(f,g)\|=(\|f\|_2^3+\|g\|_2^3)^{1/3}$. Show that $X$ is Banach space and describe $X^*$.

My Work:

We have to show that $X$ is complete. So, must prove that every absolutely convergent series is convergent. Suppose $\displaystyle \sum_{n=1}^\infty \|(f_n,g_n)\|$ converges. Then $\|(f_n,g_n)\|\rightarrow 0$ as $n\rightarrow 0$. That is $(\|f_n\|_2^3+\|g_n\|_2^3)^{1/3}\rightarrow 0$ as $n\rightarrow 0$. This implies $(\|f_n\|_2^3+\|g_n\|_2^3)\rightarrow 0$ as $n\rightarrow 0$. Hence $(f_n,g_n)\rightarrow 0$ as $n\rightarrow 0$. After that I was stuck. I want to show that $\displaystyle \sum_{n=1}^\infty (f_n,g_n)$ converges. can somebody please help me to prove it?

Is there anything I can do here with the fact that $L^2(\mu)$ complete?

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    $\begingroup$ Note the 2 on the $\|\cdot\|_2$. Show that if $(f_n,g_n)$ is Cauchy, then so are $f_n$ and $g_n$ separately. Then $f_n\to f$ and $g_n \to g$ for some $f,g$, hence $(f_n,g_n) \to (f,g)$. $\endgroup$
    – copper.hat
    Mar 5, 2015 at 19:45
  • $\begingroup$ Thanks. Regarding $X^*$, since the dual of $L^2$ is $L^2$ itself, does it imply that $X^*=X$? $\endgroup$
    – Extremal
    Mar 5, 2015 at 20:07
  • $\begingroup$ Yes, but some care is needed. Note that the norm above and the more 'usual' norm $\|(f,g)\|_2 = \sqrt{\|f\|^2+\|g\|^2}$ are equivalent, so the set of continuous functionals is the same. $\endgroup$
    – copper.hat
    Mar 5, 2015 at 20:14

1 Answer 1

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For any $1<p<\infty$ one may define $p$-sum $E\oplus_p F$ of Banach spaces $E$ and $F$ as the linear space of tuples $(x,y)\in E\oplus F$ with norm $$ \Vert(x,y)\Vert_p=(\Vert x\Vert^p+\Vert y\Vert^p)^{1/p} $$ One may show that the latter space is complete and even more that $$ (E\oplus_p F)^*=E^*\oplus_{p^*}F^* $$ where $p^*=p/(p-1)$. For your particular case $p=3$, $E=F=L_2(\mu)$. It is remains to note that $L_2(\mu)^*=L_2(\mu)$ and combine previous results.

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