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Prove the dual space of $\ell^p$ is isomorphic to $\ell^q$ if $\frac{1}{q}+\frac{1}{p}=1$ ($1<p<\infty$)

Define a map $J:\ell^q \to (\ell^p)'$ such that $Jy(x)=\sum_{k=1}^\infty x_ky_k,x\in \ell^p,y\in \ell^q$

I have verified that $Jy\in (\ell^p)'$, $J$ is linear and $\lVert Jy \rVert\leq \lVert y \rVert_q$.

How to show $\lVert Jy \rVert \geq \lVert y \rVert_q$ and $J$ is surjective?

Martin Sleziak
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Kenneth.K
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2 Answers2

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Surjectivity of the map $J$.

Let $\varphi\in (\ell^p)'$, with $\|\varphi\|_*=1$, and set $y_n=\varphi(e_n)$, where $e_n=(0,\ldots,0,1,0,\ldots)$, with exactly one $1$ in the $n$th place. We shall show that $\{y_n\}\in \ell^q$, $\|\{y_n\}\|_q=1$ and $\varphi(\{x_n\})=\sum x_ny_n$.

I. $\{y_n\}\in \ell^q$. For every $x_1,\ldots,x_n$, with $|x_1|^p+\cdots+|x_n|^p\le 1$, we have $\|(x_1,\ldots,x_n,0,0,\ldots)\|_p\le 1$ and hence $$ 1\ge \varphi(x_1e_1+\cdots+x_ne_n)=x_1y_1+\cdots+x_ny_n. $$ But $$ \sup_{|x_1|^p+\cdots+|x_n|^p\le 1}x_1y_1+\cdots+x_ny_n=\big(|y_1|^q+\cdots+|y_n|^q\big)^{1/q}, \tag{1} $$ and hence $$ |y_1|^q+\cdots+|y_n|^q\le 1, $$ and since this holds for every $n$, then $\|\{y_n\}\|_q\le 1$ and $\{y_n\}\in \ell^q$.

II. $\varphi(\{x_n\})=\sum x_ny_n$. Let $x=\{x_n\}\in\ell^p$ and set $x^n=(x_1,x_2,\ldots,x_n,0,0,\ldots)$. Then $\|x^n-x\|\to 0$ and hence $\varphi(x^n-x)\to0$. But $$ \varphi(x^n)=\sum_{k=1}^n x_ky_k, $$ and as the series $\sum_{n=1}^\infty x_ny_n$, converges then $$ \varphi(x)=\lim_{n\to\infty}\varphi(x^n)=\sum_{n=1}^\infty x_ny_n=(x,y). $$ The fact that $\|y\|_q=1$ can be readily shown.

Proof of (1). Clearly, (Hölder) $$ |x_1y_1+\cdots+x_ny_n|\le \left(|x_1|^p+\cdots+|x_n|^p\right)^{1/p} \left(|y_1|^q+\cdots+|y_n|^q\right)^{1/q}\le \left(|y_1|^q+\cdots+|y_n|^q\right)^{1/q} $$ It is easily shown that equality is obtained for $$ x_i=\frac{|y_i|^{q/p}\mathrm{sgn}(y_i)}{\left(|y_1|^q+\cdots+|y_n|^q\right)^{1-1/q}}, \quad i=1,\ldots,n. $$

Yiorgos S. Smyrlis
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  • why the supremum of $x_1y_1+\dots+x_ny_n$ is $(|y_1|^q+\dots+|y_n|^q)^{1/q}$? I know by Minkowski's inequality, the supremum is less than the right, but why could the equality be achieved? – Kenneth.K Feb 23 '17 at 18:13
  • Short answer: That's the case in $\mathbb R$ with $p-$norm. On the other hand, take a look at my edited answer. – Yiorgos S. Smyrlis Feb 23 '17 at 23:26
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Given $y=\{y_k\}\in\ell^q$ nonzero, consider the sequence $x=\{e^{-i\theta_k}\|y\|_q^{1-q}|y_k|^{q/p}\}$, where $\theta_k=\arg(y_k)$ (define $\theta_k$ arbitrarily if $y_k=0$). Then $x\in\ell^p$ with $\|x\|_p=1$ (it's not too hard to see), and \begin{align} Jy(x)&=\sum_{k=1}^\infty y_ke^{-i\theta_k}\|y\|_q^{1-q}|y_k|^{q/p}\\ &=\|y\|_q^{1-q}\sum_{k=1}^\infty|y_k|^{1+q/p}\\ &=\|y\|_q^{1-q}\sum_{k=1}^\infty|y_k|^q \\ &=\|y\|_q. \end{align} Thus $\|Jy\|\geq|Jy(x)|=\|y_p\|$.

To show that $J$ is surjective, fix $f\in(\ell^p)^*$ and look at what it does to the (Schauder) basis vectors $e_n=\{\delta_{k,n}\}$ for each $n\in\mathbb{N}$. This gives you a sequence $y=\{y_k\}$, and show that this sequence is in $\ell^q$ and $Jy=f$.

Aweygan
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  • why the p-norm of $x$ is 1? $\sum_k e^{-pi\theta_k}\lVert y\rVert_q^{p-pq}|y_k|^q=1$? – Kenneth.K Feb 23 '17 at 16:39
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    @Kenneth.K You take modulus of the entries, so $e^{-i\theta_k}$ drops out. After that, mess arround with the equation $\frac{1}{p}+\frac{1}{q}=1$ to get some formulas that help simplify this expression. – Aweygan Feb 23 '17 at 16:43
  • To prove it's surjective, I have $y_k = f(e_k)$. Then $\lVert f \rVert \geq |f(e_k)|=|y_k|$. Hence $(\sum |y_k|^q)^{1/q} \leq (\sum \lVert f \rVert^q)^{1/q}$. Is this right? – Kenneth.K Feb 23 '17 at 18:04