What is the supremum of $\frac{\sum_{i=1}^\infty |x_i|2^{-i}}{\|x\|_p}$?

Question

I have been trying to find the upper limit of this of the of elements $$\frac{\sum_{i=1}^\infty |x_i|2^{-i}}{\|x\|_p}$$

such that $x \in l^p$, i.e $\sum_{i=1}^\infty |x_i|^p < \infty$ and $ 1<p< \infty$.

Some other previous results are:

Given $f(x)= \sum_{n=1}^\infty x_n2^{-n}$. Then if $x \in l^\infty$ any we see $\|f\|_{(l^\infty)^*}=\sup_{x \in l^\infty -{0}}\sum_{n=1}^\infty\frac{ x_n2^{-n}}{\|x\|}$. However, each summand is less or equal to 1. Hence, to maximise this we can have $x =(1,1,....)$ the constant sequence and the norm of functional is $0.5/(1-0.5)=1$ (geometric series).

For $f \in (l^1)^*$, we just need to observe, $$\|f\|_{(l^1)^*}\leq \sum_{n=1}^\infty \frac{|x_n|2^{-n}}{\sum_{k=1}^\infty|x_k|} \leq \sum_{n=1}^\infty \frac{|x_n|2^{-1}}{\sum_{k=1}^\infty|x_k|}=2^{-1} $$ And as $x=(1,0,0,...)$ satisfies $\frac{f(x)}{\|x\|_{l^1}}=\frac{1}{2}$. Then we are done.

My belief is that the supremum $\frac{\sum_{i=1}^\infty |x_i|2^{-i}}{\|x\|_p} = \frac{1}{2}^{\frac{1}{p}}$ or $1$.

Please can someone help point me in the correct direction.

Answer 1

Hint: apply Hölder's inequality in order to get $$ \frac{\sum_{i=1}^\infty |x_i|2^{-i}}{\|x\|_p}\leqslant \left(\sum_{i=1}^\infty 2^{-iq}\right)^{1/q}, $$ where $q$ is the conjugate exponent of $p$, that is, $1/p+1/q=1$. The equality is also reached by a good choice of $x_i$.

Answer 2

Hints: (i) By definition this supremum is just $||L||$, where $L:\ell_p\to\Bbb C$ is defined by $$Lx=\sum x_n2^{-n}.$$

(ii) $(\ell_p)^*=\ell_{p'}$.