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On the Hilbert space $l^2$, let $f$ be the functional defined by $$f(x):= \sum_{j=1}^\infty \alpha_j \xi_j$$ for each $x:=(\xi_j)_{j=1}^\infty$ in $l^2$, where $a:= (\alpha_j)_{j=1}^\infty$ is a fixed element in $l^2$.

While this functional is linear and bounded, how to compute the norm?

What is the situation if instead of $l^2$ we take $l^p$ for an arbitrary but fixed $p \geq 1$?

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Take your favourite textbook on functional analysis, and there you will find the description of dual to $\ell^p$ or $L^p$, including the norm. – Alexander Shamov Feb 16 '13 at 2:07

For the case of $\ell_2$, apply Cauchy-Schwarz to obtain $$|f(x)| = \left|\sum_{j=1}^\infty \alpha_j x_j\right| = |\langle \alpha,x\rangle| \leq \|\alpha\|\|x\|.$$ Hence $\|f\|\leq\|\alpha\|$. To show $\|f\|=\|\alpha\|$, observe $f(\alpha)=\|\alpha\|^2$.

For general $\ell_p$ spaces, Hölder's inequality will be useful.

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For the case $\ell_p$ you need $a \in \ell_q$ where $\frac1p + \frac1q = 1$. – Martin Feb 16 '13 at 12:01
What if $a$ still belongs to $\ell^p$ instead of $\ell^q$? Do we have a way of determining the norm in this case? – Saaqib Mahmuud Feb 23 '13 at 7:50

Observe that in case of the Hilbert space $l^2$ your functional is of the form $$x\mapsto \langle x, \alpha\rangle.$$ Now use Riesz's representation theorem for Hilbert spaces. The norm of this functional is $\|\alpha\|$. Note that $(l^2)^*$ and $l^2$ are isometricaly isomorphic.

There are analogous representation theorems for other spaces $l^p$ or $L^p$, $p\neq2$. Keep in mind that those spaces are not Hilbert spaces.

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Thanks, but I can't right now recall the Riesz's representation theorem for Hilbert spaces. – Saaqib Mahmuud Feb 17 '13 at 14:51

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