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Dual of a well known space

Prove that (\ell^p)' \simeq \ell^q, where \frac{1}{p}+\frac{1}{q}=1.

Prove that, in general, (L^p(\mu))' \simeq L^q(\mu).
Solution: If X is a Banach space where X is a vector space over the field \mathbb{K} and if 1\leq  p \leq \infty, then we denote \ell^p =\{(x_n) \subset X : \sum_{n \geq 0} |x_n|^p < \infty \}.
If x \in \ell^q then define f_x : \ell^p\to \mathbb{K},\   f_x(y)=\sum_{n\geq 0} x_ny_n. This functional is obviously linear, and from Holder’s inequality, we get that \sum_{n\geq 0} |x_ny_n|\leq \left(\sum_{n\geq 0} |x_n|^q\right)^{\frac{1}{q}}\left( \sum_{n\geq 0} |y_n|^p\right)^{\frac{1}{p}}< \infty . This implies that the series from the definition of f_x is absolutely convergent, and since X is a Banach space, the series is convergent, therefore f_x is well defined for any x, and \|f_x(y)\| \leq \|x\|_q \|y \|_p,\ \forall y \in \ell^p, which implies that f_x \in (\ell^p)^\prime .
Therefore, we can define \phi: \ell^q \to (\ell_p)^\prime by \phi(x)=f_x.
It is obvious that \phi is linear. If we assume that f_x=f_z,\ x,z \in \ell^q, then taking e_n=(0,...,0,1,0,...),\forall n \in \mathbb{N} where the n‘th position is 1 and the rest are 0‘s it follows that e_n \in \ell^p and f_x(e_n)=f_z(e_n), therefore x_n=z_n,\ \forall n \in \mathbb{N}, and x=z. This implies that \phi is injective.
Now, we pick g \in (\ell^p)^\prime and we see that g(y)=\sum_{n\geq 0}y_n g(e_n),\ \forall y \in \ell^p. We should prove that a=(a_0,...,a_n,...)=(g(e_0),...,g(e_n),...) \in \ell^q.
For this, we consider x_n=(|a_0|^{q-2}a_0,...,|a_n|^{q-2}a_n,0,0,...) which is in \ell^p since all the terms of the sequence x_n are 0 except a finite number of them. \|x_n\|_p=(\sum_{k=1}^n |a_k|^{(q-1)p})^{\frac{1}{p}} =(\sum_{k=1}^n |a_k|^{q})^{\frac{1}{p}}.
We have \sum_{k=1}^n |a_k| ^q=\| f(x_n) \| \leq \|f \| \|x_n \|= \|f \| (\sum_{k=1}^n |a_k|^q) ^\frac{1}{p}, therefore \left( \sum_{k=1}^n |a_k|^q \right)^{\frac{1}{q}} \leq \|f \|. Taking n \to \infty, we get that a \in \ell^q and \|a \| \leq \|f \|. Therefore g= \phi(a), and thus, \phi is surjective. This means that phi is indeed an isomorphism, and \|a \| \leq \| \phi(a) \|. Conversely, \|\phi(a)(x) \| \leq \|a\|_q \| x \|_p, by Holder’s inequality, therefore \|\phi(a)\|=\|a\|,\ \forall a \in \ell^q. Therefore, \phi is an isometry also.

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