Home > Analysis, Functional Analysis, Measure Theory, Undergraduate > Dual of a well known space

## 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.