I've really hit a brick wall trying to prove the dual of $L^1$ Is isomorphic (as a normed space) to $L^\infty$ ?
Ive been trying for a while but have got stuck. Any help would be greatly appreciated, thank
How would I prove the dual of $L^1$ Is isomorphic to $L^\infty$?
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real-analysis
linear-algebra
ordinary-differential-equations
derivatives
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2There is a natural injective morphism $\psi: L^\infty \to (L^{1})^*$ defined by $$\psi(f)(g) = \int f(x)g(x) dx$$ The hard part is the surjectivity, and this is often proved with the help of the Radon-Nykodim. – 2017-01-30
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0That should be Nikodym, in case you're looking it up. – 2017-01-30
1 Answers
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Let's say this is over a $\sigma$-finite measure space $(X,\Sigma,m)$. A bounded linear functional $\varphi$ on $L^1$ defines a real-valued function $\mu$ on sets of finite $m$-measure by $\mu(E) = \varphi(I_E)$, where $I_E$ is the indicator function of $E$.
Temporarily restrict to subsets of a set $A$ of finite $m$-measure. It is easy to show that this restriction of $\mu$ is a (signed) measure absolutely continuous with respect to $m$, and therefore by Radon-Nikodym is given by integration with an $L^1$ function $g_A$, i.e.
$$ \varphi(I_E) = \int_E g_A \; dm \ \text{for}\ E \in \Sigma \ \text{with}\ E \subseteq A$$
Then for any $f \in L^1(m)$ that is $0$ outside $A$, you can show $ \varphi(f) = \int_A g_A f \; dm$.
Moreover, it's easy to show $|g_A(x)| \le \|\varphi\|$ for $m$-almost every $x \in A$. Next, show that these are consistent: $g_A = g_B$ $m$-almost everywhere in $A \cap B$. Using the fact that $m$ is $\sigma$-finite, we can take a sequence $A_n$ and get a single $g \in L^\infty(m)$ such that $g = g_{A_n}$ on $A_n$, and $\varphi(f) = \int g f \; dm$ for all $f \in L^1(m)$.
EDIT: You do need the $\sigma$-finiteness condition. Consider $X = [0,1]$ with the $\sigma$-algebra $\Sigma$ consisting of subsets of $X$ that are either countable or co-countable (i.e. their complement is countable), and $m$ counting measure. Thus $L^1(m)$ consists of functions $f$ such that $f(x) = 0$ outside some countable set and $\sum_{x \in [0,1]} |f(x)| < \infty$. Consider the linear functional
$$ \varphi(f) = \sum_{x \in [0,1/2]} f(x) $$
which is easily seen to have norm $1$. If $\varphi(f) = \int g f \; dm$, we would need $g(x) = 1$ for $x \in [0,1/2]$ and $0$ for $x \in (1/2, 1]$, but this is not a measurable function so it's not in $L^\infty$.
Hmm. That shows that this particular mapping does not take $L^\infty(m)$ onto $(L^1(m))^*$. I don't know whether there could be some other isomorphism in this case.
EDIT: More generally, in this example for any bounded function $h$ on $[0,1]$ (no measurability required!) you have a unique bounded linear functional
$$ \phi_h(f) = \sum_x h(x) f(x) $$
The cardinality of these is $c^c$. On the other hand, each member of $L^\infty(m)$ has a constant value outside of some countable set, so the cardinality of $L^\infty(m)$ is $c$. So $(L^1(m))^*$ is much too big to be isomorphic to $L^\infty(m)$ in this case.