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Let $(X,M,\mu)$ measure space where $\mu$ is not $\sigma-$finite. What is the dual of $L^1(\mu)$ in this case? For example:

Let $X = \{a,b\}$ and define $\mu(a) = 1$, $\mu(b) = \mu(X) = \infty$, and $\mu(\emptyset) = 0$.

In this case, is $L^{\infty}(\mu)$ the dual of $L^1(\mu)$?

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    $\begingroup$ @emptymind what do you want? I don't understand "including the form of the measure that is not sigma finite that is used" means $\endgroup$ Apr 20 '20 at 0:12
  • $\begingroup$ @mathworker21 I mean including why it is created like this. $\endgroup$
    – Emptymind
    Apr 20 '20 at 0:19
  • $\begingroup$ @Emptymind I don't know what you mean? Someone asked the question $\endgroup$ Apr 20 '20 at 0:24
  • $\begingroup$ @mathworker21 I mean How the OP created this measure and why it can be considered as not sigma finite measure …. anyway you can neglect this part if you do not know why …. no problem. $\endgroup$
    – Emptymind
    Apr 20 '20 at 0:28
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    $\begingroup$ @Emptymind ok, hopefully I added enough details. see below $\endgroup$ Apr 20 '20 at 0:47
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Nope, it isn't.

You can see this by showing that, in this setting, $L^\infty \cong \mathbb{C}^2$ and $[L^1]^* \cong \mathbb{C}$ (as vectorial spaces). Since $\mathbb{C}^2$ and $\mathbb{C}$ have different dimensions, they cannot be isomorphic, so the dual of $L^1$ is not $L^\infty$.

-#-#-#-#-#-#-#-#-#-#-#-#-

To see the isomorphisms, try to look at

$$\begin{array}{rrcl} \phi: & L^\infty &\longrightarrow & \mathbb{C}^2 \\ & f &\longmapsto& (f(a),f(b)) \end{array}$$ and $$\begin{array}{rrcl} \psi: & L^1&\longrightarrow & \mathbb{C} \\ & f &\longmapsto& f(a). \end{array}$$ Two more hints are:

  • Any finite dimensional space is isomorphic to its dual;
  • In your setting, a function $f\in L^1$ is completely determined by its value in $a$.
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Question 1: How did OP create $\mu$?

Answer: By writing it down.

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Problem 2: Prove $\mu$ is not $\sigma$-finite.

Solution: If $\mu$ were $\sigma$-finite, then $X = \cup_{n=1}^\infty A_n$ with each $A_n$ measurable and $\mu(A_n)<\infty$. Some $A_n$ would have to contain $b$, but it can't since $\mu(A_n) < \infty$. So $\mu$ is not $\sigma$-finite.

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Claim 3: $\mathbb{C}^2$ and $\mathbb{C}$ are not isomorphic as vector spaces.

Proof: They have different dimensions.

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Proposition 4: $L^\infty(X) \cong \mathbb{C}^2$ as vector spaces.

Proof: Define $\Phi: L^\infty(X) \to \mathbb{C}^2$ by $\Phi(f) = (f(a),f(b))$. Note $\Phi$ indeed maps into $\mathbb{C}^2$, since $f \in L^\infty$ implies $|f(a)|,|f(b)| < \infty$ since each $a,b$ get positive measure. Clearly $\Phi$ is linear and injective. And it is clearly surjective. So it is an isomorphism.

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Lemma 5: $L^1(X) \cong \mathbb{C}$ as vector spaces.

Proof: Define $\Phi: L^1(X) \to \mathbb{C}$ by $\Phi(f) = f(a)$. Note $\Phi$ indeed maps into $\mathbb{C}$, since $|f(a)| < \infty$ since $a$ has positive measure. Clearly $\Phi$ is linear and surjective. It is injective since any $f \in L^1(X)$ must have $f(b) = 0$, since $b$ has infinite measure.

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Fact 6: Let $V$ be a finite dimensional vector space. Then $V^* \cong V$ as vector spaces.

Proof: Google.

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Corollary 7: $(L^1(X))^* \cong \mathbb{C}$ as vector spaces.

Proof: Use Lemma 5 and Fact 6.

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Theorem 8: $L^\infty(X)$ is not the dual of $L^1(X)$.

Proof: What this means is that $L^\infty(X)$ is not isomorphic to $(L^1(X))^*$ as vector spaces. If they were isomorphic, then by Proposition 4 and Corollary 7, we'd have $\mathbb{C}^2 \cong \mathbb{C}$ as vector spaces, contradicting Claim 3. $\square$

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  • $\begingroup$ Does this measure satisfy the definition of being a measure? $\endgroup$
    – Emptymind
    Apr 20 '20 at 4:34
  • $\begingroup$ How is $ \mu(X) = \infty $ even though $X$ contains 2 elements only ? ….. I think $\mu$ is not the counting measure ..... am I correct? $\endgroup$
    – Emptymind
    Apr 20 '20 at 4:36
  • $\begingroup$ it's a measure. $\mu(\{b\}) = \infty$ and thus $\mu(X) = \infty$. See en.wikipedia.org/wiki/Measure_(mathematics)#Definition . Infinite values are fine. $\mu$ is not the counting measure $\endgroup$ Apr 20 '20 at 4:45
  • $\begingroup$ can this example be used to show what I want to show in the other question? $\endgroup$
    – Emptymind
    Apr 20 '20 at 4:48
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    $\begingroup$ @Emptymind no, I doubt you're stupid. take time to think about your questions. if you have thought about them for a while (and know all of the involved definitions completely), feel free to ask me $\endgroup$ Apr 20 '20 at 15:44

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