Show that $L^{\infty}(\Omega)^* \neq L^1(\Omega)$ for any nonempty open set $\Omega \subset$ $\mathbb{R}^d$.

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Homework Question: Show that $L^{\infty}(\Omega)^* \neq L^1(\Omega)$ for any nonempty open set $\Omega \subset$ $\mathbb{R}^d$.

My Solution:

To prove that the dual space of $L^{\infty}(\Omega)$, denoted $L^{\infty}(\Omega)^*$, is not equal to $L^1(\Omega)$, we approach this by finding a continuous linear functional on $L^{\infty}(\Omega)$ that cannot be represented by an element of $L^1(\Omega)$.

Recall:

  • $L^{\infty}(\Omega)$ is the space of essentially bounded measurable functions on $\Omega$.
  • $L^1(\Omega)$ is the space of integrable functions on $\Omega$.
  • $L^{\infty}(\Omega)^*$ consists of all continuous linear functionals on $L^{\infty}(\Omega)$.

Step 1: Duality between $L^1(\Omega)$ and $L^{\infty}(\Omega)$ For functions $f \in L^1(\Omega)$ and $g \in L^{\infty}(\Omega)$, the functional $\Phi_g$ is defined as $\Phi_g(f)=\int_{\Omega} f(x) g(x) d x$.

This represents how $L^1(\Omega)$ naturally induces functionals on $L^{\infty}(\Omega)$.

Step 2: Seeking a Counterexample By the Hahn-Banach Theorem, $L^{\infty}(\Omega)^*$ contains more than just the functionals induced by $L^1(\Omega)$. We seek a functional in $L^{\infty}(\Omega)^*$ not expressible in the form of $\Phi_g$ for any $g \in L^1(\Omega)$.

Step 3: Constructing a Counterexample Consider a measure $\mu$ on $\Omega$ that is singular with respect to Lebesgue measure. Define a functional $\Phi$ on $L^{\infty}(\Omega)$ by $$ \Phi(f)=\int_{\Omega} f d \mu \text {. } $$

This functional is linear and well-defined on $L^{\infty}(\Omega)$, but cannot be represented by any $g \in L^1(\Omega)$, as $\mu$ is singular.

Conclusion Since we have identified a functional in $L^{\infty}(\Omega)^*$ that cannot be represented by any element in $L^1(\Omega)$, it follows that $L^{\infty}(\Omega)^* \neq L^1(\Omega)$. This result highlights the complexity of dual spaces in functional analysis, demonstrating that $L^{\infty}(\Omega)^*$ is strictly larger than $L^1(\Omega)$.

-- I would love if you can proofread my solution and let me know if I have done everything correctly! :)