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Full Question:

The map $L^{1}(A) \to \mathbb{C}$, $f \mapsto \int gf$ is linear, and continuous when $g \in L^{\infty}(A)$. Assuming surjectivity, show $L^{1}(K)^*$ is isometrically isomorphic to $L^{\infty}(K)$ for $K \subseteq \mathbb{R}$ is compact.

My attempt:

Let $K$ be a compact subset of $\mathbb{R}$ and define $\phi : L^{1}(K) \to \mathbb{C}$ where $f \mapsto \int gf$ be linear and continuous with $g \in L^{\infty}(K)$. Then, the inequality: $$\|\ \phi \|\ = \left| \int_K fg \right| \leq \int_K |f| |g| \leq \sup_{x \in K \space\ \text{a.e.}} |f(x)| \int_K |g| = \|\ f\|_{L^{\infty}(K)} \|\ g \|_{L^{1}(K)}$$ establishes that $\|\ \phi \|\ \leq \|\ f \|_{L^{\infty}(K)}$.

I'm having trouble obtaining $ \|\ \phi \|\ \geq \|\ f \|_{L^{\infty}(K)}$ in order to show equality to show the isometry part. I can complete the rest of the proof from there for the isomorphism. Any help on establishing the second inequality would be much appreciated.

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    What measure are you working with? If it's Lebesgue measure, then is the measure of $K$ strictly positive? If we allow the measure of $K$ to be zero (which is certainly possible), bad things can happen. – 2017-02-07
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    @Aweygan - if $K$ has measure $0$, then the map is not surjective, contrary to the given assumption. So we can assume that $K$ does not have measure $0$. – 2017-02-08
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    In your proof, you have reversed the spaces of $f$ and $g$. It is $f \in L^1(K)$ and $g \in L^\infty(K)$, not the other way around. – 2017-02-08
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    Also, your proof assumes there is some $f\in L^1$ for which $\phi$ obtains its norm, which is not always possible (see James' theorem) – 2017-02-08

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Start by being a little bit more clear which maps you are talking about: For each $g \in L^\infty(K)$, define $$\phi_g : L^1(K) \to \Bbb C : f \mapsto \int_K gf$$ As is given in the question, for each $g$, $\phi_g$ is continuous and linear, so $\phi_g \in L^1(K)^*$. The isomorphism we are after is $$\Phi : L^\infty(K) \to L^1(K)^* : g \mapsto \phi_g$$ You are given that $\Phi$ is surjective, so you only need to show that it is linear and an isometry.

By definition, $$\|\phi_g\| = \sup\{\left|\phi_g(f)\right| : f \in L^1(K), \|f\| = 1\}\\= \sup\left\{\left|\int fg\right| : f \in L^1(K), \|f\| = 1\right\}$$ Now as you've calculated, $\left|\int fg\right| \le \|g\|\|f\| = \|g\|$. Hence $\|\phi_g\| \le \|g\|$.

To go the other way, you need show that there are $f \in L^1(K)$ with $\|f\| = 1$ that make $\left|\int fg\right|$ as close to $\|g\|$ as you like.

For that, consider the set $$Q_\epsilon = \{k \in K : \|g\|\ - |g(k)| < \epsilon\|g\| \}$$ for $\epsilon > 0$.

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    So I want to show that $| \|\ g \|\ - |\int fg| | < \epsilon$. To accomplish this: $| \|\ g \|\ - | \int_K fg | | \leq | \|\ g \|\ - \sup_{k \in K} |f(k)| \int_{K} |g| | = | \|\ g \|\ - \|\ f \|\ \|\ g \|\ | = | \|\ g \|\ ( 1 - \|\ f \|\ ) | $. But, $ \|\ f \|\ = 1$, so this is $0$ so its less than $\epsilon$? How do I use $Q_{\epsilon}$? – 2017-02-08
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    You are treating $f$ as if it were in $L^\infty$ and $g$ as if it were in $L^1$ again. Remember, it is the other way around. Even if they were, that first inequality is false. – 2017-02-09
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    You are not looking to prove that $| \|\ g \|\ - |\int fg| | < \epsilon$ holds for every $f$, but rather that there exists some $f$ for which it is true. The hint is that you can use $Q_\epsilon$ to construct such an $f$. Consider the characteristic function of $\overline Q_\epsilon$. It isn't quite what you need, but it doesn't take much to get the rest of the way. – 2017-02-09