An Elementary Duality Result

The following problem is a fairly straightforward exercise that seems to come up fairly often on qualifying exams at UCR, and I think that I finally have a nicely argued proof. I am going to put it up here mostly for my own edification. Please note that the following is intended largely for an audience of advanced undergraduate or graduate students. I will be making no effort to give all of the required background, and assume some knowledge of the theory of Lebesgue integration and $$L^p$$-spaces.

Exercise: For each $$g\in L^1([0,1])$$, define a bounded linear functional $$\varphi_g$$ by $\varphi_g(f) = \int fg.$ We know that the map $$g\mapsto \varphi_g$$ is an isometric injection of $$L^1([0,1])$$ into $$(L^\infty([0,1]))^*$$. Show that this map is not surjective.

Explanation: Basically, we have the result that for any $$1 < p < \infty$$, $$(L^p)^*$$—the dual of $$L^p$$—is isometric to $$L^q$$, where $$1/p + 1/q = 1$$. That is, with a little bit of handwaving about identifying spaces via isometries, we can state that if $$p\in(0,\infty)$$ and $$1/p + 1/q = 1$$, then $$(L^p)^*$$ is $$L^q$$. What we want to show is that this result does not (generally) hold for $$p=\infty$$, i.e. that $$(L^\infty)^*$$ is almost always strictly larger than $$L^1$$.

Solution: The goal is to explicitly construct bounded linear functional $$\varphi:L^\infty([0,1])\to\mathbb{R}$$ such that $$\varphi\ne\varphi_g$$ for any $$g\in L^1([0,1])$$. We start on a somewhat smaller space: define $$\varphi:C([0,1]) \to \mathbb{R}$$ by $$\varphi(f) = f(0)$$. We make the claims that (1) $$\varphi$$ is linear, and (2) $$\varphi$$ is dominated by the $$L^\infty$$ norm, i.e. if $$f\in C([0,1])$$, then $$\varphi(f)\le\|f\|_{\infty}$$.

Claim (1) is straight-forward: let $$f,g\in C([0,1])$$ and take $$\alpha\in\mathbb{R}$$ to be an arbitrary scalar. Then $\varphi(f+g) = (f+g)(0) = f(0) + g(0) = \varphi(f) + \varphi(g),$ and $\varphi(\alpha f) = (\alpha f)(0) = \alpha f(0) = \alpha \varphi(f),$ therefore $$\varphi$$ is a linear functional on $$C([0,1])$$, as claimed.

Claim (2) requires us to recall that the $$L^\infty$$ norm is equal to the uniform norm on the space of continuous functions. Moreover, the domain $$[0,1]$$ is compact, and so the uniform norm of a continuous function is the maximum magnitude that the function attains. Hence for any $$f\in C([0,1])$$, we have $\varphi(f) = f(0) \le |f(0)| \le \max_{x\in[0,1]} |f(x)| = \|f\|_u = \|f\|_{\infty}.$

With both claims proved, we apply the Hahn-Banach Theorem. Namely, we can extend $$\varphi$$ to a bounded linear functional $$\overline{\varphi}$$ on $$L^{\infty}([0,1])$$ such that $$\overline{\varphi}(f) = f(0)$$ for any $$f\in C([0,1])$$. Since there is no ambiguity, we will drop the overline and simply write $$\varphi := \overline{\varphi}$$.

It remains to show that there is no $$g\in L^1([0,1])$$ such that $$\varphi = \varphi_g$$. For contradiction, suppose that such $$g$$ does exist, and consider the sequence of functions $$\{f_n\}\subset C([0,1])$$ defined by $$f_n(x) = \max\{1-nx,0\}$$. On the one hand, note that $\varphi\left(\lim_{n\to\infty} f_n\right) = \lim_{n\to\infty} \varphi(f_n) = \lim_{n\to\infty} f_n(0) = 1.$ Hence by the construction of $$\varphi_g$$, we have $\lim_{n\to\infty} \int f_n g = 1. \tag{*}$

On the other hand, note that $$f_n$$ converges pointwise to the zero function. Moreover, $$f_n(x) \le 1$$ for all $$x\in[0,1]$$, and so $$(f_n g)(x) \le g(x)$$ for all such $$x$$. It then follows from the Dominated Convergence Theorem that $\lim_{n\to\infty} \int f_n g = \int \lim_{n\to \infty} f_n g = \int 0 = 0.$ This contradicts $$(*)$$, and so there cannot be any $$g\in L^1([0,1])$$ such that $$\varphi = \varphi_g$$, which completes the proof.

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