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.