Show that \(L^2([0,1])\subseteq L^1([0,1])\) and that \(L^2([0,1])\) is dense in \(L^1([0,1])\).
Let \(\Lambda\) be a continuous linear functional on \(L^1([0,1])\).
Using the Riesz representation theorem for \(L^2([0,1])\), prove that there is \(g\in L^2([0,1])\) such that \[ \Lambda(f)=\int_0^1 f(x)\overline{g(x)}\,dx \qquad(f\in L^2([0,1])). \]
Prove that \(g\in L^\infty([0,1])\), that the same formula holds for every \(f\in L^1([0,1])\), and that \[ \|g\|_\infty=\|\Lambda\|_{(L^1)^*}. \]
Solution
1. Inclusion and density.
By Cauchy–Schwarz, \[ \|f\|_1 =\int_0^1|f|\cdot1 \le \|f\|_2\|1\|_2 =\|f\|_2. \] Hence \(L^2\subseteq L^1\) continuously.
Bounded measurable functions lie in \(L^2([0,1])\), and bounded simple functions are dense in \(L^1([0,1])\). Therefore \(L^2([0,1])\) is dense in \(L^1([0,1])\).
2. Restrict \(\Lambda\) to \(L^2\) and apply Riesz.
For \(f\in L^2\), \[ |\Lambda(f)|\le \|\Lambda\|\,\|f\|_1 \le \|\Lambda\|\,\|f\|_2. \] Thus \(\Lambda|_{L^2}\) is a bounded linear functional on the Hilbert space \(L^2([0,1])\). By the Riesz representation theorem, there is a unique \(g\in L^2([0,1])\) such that \[ \Lambda(f)=\int_0^1 f\overline g \qquad(f\in L^2). \]
3. Prove that \(g\in L^\infty\) and \(\|g\|_\infty\le\|\Lambda\|\).
Let \(M:=\|\Lambda\|\). Suppose for contradiction that \[ m(E)>0, \qquad E:=\{x:|g(x)|>M+\varepsilon\} \] for some \(\varepsilon>0\). Since \(E\subset[0,1]\), the function \[ h(x):=\frac{1}{m(E)}\frac{g(x)}{|g(x)|}\mathbf1_E(x) \] belongs to both \(L^1\) and \(L^2\), and \(\|h\|_1=1\). Using the \(L^2\) representation, \[ \begin{aligned} |\Lambda(h)| &=\left|\frac1{m(E)}\int_E |g(x)|\,dx\right|\\ &>M+\varepsilon, \end{aligned} \] contradicting \(|\Lambda(h)|\le M\|h\|_1=M\). Hence \[ |g|\le M\quad\text{a.e.}, \] so \(g\in L^\infty\) and \[ \|g\|_\infty\le\|\Lambda\|. \]
4. Extend the representation to every \(L^1\) function.
Let \(f\in L^1\). Choose \(f_n\in L^2\) with \[ \|f_n-f\|_1\to0. \] Continuity of \(\Lambda\) gives \[ \Lambda(f_n)\to\Lambda(f). \] Since \(g\in L^\infty\), \[ \left|\int (f_n-f)\overline g\right| \le \|f_n-f\|_1\|g\|_\infty\to0. \] Therefore \[ \Lambda(f)=\int_0^1 f\overline g \] for every \(f\in L^1\).
5. Prove equality of the norms.
The representation and Hölder’s inequality give \[ |\Lambda(f)|\le\|f\|_1\|g\|_\infty, \] so \[ \|\Lambda\|\le\|g\|_\infty. \] Combined with Step 3, \[ \|g\|_\infty\le\|\Lambda\|, \] we obtain \[ \boxed{\|g\|_\infty=\|\Lambda\|.} \]