Let \(C([0,1])\) be the space of continuous real-valued functions on \([0,1]\).
Prove that \(C([0,1])\) is complete under the uniform norm \[ \|f\|_\infty:=\sup_{x\in[0,1]}|f(x)|. \]
Prove that \(C([0,1])\) is not complete under the \(L^1\) norm \[ \|f\|_1:=\int_0^1|f(x)|\,dx. \]
Solution
1. Completeness in the uniform norm.
Let \((f_n)\) be Cauchy in \(\|\cdot\|_\infty\). For each \(x\in[0,1]\), the real sequence \((f_n(x))\) is Cauchy, so define \[ f(x):=\lim_{n\to\infty}f_n(x). \] Given \(\varepsilon>0\), choose \(N\) such that \[ \|f_n-f_m\|_\infty<\varepsilon \qquad(n,m\ge N). \] Fix \(n\ge N\) and let \(m\to\infty\). Then \[ |f_n(x)-f(x)|\le\varepsilon \] for every \(x\), hence \[ \|f_n-f\|_\infty\le\varepsilon. \] Thus \(f_n\to f\) uniformly. A uniform limit of continuous functions is continuous, so \(f\in C([0,1])\) and the space is complete.
2. Failure of completeness in the \(L^1\) norm.
Let \[ g=\mathbf1_{[1/2,1]}. \] For \(n\ge2\), define \(f_n\in C([0,1])\) by \[ f_n(x)= \begin{cases} 0,&x\le \frac12-\frac1n,\\ \frac n2\left(x-\frac12+\frac1n\right),&\frac12-\frac1n<x<\frac12+\frac1n,\\ 1,&x\ge \frac12+\frac1n. \end{cases} \] Then \(0\le f_n\le1\) and \(f_n=g\) outside an interval of length \(2/n\). Therefore \[ \|f_n-g\|_1\le\frac2n\to0. \] Hence \((f_n)\) is Cauchy in the \(L^1\) norm.
If \(C([0,1])\) were complete in that norm, there would exist \(h\in C([0,1])\) with \[ \|f_n-h\|_1\to0. \] But limits in \(L^1\) are unique up to almost-everywhere equality, so \(h=g\) almost everywhere. A continuous function equal almost everywhere to \(g\) must be \(0\) on \([0,1/2)\) and \(1\) on \((1/2,1]\), contradicting continuity at \(1/2\). Thus no such \(h\) exists.
Therefore \(C([0,1])\) is not complete under \(\|\cdot\|_1\).