Let \(V\) be the vector space of absolutely continuous functions on \([0,1]\) endowed with the norm \[ \|f\|:=\int_0^1|f(t)|\,dt+\int_0^1|f'(t)|\,dt,\qquad f\in V. \] Show that \((V,\|\cdot\|)\) is a Banach space. You do not need to prove that \(\|\cdot\|\) is a norm.
Solution
1. Show that the norm controls the value at the left endpoint.
Let \(f\in V\). For every \(x\in[0,1]\), absolute continuity gives \[ f(0)=f(x)-\int_0^x f'(t)\,dt. \] Hence \[ |f(0)|\le |f(x)|+\int_0^1|f'(t)|\,dt. \] Integrating this inequality in \(x\) over \([0,1]\) yields \[ |f(0)| \le \|f\|_{L^1}+\|f'\|_{L^1} = \|f\|. \]
2. Extract limits of the initial values and derivatives.
Let \((f_n)\) be Cauchy in \(V\). Then \[ \|f_n'-f_m'\|_1\le \|f_n-f_m\|, \] so \((f_n')\) is Cauchy in \(L^1([0,1])\). Since \(L^1\) is complete, there exists \(g\in L^1([0,1])\) such that \[ f_n'\to g \quad\text{in }L^1. \]
By Step 1, \[ |f_n(0)-f_m(0)| \le \|f_n-f_m\|, \] so \((f_n(0))\) is Cauchy in \(\mathbb R\). Let \[ f_n(0)\to c. \]
3. Reconstruct the candidate limit.
Define \[ f(x):=c+\int_0^x g(t)\,dt. \] Then \(f\) is absolutely continuous on \([0,1]\) and \[ f'=g \] almost everywhere.
Also, since \[ f_n(x)=f_n(0)+\int_0^x f_n'(t)\,dt, \] we have for every \(x\in[0,1]\), \[ |f_n(x)-f(x)| \le |f_n(0)-c|+\|f_n'-g\|_1. \] Therefore \[ \|f_n-f\|_\infty \le |f_n(0)-c|+\|f_n'-g\|_1 \longrightarrow0. \] In particular, \[ \|f_n-f\|_1\to0. \] Together with \(f_n'\to g=f'\) in \(L^1\), this gives \[ \|f_n-f\| = \|f_n-f\|_1+\|f_n'-f'\|_1 \longrightarrow0. \] Thus every Cauchy sequence in \(V\) converges in \(V\), so \[ \boxed{(V,\|\cdot\|)\text{ is a Banach space}.} \]