Let \(X\) be the set of functions \(f:[0,1]\to\mathbb R\) for which \[ \|f\|:=|f(0)|+\sup_{x\ne y}\frac{|f(x)-f(y)|}{|x-y|^{1/5}}<\infty. \] Prove that \((X,\|\cdot\|)\) is a Banach space.
Solution
Write \[ [f]_{1/5}:=\sup_{x\ne y}\frac{|f(x)-f(y)|}{|x-y|^{1/5}}. \] Then \[ \|f\|=|f(0)|+[f]_{1/5}. \]
1. \(X\) is a vector space and \(\|\cdot\|\) is a norm.
If \(f,g\in X\) and \(a,b\in\mathbb R\), then \[ [af+bg]_{1/5}\le |a|[f]_{1/5}+|b|[g]_{1/5}<\infty, \] so \(af+bg\in X\). The seminorm \([\cdot]_{1/5}\) is homogeneous and subadditive, hence so is \(\|\cdot\|\).
If \(\|f\|=0\), then \(f(0)=0\) and \([f]_{1/5}=0\), so \(f(x)=f(y)\) for all \(x,y\). Thus \(f\) is constant, and since \(f(0)=0\), one has \(f\equiv0\). Therefore \(\|\cdot\|\) is a norm.
2. The norm controls uniform convergence.
For every \(f\in X\) and \(x\in[0,1]\), \[ |f(x)|\le |f(0)|+|f(x)-f(0)| \le |f(0)|+[f]_{1/5}|x|^{1/5} \le \|f\|. \] Hence \[ \|f\|_\infty\le\|f\|. \]
3. \(X\) is complete.
Let \((f_n)\) be Cauchy in \(\|\cdot\|\). By 2, it is uniformly Cauchy, so there is a function \(f:[0,1]\to\mathbb R\) such that \[ f_n\to f \] uniformly.
Fix \(\varepsilon>0\). Since \((f_n)\) is Cauchy in \(\|\cdot\|\), there exists \(N\) such that for all \(m,n\ge N\), \[ |f_n(0)-f_m(0)|+[f_n-f_m]_{1/5}<\varepsilon. \] Fix \(n\ge N\) and let \(m\to\infty\). Uniform convergence gives \[ |f_n(0)-f(0)|\le\varepsilon, \] and for every \(x\ne y\), \[ |(f_n-f)(x)-(f_n-f)(y)| =\lim_{m\to\infty}|(f_n-f_m)(x)-(f_n-f_m)(y)| \le\varepsilon |x-y|^{1/5}. \] Therefore \[ [f_n-f]_{1/5}\le\varepsilon, \] so \[ \|f_n-f\|\le2\varepsilon. \] In particular \(f_n\to f\) in \(\|\cdot\|\). Since \(f=f_n-(f_n-f)\) and both terms have finite Hölder seminorm, \(f\in X\). Thus every Cauchy sequence converges in \(X\), and \(X\) is Banach.