Let \(p\in[1,\infty)\), and suppose that \(f:\mathbb{R}\to\mathbb{R}\) is an absolutely continuous function such that \(\|f'\|_p<\infty\).
Prove that \(f\) is Hölder continuous with exponent \(\dfrac{p-1}{p}\), i.e. there is a constant \(C\) such that \[ |f(x)-f(y)|\le C|x-y|^{\frac{p-1}{p}}. \]
Write down the corresponding statement when \(p=\infty\) and prove your assertion.
Solution
1. Prove the estimate for finite \(p\).
Let \(x<y\). Since \(f\) is absolutely continuous, \[ f(y)-f(x)=\int_x^y f'(t)\,dt. \] Hence \[ |f(y)-f(x)| \le \int_x^y |f'(t)|\,dt. \] If \(1<p<\infty\), let \(q=p/(p-1)\). Hölder’s inequality gives \[ \begin{aligned} |f(y)-f(x)| &\le \left(\int_x^y |f'(t)|^p\,dt\right)^{1/p} \left(\int_x^y1^q\,dt\right)^{1/q}\\ &\le \|f'\|_p\,|y-x|^{1/q}\\ &= \|f'\|_p\,|y-x|^{(p-1)/p}. \end{aligned} \] Thus \(f\) is Hölder continuous of exponent \((p-1)/p\), with \(C=\|f'\|_p\).
When \(p=1\), the same fundamental-theorem estimate gives \[ |f(y)-f(x)| \le \int_x^y|f'(t)|\,dt \le \|f'\|_1, \] which is precisely the exponent-\(0\) form of the displayed estimate.
2. Handle \(p=\infty\).
Assume \(f'\in L^\infty(\mathbb R)\). Then for \(x<y\), \[ \begin{aligned} |f(y)-f(x)| &\le \int_x^y |f'(t)|\,dt\\ &\le \|f'\|_\infty |y-x|. \end{aligned} \] Therefore \(f\) is Lipschitz continuous, i.e. Hölder continuous of exponent \(1\), with Lipschitz constant at most \(\|f'\|_\infty\).