Kolgomorov's inequality

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Kolmogorov's inequality is an inequality which gives a relation among a function and its first and second derivatives. Kolmogorov's inequality states the following:

Let f \colon \mathbb{R} \rightarrow \mathbb{R} be a twice differentiable function on \mathbb{R} such that f\, and f'' \, are bounded on \mathbb{R}. Denote

M_0 = \sup_{x\in\mathbb{R}} |f(x)|,\ M_1 = \sup_{x\in\mathbb{R}} |f'(x)|,\ M_2 = \sup_{x\in\mathbb{R}} |f''(x)|.

Then, f' \,\! is bounded on \mathbb{R} and M_1 \le \sqrt{2M_0M_2}.

[edit] Proof

The proof of this inequality uses Taylor's theorem.

Let a \in \mathbb{R}_+^*, x \in \mathbb{R}. Apply the Taylor-Lagrange Inequality to f \,\! on the intervals [x-a,x] \,\! and [x,x+a] \,\! and obtain


\begin{cases}
|f(x-a)-(f(x)-af'(x))| \le \frac{a^2}{2}M_2\\
|f(x+a)-(f(x)+af'(x))| \le \frac{a^2}{2}M_2.
\end{cases}

from which

|f(x+a)-f(x-a)-2af'(x)| \,\!

\begin{alignat}{2}
&=|(f(x+a)-(f(x)+af'(x)))-(f(x-a)-(f(x)-af'(x)))|\\
&\le a^2M_2,\\
\end{alignat}

so that

|2af'(x)| \le |f(x+a)-f(x-a)|+a^2M_2 \le 2M_0+a^2M_2.

Hence,

M_1 \le \frac{M_0}{a}+\frac{1}{2}aM_2 \le \sqrt{2M_0M_2},

where we have used the AM-GM inequality in the last step.

[edit] References

  • Serge Francinou, Hervé Gianella, Serge Nicolas (2003). Exercices de Mathématiques Oraux X-ENS. Cassini, Paris. ISBN 2-8425-032-X.