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Given $f:\mathbb{R}\rightarrow \mathbb{R}$ continuous and a fixed $\delta$, define $$f_{\delta}(x):=\int_{x-\delta}^{x+\delta}f(\xi)\,\mathrm{d}\xi.$$

$f_{\delta}$ behaves like the average of $f$ in a short interval $(x-\delta,x+\delta)$.

Apparently it is not linear, but it should be Lipschitz continuous.

Without loss of generality, we may assume $x<y$.

From mean value theorem,

$$\frac{|f_{\delta}(y)-f_{\delta}(x)|}{|y-x|}=f(s),$$ for some $s\in[x,y]$.

However this only shows local Lipschitz continuity. Since $f$ is defined in $\mathbb{R}$, it may not be bounded.

How to achieve global one?

newbie
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2 Answers2

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Take $f(x) = 3x^2$ so that $f_\delta (x) = (x + \delta)^3 - (x-\delta)^3 = 6 \delta x^2 + 2\delta^3$ which is not even uniformly continuous on $\mathbb{R}$, so definitely not Lipschitz.

Chris Janjigian
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In general it is not globally Lipschitz, but it is locally Lipschitz. Consider your function in a bounded interval, e.g., in $[0,1]$. Then you try to bound $|f_\delta(x)-f_\delta(y)|$ by a multiple of $|x-y|$ when $|x-y|<\delta$. For $|x-y|\geq\delta$, you can use the trivial bound $$ |f_\delta(x)-f_\delta(y)| \leq |x-y|\cdot\delta^{-1}\max_{0\leq z\leq 1}|f_\delta(z)-f_\delta(z+\delta)|. $$ The meaning of this is that the locally Lipschitz property depends only on how the function varies locally, and it does not care how the far away points relate to each other.

As for the globally Lipschitz property, it is no longer only about local regularity, but it is also about global growth. By a local averaging procedure you cannot, in general, change the growth behaviour of a function.

timur
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