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If for a function $f:\mathbb{R}\rightarrow\mathbb{R}$ it is given that it satisfies a Lipschitz condition

$\big|f(x)-f(y)\big| \le L\big|x-y\big|$, for all $x,y\in\mathbb{R}$,

can we say anything about the bound of $|f(x)|$. Can it be guaranteed to satisfy $|f(x)| \le a|x| + b$, for some $a,b>0$? Intuitively it should, as the derivative of the function is finite at all points, we should be able to find a line (no matter how steep) such that $f(x)$ is under the line everywhere. But how do I prove it?

Also, what about the inverse: Given that $|f(x)| \le a|x| + b$ is satisfied for all $x\in\mathbb{R}$, can we say anything about the Lipschitz condition, either locally or globally?

Panagiotis Panagi
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1 Answers1

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Using MVT won't quite work since Lipschitz functions need not be differentiable $f(x)=|x|$ for example.

But, your intuition is correct - globally Lipschitz does imply linear growth. If $|f(x)-f(y)|\leq L|x-y|$ for all $x, y\in\mathbb{R}$, we may as well let $y=0$ and $x$ be arbitrary, then we get $$|f(x)-f(0)|\leq L|x|$$ so by the triangle inequality $$|f(x)|\leq |f(0)|+L|x|$$ notice that $f(0)$ is just a constant, so $f$ is indeed of linear growth.

The converse however is not correct; for example take $f(x)=1, x\in\mathbb{Q}$ and $0$ otherwise. This function is discontinuous everywhere, yet $|f(x)|\leq 1+|x|$ so it is still of linear growth.

Tyr Curtis
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