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We are given a function $f: \mathbb R^d \to \mathbb R$. For simplicity we can assume that $f$ is smooth and compactly supported. Is the Sobolev norm of order $\frac{1}{2}$ strong enough to prove an inequality like this $$|f(a)-f(b)| \leq \| f \|_{H^{\frac 1 2}}$$ or a variant, where the left side depends only on the difference in function values and the right hand side on the norm.

"Jumps" is not the exact term, since $H^{\frac 1 2}$-functions can't have jump discontinuities, but is the norm able to measure, by how much the function values change?

For $H^s$ and $0 \leq s < \frac 1 2$ this is not the case. Characteristic functions of bounded sets lie in $H^s$ and when the measure of the set goes to 0, so does the $H^s$-norm. On the other hand, this is true for the $H^1$-norm, because of $f(b) - f(a) = \int_a^b f(t)dt$.

At which order does the behaviour switch?

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up vote 1 down vote accepted

You need s>d/2 for this to hold. You will find this in any text on Sobolev spaces.

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Thank you. I had forgotten about this. – Martins Bruveris Mar 30 '11 at 8:41

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