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Consider $f(x) = x^2$ and $g(x) = |x|$. Both graphs have an upward open graph, but $g(x) = |x|$ is "sharper". Is there a way to measure this sharpness?

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Sure: look at the derivative. If the derivative has a discontinuity, you'll see a sharp point. –  Clive Newstead Jan 7 '13 at 17:45
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2 Answers 2

This may be somewhat above your pay grade, but a "measure" of a discontinuity of a function at a point may be seen in a Fourier transform of that function. For example consider the function

$$f(x) = \exp{(-|x|)} $$

which is proportional to a Lorentzian function:

$$\hat{f}(w) = \frac{1}{1+w^2} $$

(I am ignoring constants, etc., which are not important to this discussion.) Note that $\hat{f}(w) \approx 1/w^2 (w \rightarrow \infty)$. The algebraic, rather than exponential, behavior at $\infty$ is characteristic of a type of discontinuity. In this case, there is a discontinuity in the derivative. For a straight discontinuity, there is a $1/w$ behavior at $\infty$. (Note the step function and its transform which is proportional to $1/w$ at $\infty$.) For a discontinuity in the 2nd derivative, there is a $1/w^3$ behavior at $\infty$. And a discontinutiy in the $k$th derivative of $f(x)$ translates into a $1/w^{k+1}$ behavior of the Fourier transform at $\infty$.

No, I do not have a proof of this, so I am talking off the cuff from my experiences some moons ago. But I am sure this is correct for the functions we see in physics.

Also note that I define the Fourier Transform here as

$$\hat{f}(w) = \int_{-\infty}^{\infty} dx \: f(x) \exp{(-i w x)} $$

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I think the notion which fits the bill is Hölder continuity.

Probably your example is not the most illustrative one (since $f$ is infinitely differentiable at 0). A function $f$ is $C^{k,\alpha}$ Hölder continuous (with $k\in\mathbb{N}$ and $\alpha\in[0,1]$) if it is $k$-times differentiable and for its $k$th derivative $f^{(k)}$ there exists a constant $C$ such that $$ \sup_{x\neq y}\frac{|f(x) - f(y)|}{|x-y|^\alpha}\leq C. $$

Anyway, $g(x) = |x|$ is in the space $C^{0,1}$, i.e. not differentiable but Lipschitz continuous. Sharper cusps, like $h(x) = \sqrt{|x|}$ have lower Hölder regularity ($C^{0,1/2}$ in this case).

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I suppose you forgot an exponent $\alpha$? –  Eckhard Jan 7 '13 at 19:51
    
Ooops! Corrected... –  Dirk Jan 7 '13 at 21:25
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