Is there any way, to make continuous function with some sharp edges smooth function? for example if i consider a function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x)=|x|$, this function is continuous but NOT smooth since it has a sharp edge at "0". can we make functions of such kind smooth?
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1It is not clear what you mean by "making the function smooth". Just so we are on the same page, the graph of a non-smooth function need not have sharp edges, consider $ f(x) = |x|^3 $. – Ege Erdil May 11 '16 at 13:25
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1How smooth are we talking? What derivative has to be continuous? And... what do you mean by "make smooth"? If you change it, it's not the same function anymore. But you can create any number of smooth function sequences with this function as a limit. – orion May 11 '16 at 13:28
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Over a finite interval create its fourier series to as many terms as you are happy with. – Paul May 11 '16 at 13:41
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Sharp corners in absolute values can be made smooth with $|x|=\lim_{k\to\infty}\ln(e^{kx}+e^{-kx})/k$. – orion May 11 '16 at 13:52
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i just want to know is there any way to remove that sharp edge and make function infinitely many times differentiable that is smooth $C^{\infty}$ – maths_freak_007 May 11 '16 at 14:08
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Answers may depend on the need. For optimization purposes, the $x\to |x|$ function (corresponding to the $\ell_1$ norm, useful for robust or sparse restoration) can be smoothed at the apex with techniques related to the Huber loss function. Examples are: $$L_\delta (x) = \delta^2 \left(\sqrt{1+(x/\delta)^2}-1\right)$$ or $$ L_\delta (a) = \begin{cases} \frac{1}{2}{x^2} & \text{for } |x| \le \delta, \\ \delta (|x| - \frac{1}{2}\delta), & \text{otherwise.} \end{cases} $$ Resultingly, the function behaves like $x^2$ around $0$, and inherits its smmoothness. Many more have been used in the optimization literature (related to the technique of majorization).
Laurent Duval
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