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Given a continuous piecewise linear function from the reals to the reals, can it be expressed as a sum of terms of form $k | ax+b |$ ? [assuming a finite number of pieces]

If some piecewise functions are not so expressible, is there a simple description of the cases where there is such an expression?

Following John's hint:

Let $f(x)=\sum_{i=0}^Nk_i|a_ix+b_i|$

Then on the rightmost piece, the derivative of $f$ is $A=\sum_{i=0}^Nk_ia_i$ while on the leftmost piece it is $-A$. This means we cannot obtain piecewise linear functions which increase at both ends (or decrease at both). I still wonder about if there is a good description of those which can be obtained.

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    $\begingroup$ Are you allowed to add additional terms of the form $ax+b$? Otherwise I don't think you can ever construct a 2 piece function that isn't symmetric about the corner $\endgroup$ Commented Aug 29 at 20:28
  • $\begingroup$ What exactly do you want to achieve? With the right transformations, I don't see any issue. However, you must ensure that the transformation is performed in a way that the function remains continuous. If you have a function in mind, please share it. $\endgroup$ Commented Aug 29 at 20:28
  • $\begingroup$ @CroCo I only mean a sum of terms k|ax+b|. I am not using "transforms". I specified in the question what I wanted to achieve. $\endgroup$ Commented Aug 29 at 21:26

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My hint is too long for a comment, so bear with me if I write it here in an effort to guide you in answering your own question.

Let $ f(x)=\sum_{i=0}^Nk_i|a_ix+b_i|$.

Can you show that either $\lim_{x\to\infty} f(x)=\infty$ or $\lim_{x\to\infty} f(x)=-\infty$ and the same for $f(-x)$?

Now, try to show that $\left|k|ax+b|-k|-ax+b|\right|$ is bounded.

Then show that $g(x)=|f(x)-f(-x)|$ is bounded.

Now show

$$ \lim_{x\to\infty} f(x)=\lim_{x\to\infty} f(-x)$$

Now suppose that you have a finitely piecewise continuous function from the reals onto the reals which is increasing on its leftmost interval and increasing on its rightmost interval. Can it be expressed in the form of the function $f(x)$ above?

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  • $\begingroup$ This looks promising, thanks. $\endgroup$ Commented Aug 30 at 1:48
  • $\begingroup$ A piecewise linear function can be expressed as a sum of Heavide functions. They are a lot of examples in fr.scribd.com/document/380941024/… $\endgroup$ Commented Aug 30 at 9:06
  • $\begingroup$ Yes, one would write it as a Heaviside function for example, if you wanted to find the Laplace transform. $\endgroup$ Commented Aug 30 at 13:23
  • $\begingroup$ @JJacquelin I looked at your source. Thank you. My question as posed has already been answered here. $\endgroup$ Commented Aug 30 at 19:44
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If $f$ has $n$ singular points $x_1,\dots,x_n$ with $\lim_{x\to x_i^-}f'(x)=\alpha_i\neq\beta_i=\lim_{x\to x_i^+}f'(x)$ then you can remove all singular points with $$g(x)=f(x)-\sum_{i=1}^n\frac{\beta_i-\alpha_i}2|x-x_i|$$

You can check immediately if $g$ is constant, by the necessary and sufficient condition that the slope of $f$ beyond its greatest singular point is equal to $\frac 12\sum_i(\beta_i-\alpha_i)$. And you can check that the constant is zero with a direct calculation at any real.

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  • $\begingroup$ I was only looking at continuous piecewise linear functions [so no singularities] and wondered about expressing them via absoloute value sums. --- or were you considering f' ? And what is $x_0$? Typo for $x_i$? $\endgroup$ Commented Aug 30 at 10:13
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    $\begingroup$ By singularity, I mean "not differentiable". But yes, I forgot to write the prime symbol. $\endgroup$ Commented Aug 30 at 10:25

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