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Suppose I have an arbitrary FINITE function from $f: \mathbb{R} \rightarrow \mathbb{R}$. What could be the possible ways to measure the non-linearity of this function?

By FINITE, I mean $f(x) < \infty, \, \forall x \in \mathbb{R}$

Link to relevant material is highly appreciated. Thanks.

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    By you definition, every function $f \colon \mathbb R \to \mathbb R$ is "FINITE". – 2017-01-20
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    Do you want that $f$ is bounded, i.e. $f(x) \leq M$ for all $x$ and some $M$? Like would $f(x) = x$ be finite? – 2017-01-20
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    A function is a linear polynomial $f(x)=ax+b$ if and only if $f''(x)=0$ everywhere. So the value of $f''(x)^2$ can be used as a measure of nonlinearity. – 2017-01-20
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    Of course, good idea for twice differentiable functions. What to do with, for instance, broken lines? – 2017-01-20
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    Sorry, I wanted to convey that I am not considering functions like $\tan(x)$ which is $\infty$ or undefined at $x=\frac{\pi}{2}$. And yes $f''(x)^2$ can probably be used as a measure of nonlinearity given that the set of functions that we are considering are twice differentiable everywhere. – 2017-01-21

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The graph of a linear function is, of course, a straight line. If it passes through $0$, the function is a linear map of the form $x\mapsto ax$. If the origin does not lie on a line, we have an affine function $x\mapsto ax+b$. Let me describe this case. Let $f$ be an affine function. If could be easily checked that the following equation is fulfilled: $$f\bigl(tx+(1-t)y\bigr)=tf(x)+(1-t)f(y)$$ for any $x,y\in \Bbb R$ and any $t\in\Bbb R$. If we consider $t\in [0,1]$, we describe the segment with endpoints $\bigl(x,f(x)\bigr)$, $\bigl(y,f(y)\bigr)$. On any such segment we could measure how much $f$ differs from this chord. So, fix $x

What to do, if we consider the whole $\Bbb R$ as a domain? Maybe (possibly infinite) supremum taken over all $x,y\in\Bbb R$? I think about

$$ \sup\left\{\left|tf(x)+(1-t)f(y)-f\bigl(tx+(1-t)y\bigr)\right|\colon t\in[0,1],x,y\in\Bbb R\right\} $$

If it is infinite (as for parabola), the fuction is strongly nonlinear. If it is finite, then we have $$\left|tf(x)+(1-t)f(y)-f\bigl(tx+(1-t)y\bigr)\right|\le\varepsilon$$ for some $\varepsilon>0$, all $x,y\in\Bbb R$ and all $t\in[0,1]$. As I have proved in 1995 (together with Nikodem), in this case there is an affine function $\alpha(x)=ax+b$ close to $f$: $$|f(x)-(ax+b)|\le\frac{\varepsilon}{2}$$ for any $x\in\Bbb R$.

The reference to the paper: http://dx.doi.org/10.1007/BF01827935 It could be freely viewed at https://eudml.org/doc/182478

The 2nd named author is me.

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    Great answer. Thanks. Now I wonder whether it's possible to obtain a bounded function with maximum non-linearity in an interval based on your metric of non linearity i.e. the supremum value. So may be a formulation of the problem would be to find $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x) \leq M \, \, \forall x \in [a,b]$ and the supremum in the interval $[a,b]$ (as you defined) is maximized. – 2017-01-21
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    Also, $f$ must be continuous. – 2017-01-21
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    @expiTT--I--1z0, not necessarily. The Dirichlet function is close to $\frac{1}{2}$ function. Any bounded function fulfils the described condition and it is close to the constant function lying between lower $(m)$ and upper ($M$ ) bound. The optimal constant function is $\frac{m+M}{2}$. – 2017-01-21