By Lonidard

2017-06-21 10:40:04 8 Comments

I am trying to understand how many functions $f:\mathbb R\to \mathbb R$ such that $$\int f(g(x))\mathrm d x=f\left ( \int g(x)\mathrm d x \right) \tag{$\star$}$$ for every $g:\mathbb R \to \mathbb R$ exist. We shall refer to a function satisfying $(\star)$ as an I-linear function.

It is easy to show that the identity function is I-linear. Moreover, the space of I-linear functions forms a vector space because if $f_1,f_2$ are I-linear $$ \begin{align}(\alpha f_1+\beta f_2) \left (\int g(x)\mathrm d x \right ) &= \alpha f_1 \left (\int g(x)\mathrm d x \right)+\beta f_2 \left ( \int g(x)\mathrm d x \right) \\ &= \alpha \int f_1(g(x))\mathrm d x+\beta \int f_2(g(x))\mathrm d x\\ &= \int (\alpha f_1+\beta f_2)(g(x))\mathrm d x \end{align}$$

As a consequence, all scalar multiples of identity are I-linear functions. This claim is essentially equivalent to stating the obvious $\int \alpha f=\alpha \int f$.

Observe that if $f$ is both I-linear and differentiable, by setting $g(x)=e^x$ and differentiating $(\star)$ one obtains $$f(e^x)=e^xf'(e^x+C)=\frac{\mathrm d}{\mathrm d x}f(e^x+C)$$ which looks almost like a differential equation.


  1. Under no regularity hypotheses on $f$, is it possible to show that if $f$ is I-linear, then $f(x)=\alpha x$ for some $\alpha$? To do so, it would be sufficient to show that the vector space of I-linear functions is $1$-dimensional.
  2. If the claim is false under no regularity hypotheses, is it possible to add these to prove our claim?


@Benoit Sanchez 2017-06-21 11:07:47

The problem is not fully specified. You must say for what class of functions $g$ the equality is supposed to hold. I suppose this class contains at least piecewise constant functions.

Then $(\star)$ implies that $f$ is a linear map. No need to assume anything about the regularity of $f$.


Take for example a function $g$ that takes value $a$ on an interval of length 1 and $b$ on an interval of length 1 (intervals are disjoint) and $0$ anywhere else. Then :

  • $\int f(g)=f(a)+f(b)$
  • $f(\int g)=f(a+b)$

Thus $f(a+b)=f(a)+f(b)$


Now take $g$ that is $a$ on an interval of length $\lambda$ ($0$ anywhere else) :

  • $\int f(g)=\lambda f(a)$
  • $f(\int g)=f(\lambda a)$

But it only works for non negative $\lambda$ (it is the length of an interval). Finally we need to prove $f(-x)=-f(x)$. It is a consequence of the previous paragraph with $a=-b$.

So : $f(\lambda a)=\lambda f(a)$

Any function that is a solution is a linear map. The only linear maps on $\mathbb{R}$ are $x\mapsto \alpha x$.

Note that the proof for additivity is "overkill" (not really needed) when we are in $\mathbb{R}$. But this allows to generalize the result when $f$ is $\mathbb{R^p}\to \mathbb{R^q}$.

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