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.

**Questions**:

- 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.
- If the claim is false under no regularity hypotheses, is it possible to add these to prove our claim?

### Related Questions

#### Sponsored Content

#### 0 Answered Questions

### Theorem 6.12 (a) in Baby Rudin: $\int_a^b \left( f_1 + f_2 \right) d \alpha=\int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha$

**2017-06-18 08:37:05****Saaqib Mahmood****354**View**7**Score**0**Answer- Tags: real-analysis integration analysis proof-verification definite-integrals

#### 1 Answered Questions

### [SOLVED] Theorem 6.15 in Baby Rudin: Is this result valid for vector-valued functions?

**2017-07-25 18:54:16****Saaqib Mahmood****213**View**1**Score**1**Answer- Tags: real-analysis integration analysis proof-verification definite-integrals

#### 0 Answered Questions

### Theorem 6.19 in Baby Rudin: Does this theorem hold for vector-valued functions too?

**2017-07-26 06:02:41****Saaqib Mahmood****98**View**1**Score**0**Answer- Tags: real-analysis integration analysis proof-verification definite-integrals

#### 0 Answered Questions

### Theorem 6.16 in Baby Rudin: Is this theorem also valid for vector valued functions?

**2017-07-25 19:15:36****Saaqib Mahmood****125**View**1**Score**0**Answer- Tags: real-analysis integration analysis definite-integrals

#### 0 Answered Questions

### Integral of product of two inverse regularized incomplete beta functions

**2016-10-16 06:05:50****Abhishek Halder****207**View**2**Score**0**Answer- Tags: integration definite-integrals special-functions hypergeometric-function beta-function

#### 1 Answered Questions

### [SOLVED] For which $\alpha$, $\beta$ does $\int\limits_1^{\infty} x^{\alpha} \cdot (\ln x)^\beta dx$ converge?

**2016-07-16 16:12:13****Pixar****327**View**2**Score**1**Answer- Tags: integration convergence improper-integrals parametrization

#### 0 Answered Questions

### Closed form for $\int\frac{\left((x + i) \beta\right)^\beta x^{\beta - 2}}{(x^2 + 1)^\beta} \exp\left(-\frac{\alpha}{A x}\right) \, \mathrm{d} x$

**2016-05-31 18:10:52****TimD****46**View**1**Score**0**Answer- Tags: integration ordinary-differential-equations indefinite-integrals

#### 1 Answered Questions

### [SOLVED] Schwartz functions & differentiation under the integral sign.

**2015-02-06 08:17:34****providence****197**View**4**Score**1**Answer- Tags: real-analysis integration functional-analysis pde

#### 2 Answered Questions

### [SOLVED] homogeneous linear differential equation question

**2012-03-21 16:20:04****Comic Book Guy****93**View**1**Score**2**Answer- Tags: integration integral-equations

#### 2 Answered Questions

### [SOLVED] On sort-of-linear functions

**2011-01-02 23:15:26****Riley E****2216**View**12**Score**2**Answer- Tags: functions functional-equations

## 1 comments

## @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 $f$.

AdditivityTake 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 :

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

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

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 $\mathbb{R^p}\to \mathbb{R^q}$.