Let $\mathcal{F}$ be the set of all functions $f: \mathbb{R} \rightarrow \mathbb{R}$.

Consider an operator $\mathcal{O}: \mathcal{F} \rightarrow \mathcal{F}$ such that:

  1. $\mathcal{O}( f_1 + f_2) = \mathcal{O}(f_1) + \mathcal{O}(f_2) \ \text{for all } f_1, f_2 \in \mathcal{F}$;

  2. $\mathcal{O}( f_1 \cdot f_2) = \mathcal{O}(f_1) \cdot f_2 + \mathcal{O}(f_2)\cdot f_1 \ \text{for all } f_1, f_2 \in \mathcal{F}$;

  3. $\mathcal{O}( \mathbb{1}) = \mathbb{0} $.

Question. Find all operators satisfying the above properties.

Notation. The sum $f := f_1 + f_2$ is defined as $f(x) := f_1(x) + f_2(x)$ for all $x \in \mathbb{R}$. The product $f := f_1 \cdot f_2$ is defined as $f(x) := f_1(x) f_2(x)$ for all $x \in \mathbb{R}$. $\mathbb{1}$ denotes the constant function $f(x) = 1$ for all $x \in \mathbb{R}$. $\mathbb{0}$ denotes the constant function $f(x) = 0$ for all $x \in \mathbb{R}$.

Comments. Trivially, $\mathcal{O}(f) := \mathbb{0}$ satisfies the properties. Also the derivative, i.e. $\mathcal{O}(f) := df/dx$ does, whenever $\mathcal{F}$ is the set of differentiable functions. I am not able to prove if those are the only ones, even for this particular $\mathcal{F}$.

  • 5
    $\begingroup$ Well, any scalar product of the derivative operator also works, and it already includes the zero operator. Yet there's already a problem: this operator isn't defined for all of $\,\mathcal F\,$...You'd need to take the subalgebra of differentiable real functions. $\endgroup$ – DonAntonio Apr 3 '13 at 14:07
  • $\begingroup$ I agree. I inserted this in the Comments. But, still, if $\mathcal{F}$ is the set of differentiable functions, is the null operator the only one satisfying the properties? $\endgroup$ – user693 Apr 3 '13 at 14:16
  • 3
    $\begingroup$ It seems to me that if you define an arbitrary value for $\mathcal O(x)$, then you can extend the operator uniquely to polynomials (and then by density to differentiable functions). $\endgroup$ – Angelo Lucia Apr 3 '13 at 14:19
  • 2
    $\begingroup$ An idea: note the operator $\,\mathcal O\,$ is linear so we can try to show that it coincides with some scalar product of the differential operator on some basis of the linear space $\,\mathcal F\,$ ... $\endgroup$ – DonAntonio Apr 3 '13 at 14:20
  • $\begingroup$ Angelo's idea looks promising...nice! +1 $\endgroup$ – DonAntonio Apr 3 '13 at 14:20

Throughout, we shall use the same abuse of notation: If $c$ is understood to be a constant, then $\mathcal{O}(c)$ means the operator $\mathcal{O}$ applied to the function $f(x)=c \, \forall x \in \mathbb{R}$. We shall also use $\mathcal{O}(f)_{|x=a}$, which represents the value of the function $\mathcal{O}(f)$ when evaluated at $x=a$.

Note that by property 1, $\mathcal{O}(f)=\mathcal{O}(f+0)=\mathcal{O}(f)+\mathcal{O}(0)$, so $\mathcal{O}(0)=0$.

Step 1: Consider the indicator function $1_A$ for a set $A \subseteq \mathbb{R}$, $1_A(x)=\begin{cases} 1 & \text{if} \, x \in A \\ 0 & \text{if} \, x \not \in A \end{cases}$, then $\mathcal{O}(1_A)=0$.

Proof: Consider the indicator function for the complement of $A$, $1_{A^c}(x)=1-1_A(x)$.

Then by property 1 and 3, $$0=\mathcal{O}(1)=\mathcal{O}(1_A+1_{A^c})=\mathcal{O}(1_A)+\mathcal{O}(1_{A^c})$$

Using this and property 2, we get

$$0=\mathcal{O}(0)=\mathcal{O}(1_A \cdot 1_{A^c})=\mathcal{O}(1_A) \cdot 1_{A^c}+\mathcal{O}(1_{A^c}) \cdot 1_A=\mathcal{O}(1_A) \cdot (1_{A^c}-1_A)$$

Since $1_{A^c}-1_A$ is never $0$, this implies that $\mathcal{O}(1_A)=0$.

Step 2: $\mathcal{O}(f)_{|x=a}=\mathcal{O}(f(a))_{|x=a}$.

Proof: Fix a constant $a$. Consider the indicator function $1_{\{a\}}$, and any function $f$. Then $(f(a) \cdot 1_{\{a\}})(x)=(f \cdot 1_{\{a\}})(x)=\begin{cases} f(a) & \text{if} \, x=a \\ 0 & \text{if} \, x \neq a \end{cases}$. By property 2,

$$\mathcal{O}(f \cdot 1_{\{a\}})=\mathcal{O}(f) \cdot 1_{\{a\}}+\mathcal{O}(1_{\{a\}}) \cdot f=\mathcal{O}(f) \cdot 1_{\{a\}}$$ $$\mathcal{O}(f(a) \cdot 1_{\{a\}})=\mathcal{O}(f(a)) \cdot 1_{\{a\}}+\mathcal{O}(1_{\{a\}}) \cdot f(a)=\mathcal{O}(f(a)) \cdot 1_{\{a\}}$$

Thus $\mathcal{O}(f) \cdot 1_{\{a\}}=\mathcal{O}(f(a)) \cdot 1_{\{a\}}$, so $\mathcal{O}(f)_{|x=a}=\mathcal{O}(f(a))_{|x=a}$.

Step 3: An operator $\mathcal{O}$ satisfies the given properties for all functions if and only if $\mathcal{O}$ satisfies both $\mathcal{O}(f)_{|x=a}=\mathcal{O}(f(a))_{|x=a}$ and the given properties for all constant functions.

Proof: Clearly if the given properties hold for all functions, they hold for all constant functions as well. Morever by step 2, $\mathcal{O}$ satisfies $\mathcal{O}(f)_{|x=a}=\mathcal{O}(f(a))_{|x=a}$.

Now suppose we have an operator $\mathcal{O}$ satisfying both $\mathcal{O}(f)_{|x=a}=\mathcal{O}(f(a))_{|x=a}$ and the given properties for all constant functions. We shall show that it satisfies the given properties for all functions. Property 3 is trivial.

Consider property 1 with 2 arbitrary functions $f_1, f_2$. To show that the property holds, we show that at each value $a$ of $x$, the equation holds.


Similarly, for property 3, we have

\begin{align} \mathcal{O}(f_1 \cdot f_2)_{|x=a}=\mathcal{O}(f_1(a) \cdot f_2(a))_{|x=a}& =\mathcal{O}(f_1(a))_{|x=a}\cdot f_1(a)+\mathcal{O}(f_2(a))_{|x=a} \cdot f_2(a) \\ &=\mathcal{O}(f_1)_{|x=a}\cdot f_1(a)+\mathcal{O}(f_2)_{|x=a} \cdot f_2(a) \end{align}

Therefore the given properties hold for all functions.

Step 4: Characterising all (infinitely many) operators satisfying the given properties.

By above, we need only determine the image of constant functions under the operator, and the image of all functions is then determined by the relation $\mathcal{O}(f)_{|x=a}=\mathcal{O}(f(a))_{|x=a}$.

For any real number $a$, define a function $f_a: \mathbb{R} \to \mathbb{R}$ such that $f_a(c)=\mathcal{O}(c)_{|x=a}$. Our task then reduces to finding all functions $f_a$ for each $a$, such that:

$$f_a(x+y)=f_a(x)+f_a(y) \, \forall x, y \in \mathbb{R}$$ $$f_a(xy)=f_a(x)y+f_a(y)x \, \forall x, y \in \mathbb{R}$$ $$f_a(1)=0$$

Let $\mathbb{S}$ be the (infinite) set of functions satisfying the above properties. (Note that if a function is a solution for some $a$, it is also a solution for all $a$.)

Now consider any operator $\mathcal{T}: \mathbb{R} \to \mathbb{S}$, and this corresponds to an operator $\mathcal{O}$ satisfying the given properties, via the relation $$\mathcal{O}(f)_{|x=a}=\mathcal{O}(f(a))_{|x=a}=\mathcal{T}(a)_{|x=f(a)}$$

Final Step: Determining $\mathbb{S}$.

$\mathbb{S}$ is the set of functions $f(x): \mathbb{R} \to \mathbb{R}$ such that $f(x+y)=f(x)+f(y), f(xy)=f(x)y+f(y)x \, \forall x, y \in \mathbb{R}$, and $f(1)=0$.

Now $f$ satisfies the Cauchy equation, and along with $f(1)=0$, we can conclude that $f(q)=0 \, \forall q \in \mathbb{Q}$. Thus $f(qx)=qf(x)+xf(q)=qf(x) \, \forall q \in \mathbb{Q}$.

It is straightforward to prove by induction (I'm not going to prove it here, since my post is already so long.) that $f(x^n)=nx^{n-1}f(x) \, \forall n \in \mathbb{Z}$. Therefore

$$f(\sum_{i=m}^{n}{a_ix^i})=\sum_{i=m}^{n}{ia_ix^{i-1}}f(x) \, \forall m, n \in \mathbb{Z}, m \leq n, \, \forall a_i \in \mathbb{Q}, \forall x \in \mathbb{R}$$

This immediately implies that $f(x)=0$ for all algebraic numbers $x$.

For the last part, I will just give a sketch. (Hope I did not make any mistakes)

We proceed in a manner similar to that for the Cauchy equation. By the axiom of choice, there exists an infinite set $P=\{\alpha_1, \alpha_2, \ldots\}$ such that $P$ is algebraically independent over $\mathbb{Q}$ and the elements of $P$ generate the whole of $\mathbb{R}$. (Note that trivially by Lindermann-Weierstrass Theorem no such finite set exists) Now each function $f \in \mathbb{S}$ corresponds to a (infinite) sequence $f(\alpha_1), f(\alpha_2), \ldots$ of real numbers, and each distinct sequence corresponds to a unique $f \in \mathbb{S}$.

  • $\begingroup$ So I don't get the Message of this statements? Are all admissible operators like $\mathcal{O}(f) = c f'$ for some $c \in \mathbb{R}$? $\endgroup$ – user693 Apr 4 '13 at 8:10
  • 1
    $\begingroup$ @Adam no, in fact $\mathcal{O}=cf'$ doesn't work. It might seem to work for differentiable functions, but the set $\mathcal{F}$ we are dealing with here includes functions which are not differentiable. In particular, because the indicator function $1_{a}$ is not differentiable at $x=a$, but its image under $\mathcal{O}$ must be $0$ at $x=a$, applying the properties at $x=a$ using this function shows that the image of $f$ evaluated at $x=a$ only depends on the value of $f(a)$, and not the neighbourhood of $x=a$. In particular, $\mathcal{O}$ cannot be a scalar multiple of the derivative. $\endgroup$ – Ivan Loh Apr 4 '13 at 13:55
  • $\begingroup$ I see, thanks. So what is the Message of your comment? If our domain $\mathcal{F}$ includes \textit{all} functions, what is an example of admissible Operator $\mathcal{O}$? $\endgroup$ – user693 Apr 4 '13 at 13:58
  • $\begingroup$ @Adam You mean apart from the trivial $\mathcal{O}(f)=0$? Well it is a bit hard to describe. For example, consider the field extension $A$ generated by $\pi$ over $\mathbb{Q}$. Now put $\mathcal{O}(c)=0$ for $c \not \in A$ and for algebraic $c$, and put $\mathcal{O}(\pi)=1$. Now for every element $k$ in $\mathbb{A}$, we have $k=\frac{P_1(\pi)}{P_2(\pi)}$ for some polynomials $P_1, P_2 \in \mathbb{Q}[x]$, so put $\mathcal{O}(k)=\frac{d}{dx}(\frac{P_1}{P_2})_{|x=\pi}$. Finally for any non-constant function $f$, $\mathcal{O}(f)$ is the function that gives $\mathcal{O}(f(a))_{|x=a}$ at $x=a$. $\endgroup$ – Ivan Loh Apr 4 '13 at 18:48
  • $\begingroup$ @Adam So, continuing from the above example, we would have e.g. if $f(x)$ is the identity function, $f(x)=x \, \forall x \in \mathbb{R}$, then $$\mathcal{O}(f)(x)=\begin{cases} 0 & \text{if} \, x \not \in A \\ \frac{d}{dx}(\frac{P_1(x)}{P_2(x)})_{|x=\pi} & \text{if} \, x=\frac{P_1(\pi)}{P_2(\pi)} \end{cases}$$ $\endgroup$ – Ivan Loh Apr 4 '13 at 18:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.