Additive but not Scalable f: R-->R

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In summary, we can find a function ##f: \mathbb{R} \rightarrow \mathbb{R}## that satisfies the condition ##f(x+y)=f(x)+f(y)## but is not linear. However, it is also possible to find a function ##g: \mathbb{R} \rightarrow \mathbb{R}## that satisfies the condition ##g(cx)=cg(x)## for all real constants ##c## and all real values ##x##, but ##g(x+y) \neq g(x)+g(y)## for all ##x,y##. This function is highly pathological and its graph is dense in the plane. This is proven by showing that the set of linear combinations
  • #1
WWGD
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Hi, we know that using choice * that we can find a function ## f: \mathbb R \rightarrow \mathbb R ## so that ## f(x+y)=f(x)+f(y) ## , but f is not linear. Is the "other way" possible, i.e., can we have :
## g: \mathbb R \rightarrow \mathbb R ## so that, for all Real constants c, all Real values x , we have

## g(cx)=cg(x) ## , but ## g(x+y ) \neq g(x)+g(y) ## for all x,y ?
* I guess I am pro-choice ;).
 
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  • #2
Assume that the condition ##g(cx) = cg(x)## holds for all real ##c,x##. Choose ##c=t## and ##x=1##. Then ##g(t) = g(cx) = cg(x) = tg(1)## for all real ##t##, so ##g## is linear with slope ##g(1)##.
 
  • #3
Ah, yes, simpler than I thought, thanks.
 
  • #4
I meant your solution is simple and nice, thanks.
 
  • #5
There is a cool problem in Rudin's Real and Complex Analysis that is related to your question.

Consider the real numbers as a vector space over the rational numbers and let f be a Q linear map of the reals into the reals.

- Show that if f is continuous then it is multiplication by a constant
- Show that if it is not continuous, then its graph is dense in the plane.

BTW: Your additive function is automatically linear over the rational numbers.

If you are pro choice, how do you feel about the Continuum Hypothesis?
 
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  • #6
@lavinia - that is a nice problem. "The graph is dense in the plane" means the function is highly pathological, so one might expect the solution to be difficult. But it turns out to be quite elementary. First, if we let ##c = f(1)##, then ##f(x) = f(x \cdot 1) = xf(1) = cx## for all rational ##x##. Suppose there is an irrational ##x## with ##f(x) = y \neq cx##. Pick any disk centered at a point ##(a,b)## with radius ##r > 0##, and show that there exist rationals ##p,q## such that the point ##(px + q, f(px + q)) = (px + q, py + cq)## lies inside the disk, i.e.
$$(a - (px + q))^2 + (b - (py + cq))^2 < r^2$$
Now ##g(p,q) = (a - (px + q))^2 + (b - (py + cq))^2## is a continuous function of ##p## and ##q##, and we can find a real solution ##(u,v)## such that ##g(u,v) = 0## by solving ##a = ux + v## and ##b = uy + cv## to obtain
$$u = \frac{b - ca}{y-cx} \> \text{ and } \> v = a - \left(\frac{b - ca}{y-cx}\right) x$$
The denominator ##y-cx## is nonzero because ##y \neq cx##.

Since ##g## is continuous, there is a ##\delta##-neighborhood ##N_\delta(u,v)## centered at ##(u,v)## such that every point ##(p,q) \in N_\delta(u,v)## satisfies ##g(p,q) < r^2##. There are infinitely many rational points in ##N_\delta(u,v)##, so all we have to is choose one.
 
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  • #7
jbunniii said:
@lavinia - that is a nice problem. "The graph is dense in the plane" means the function is highly pathological, so one might expect the solution to be difficult. But it turns out to be quite elementary. First, if we let ##c = f(1)##, then ##f(x) = f(x \cdot 1) = xf(1) = cx## for all rational ##x##. Suppose there is an irrational ##x## with ##f(x) = y \neq cx##. Pick any disk centered at a point ##(a,b)## with radius ##r > 0##, and show that there exist rationals ##p,q## such that the point ##(px + q, f(px + q)) = (px + q, py + cq)## lies inside the disk, i.e.
$$(a - (px + q))^2 + (b - (py + cq))^2 < r^2$$
Now ##g(p,q) = (a - (px + q))^2 + (b - (py + cq))^2## is a continuous function of ##p## and ##q##, and we can find a real solution ##(u,v)## such that ##g(u,v) = 0## by solving ##a = ux + v## and ##b = uy + cv## to obtain
$$u = \frac{b - ca}{y-cx} \> \text{ and } \> v = a - \left(\frac{b - ca}{y-cx}\right) x$$
The denominator ##y-cx## is nonzero because ##y \neq cx##.

Since ##g## is continuous, there is a ##\delta##-neighborhood ##N_\delta(u,v)## centered at ##(u,v)## such that every point ##(p,q) \in N_\delta(u,v)## satisfies ##g(p,q) < r^2##. There are infinitely many rational points in ##N_\delta(u,v)##, so all we have to is choose one.
Really nice.
 
  • #8
By the way, I think that this is a sketch of your proof done by contradiction rather than directly. It gives a nice picture of what is going on - provided I have not made a mistake.

Suppose there is an open disk, D, in the plane that is not hit by the function,f. That is: the graph of f completely misses the disk. This would happen if the graph were not dense in the plane.

Since f is additive and linear over the rationals and since the rationals are dense in the reals, it must also miss all open disks that are translates of D along a line of slope c = f(1).

For the same reason, f must miss all open disks that are translates of D along a line of slope ,y/x where y= f(x) and x is the irrational such that f(x) is not equal to c.

This second line has a different slope than c. So one can translate D along lines of both slopes iteratively to remove every open disk from the image of f. But this is impossible.
 
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  • #9
@lavinia - That sounds right to me. Putting it another way, viewing ##\mathbb{R}^2## as a vector space over the rationals, the vectors ##v = (1, f(1))## and ##w = (x, f(x))## are linearly independent. The 2-dimensional subspace ##S## spanned by ##v## and ##w## is the set of linear combinations ##pv + qw##, where ##p,q\in \mathbb{Q}##. The graph of ##f## must contain ##S##. So it suffices to prove that ##S## is dense in ##\mathbb{R}^2##.

To prove this, we can define ##A : \mathbb{R}^2 \to \mathbb{R}^2## by ##A(r,s) = rv + sw##, where ##r,s \in \mathbb{R}## and ##v,w## are the vectors defined above. Since ##A## is a linear operator on ##\mathbb{R}^2##, it is continuous. It is also surjective (indeed, bijective) since ##v,w## are linearly independent over ##\mathbb{R}## (not just over ##\mathbb{Q}##). Therefore, ##\mathbb{R}^2 = A(\mathbb{R}^2) = A(\text{cl}(\mathbb{Q}^2)) \subset \text{cl}A(\mathbb{Q}^2)##, where ##\text{cl}## denotes closure. In other words, ##S = A(\mathbb{Q}^2)## is dense in ##\mathbb{R}^2## as desired.
 
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  • #10
RIght. So simply put, all rational linear combinations of two linearly independent vectors are a dense subset of the plane,
 
  • #11
lavinia said:
RIght. So simply put, all rational linear combinations of two linearly independent vectors are a dense subset of the plane,
Yes, provided that the vectors are linearly independent over ##\mathbb{R}##, not just over ##\mathbb{Q}##. Example: ##(0,1)## and ##(0,\sqrt{2})## won't work even though they are linearly independent over ##\mathbb{Q}##.
 
  • #12
jbunniii said:
Yes, provided that the vectors are linearly independent over ##\mathbb{R}##, not just over ##\mathbb{Q}##. Example: ##(0,1)## and ##(0,\sqrt{2})## won't work even though they are linearly independent over ##\mathbb{Q}##.[/QU
jbunniii said:
Yes, provided that the vectors are linearly independent over ##\mathbb{R}##, not just over ##\mathbb{Q}##. Example: ##(0,1)## and ##(0,\sqrt{2})## won't work even though they are linearly independent over ##\mathbb{Q}##.
Yes. That is what I meant.
 

1. What does "additive but not scalable" mean?

"Additive but not scalable" refers to a function that satisfies the property of additivity, meaning that the output of the function when given two inputs is equal to the sum of the outputs of each individual input, but is not considered scalable because it does not exhibit the property of scalability, where the output of the function changes proportionally with the input. In other words, increasing the input by a certain amount does not result in the output increasing by the same amount.

2. Can you provide an example of an "additive but not scalable" function?

A common example of an additive but not scalable function is the floor function, denoted as f(x) = ⌊x⌋. This function takes any real number as an input and outputs the largest integer less than or equal to that number. While this function is additive, as the output of f(x + y) is equal to f(x) + f(y), it is not scalable because the output does not increase proportionally with the input. For example, f(2) = 2 but f(4) = 3, an increase of 100% in the input results in only a 50% increase in the output.

3. How is "additive but not scalable" different from a linear function?

A linear function is a function that satisfies both the properties of additivity and scalability, meaning that the output increases proportionally with the input. "Additive but not scalable" functions satisfy the property of additivity but not scalability, as the output does not increase proportionally with the input. This means that while linear functions have a constant rate of change, "additive but not scalable" functions have a variable rate of change.

4. Why is it important to understand "additive but not scalable" functions?

Understanding "additive but not scalable" functions is important in various fields, such as economics, physics, and computer science. In economics, for example, understanding the properties of a function can help in determining the relationship between inputs and outputs, and in turn, make predictions about the behavior of a system. In physics, "additive but not scalable" functions can help model non-linear systems. In computer science, understanding these functions can aid in the development of algorithms and data structures.

5. Can "additive but not scalable" functions have real-world applications?

Yes, "additive but not scalable" functions can have real-world applications. As mentioned before, these functions can help model non-linear systems, making them useful in various fields such as economics, physics, and computer science. For example, in economics, the floor function can be used to model the relationship between the quantity of goods produced and the demand for those goods. In physics, the floor function can be used to model the behavior of a particle moving in a non-linear path. In computer science, the floor function can be used in sorting algorithms to partition data into smaller, non-overlapping groups.

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