Additive but not Scalable f: R-->R

[WWGD]

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 ;).

 

[jbunniii]

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 $g(1)$.

 

[WWGD]

Ah, yes, simpler than I thought, thanks.

 

[WWGD]

I meant your solution is simple and nice, thanks.

 

[lavinia]

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?

 

[jbunniii]

@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.

 

[lavinia]

@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.

 

[lavinia]

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.

 

[jbunniii]

@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.

 

[lavinia]

RIght. So simply put, all rational linear combinations of two linearly independent vectors are a dense subset of the plane,

 

[jbunniii]

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 $\mathbb{Q}$.

 

[lavinia]

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

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.