# Additive but not Scalable f: R-->R

1. Jan 11, 2015

### 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 ;).

Last edited: Jan 11, 2015
2. Jan 11, 2015

### 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 $g(1)$.

3. Jan 11, 2015

### WWGD

Ah, yes, simpler than I thought, thanks.

4. Jan 12, 2015

### WWGD

I meant your solution is simple and nice, thanks.

5. Jan 12, 2015

### 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.

If you are pro choice, how do you feel about the Continuum Hypothesis?

Last edited: Jan 12, 2015
6. Jan 12, 2015

### 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.

Last edited: Jan 12, 2015
7. Jan 13, 2015

### lavinia

Really nice.

8. Jan 15, 2015

### 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.

Last edited: Jan 15, 2015
9. Jan 15, 2015

### 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.

Last edited: Jan 15, 2015
10. Jan 16, 2015

### lavinia

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

11. Jan 16, 2015

### jbunniii

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. Jan 16, 2015

### lavinia

Yes. That is what I meant.