$f(a+b)=f(a)+f(b)$ but $f$ is not linear

Question

Can you show me a continuous function $f \colon \mathbb{R}^n\to\mathbb{R}^m$ that satisfies $f(a+b)=f(a)+f(b)$ but is not linear?

We have that $$f(0)=f(0+0)=2f(0)\implies f(0)=0\\ f(x-x)=f(0)=f(x)+f(-x)=0\implies f(-x)=-f(x)\\ f(nx)=f(x+x+\dots+x)=f(x)+\dots+f(x)=nf(x)\quad \forall n \in \mathbb{N}$$ But $$ f(-nx)=-f(nx)=-nf(x) $$ So: $$ f(ax)=af(x) \quad \forall a \in \mathbb{Z} $$

Try it for $n=m=1$. Your argument seems to have shown that it's linear on $\Bbb Q\subset\Bbb R$. But do you know anything about $f(\sqrt2)$ or $f(\pi)$? — Ted Shifrin, Oct 03 '14 at 20:40
@mlainz: Now that it has come to your recollection, do you see how to Answer the Question yourself? — hardmath, Oct 03 '14 at 20:45
@mlainz If $f$ has to be continous, then $f(a+b)=f(a)+f(b)$, for all $a$ and $b$, suffices to say $f$ is linear. — egreg, Oct 03 '14 at 20:46
@mlainz: The short answer is that rationals are dense in the reals, so continuity suffices to preserve scalar multiplication by reals as well as by rationals. — hardmath, Oct 03 '14 at 20:51
@mlainz I think that you should state the requirement that the function be continuous as an addition to the end of your question. As things stand, you've made Jack's answer seem incorrect. — user180040, Oct 03 '14 at 21:23
For $n=m=1$ see here: http://math.stackexchange.com/questions/423492/overview-of-basic-facts-about-cauchy-functional-equation — Martin Sleziak, Oct 04 '14 at 17:00

score 58 · Accepted Answer · answered Oct 03 '14 at 20:58

58

Nobody can, because a continuous function $f \colon \mathbb{R}^n\to\mathbb{R}^m$ satisfying $$ f(x+y)=f(x)+f(y) $$ for all $x,y\in\mathbb{R}^n$ is linear.

The proof is quite easy.

$f(ax)=af(x)$ for all $a\in\mathbb{Z}$ and $x\in\mathbb{R}^n$
$f(\frac{a}{b}x)=\frac{a}{b}f(x)$, for all $\frac{a}{b}\in\mathbb{Q}$ and all $x\in\mathbb{R}^n$
$f(rx)=rf(x)$, for all $r\in\mathbb{R}$ and all $x\in\mathbb{R}^n$.

For the last step, if $r\in\mathbb{R}$, consider a sequence $q_k$ in $\mathbb{Q}$ converging to $r$. Then $q_kx$ is a sequence in $\mathbb{R}^n$ converging to $rx$ and $q_kf(x)$ is a sequence in $\mathbb{R}^m$ converging to $rf(x)$. By continuity of $f$, $$ f(rx)=f(\lim_{k\to\infty}q_kx)= \lim_{k\to\infty}f(q_kx)= \lim_{k\to\infty}q_kf(x)= rf(x) $$

answered Oct 03 '14 at 20:58

egreg

238,574

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You could cheat and change the topology so that step 3 fails, in which case Jack's answer works. – R.. GitHub STOP HELPING ICE Oct 03 '14 at 23:15
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@R.. But $\mathbb{R}^n$ is a topological vector space only in one way. ;-) – egreg Oct 03 '14 at 23:18
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@egreg, hmm what about discrete topology? – Oct 07 '14 at 21:02
@JeromeBaum There is just one (up to isomorphisms) discrete topological vector space. – egreg Oct 07 '14 at 21:06
@egreg, can't you equip $\mathbb R$ and $\mathbb R^n$ with the discrete topology (then, since $\mathbb R^n\times\mathbb R^n$ is also discrete, addition is continuous, and with discrete topology on $\mathbb R$, so is scalar multiplication), so you have an $\mathbb R$-vector space with the discrete topology, compatible with the vector space structure. Is this construction incorrect? – Oct 08 '14 at 07:50
@JeromeBaum It's not incorrect, but it's not a topological vector space in the usual sense. – egreg Oct 08 '14 at 08:07
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@egreg I'm sorry but I don't understand. Are you saying that it is not a TVS in the "usual sense" as in because we have an unusual topology on $\mathbb R$, or some other definition that excludes trivialities? (The definitions I'm aware of all seem to fit -- but I agree it is a very constructed example. :)) – Oct 09 '14 at 12:41
@JeromeBaum The thing is that, while you could choose the topology of the underlying set of vectors, the topology on the scalars $\mathbb{R}$, or $\mathbb{C}$ is assumed to be the usual one in a topological vector space. So, for the multiplication you need to consider $\mathbb{R}\times\mathbb{R}$, where the second factor has the discrete topology, but the first factor must use the usual topology of $\mathbb{R}$. – Pp.. Jan 01 '15 at 16:46
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@Pp.. from wikipedia: "A topological vector space X is a vector space over a topological field K (most often the real or complex numbers with their standard topologies) ..." (emphasis added) So it's not true that it must (according to Wikipedia on TVS). – May 19 '15 at 22:55
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@JeromeBaum Without any specification, I assume that $\mathbb{R}^n$ is a TVS with respect to the standard topology. And I don't think to be alone. :) – egreg May 19 '15 at 22:57
@egreg a real-world example might be the space of bytes/8-bit tuples equipped with taxicab metric. This is a metric space with the discrete metric, over the finite field with two elements, which itself would typically have the discrete topology applied (and this is the only metrizable topology on it). The vector space is a TVS as everything has the discrete topology and hence every map is continuous. I agree that there is an implication of R/C with the natural topology, but it's not like there are no real-world examples of TVS that are discrete. Now, why you would study these as TVS... :) – May 19 '15 at 23:02
(I love how as a TVS it basically says "a sequence of bytes converges iff it stops changing at some point." Reminds me of this (incorrect) statement: "(x_n) is a null sequence, therefore all elements are zero after some index.") – May 19 '15 at 23:06
@JeromeBaum If you have a discrete field and equip the dual of a discrete vector space with the product topology, then the continuous dual is isomorphic to the space we started with. I surprised with this a couple of members of a committee who had asked me about “duality of vector spaces”. :) – egreg May 19 '15 at 23:08
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@JeromeBaum the language used in the question, says: "I am an elementary analysis problem!" so don't stress your self to think for other topologies. – Omid Ghayour Sep 16 '16 at 10:59

score 17 · Answer 2 · answered Oct 03 '14 at 20:41

17

The general counterexample for a function satisfying $$ f(a+b)=f(a)+f(b) $$ but not being linear is to take a Hamel basis and acting with different linear maps on the elements of such a basis.

answered Oct 03 '14 at 20:41

Jack D'Aurizio

353,855

Sorry, I forgot to mention. Would such function be continuous? – mlainz Oct 03 '14 at 20:53
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Absolutely not. If you assume that $f$ is continuous and satisfies $f(a+b)=f(a)+f(b)$, then $f$ is linear as expected. – Jack D'Aurizio Oct 03 '14 at 20:54
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@mlainz: You might find it amusing to learn that the graph of any such function that is not continuous is dense in ${\mathbb R}^m \times {\mathbb R}^{n}.$ In particular, the graph of any such function from the reals to the reals that is not continuous has a graph that is dense in the plane. This means that the graph comes arbitrarily close to every point in the plane, which is way worse than the function being discontinuous at every point. – Dave L. Renfro Oct 03 '14 at 21:08
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In light of the 10 upvotes for my last comment here (my personal record?), I dug up some old posts of mine having more information for those interested in this topic. This 6 October 2006 sci.math post has a proof that the graph of a nonlinear additive function from the reals to the reals is dense in the plane. (continued) – Dave L. Renfro Oct 06 '14 at 14:55
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(continuation) This 29 October 2001 sci.math post gives a Baire $2$ function whose graph is dense in the plane. Finally, this 5 October 2006 sci.math gives various explicit constructions of wild functions. – Dave L. Renfro Oct 06 '14 at 14:56

$f(a+b)=f(a)+f(b)$ but $f$ is not linear

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