$h(x+y)+h(0)=h(x)+h(y)$ iff only if $h(x)$ is affine or constant

Question

Let $h : \mathbb{R}^n \to \mathbb{R}^n$ satisfy \begin{align} h(x+y)+h(0)=h(x)+h(y). \end{align}

Can we show that that the only functions that satisfy the above property are affine functions (i.e., $h(x)=Ax+b$) or constant functions.

Proof for $n=1$. I have a proof for $n=1$ with an extra assumption of differentiability. Take the derivative with respec to $x$ \begin{align} \frac{d}{dx} h(x+y) = \frac{d}{dx} h(x) \end{align} since, $y$ and $x$ are arbitrary this implies $h(x)$ is either a constant or and affine function or constant functions.

It is not difficult to extend this proof to $n>1$. However, I am not satisfied with it as it assumes differentiability.

My question: Can we show the conjectured result without any differentiability assumptions? Also, I am not sure if continuity must be assumed. It would be nice to know if continuity is required or not.

Answer 1

$h(x+y)+h(0)=h(x)+h(y)\\ h(1+1)+h(0)=h(1)+h(1)\\ h(2)= 2h(1)-h(0)\\ h(3)+h(0) = h(2) + h(1)= 3h(1)-h(0)\\ h(3) = 3(h(1)-h(0)) + h(0)$

Lets say $m = (h(1) -h(0))$

Now we can make the inductive proof:

$h(n) = mn +h(0)\\ h(n+1)+h(0) = h(n) + h(1)\\ h(n+1) = mn + h(0) + h(1) - h(0)\\ h(n+1) = m(n+1) + h(0)$

And we have covered the natural numbers.

negative integers:

$h(n-n) + h(0) = h(n) + h(-n)\\ 2h(0) = mn + h(0) + h(-n)\\ h(-n) = m(-n) + h(0)$

Rational numbers:

$h(\frac 1q + \frac 1q) = 2 (h(\frac 1q) - h(0)) + h(0)\\ h(\frac qq) = q (h(\frac 1q) - h(0)) + h(0)\\ h(\frac 1q) = m\frac 1q + h(0)\\ h(\frac pq) = m\frac {p}{q} + h(0)$

For any real number $x$, there is a sequence of rationals $a_n$ that converges to $x$ and the sequence $h(a_n)$ converges to $h(x) = mx + h(0)$

Answer 2

Combining Eric Wofsey's Comment and Mine

Write $f(x):=h(x)-h(0)$ for each $x$. Let $\iota_i:\mathbb{R}\to\mathbb{R}^n$ and $\pi^i:\mathbb{R}^n\to\mathbb{R}$ be the $i$-th standard injection and the $i$-th standard projection $t\mapsto t\,e_i$ and $\left(t^1,t^2,\ldots,t^n\right)\mapsto t^i$, respectively. Here, $e_i$ is the $i$-th standard basis vector of $\mathbb{R}^n$. Then, we can easily seen that each function of the form $f^j_i:=\pi^j\circ f \circ \iota_i$ satisfies Cauchy's functional equation. Furthermore, $f$ is completely and uniquely determined by the family of functions $f^j_i$ (and vice versa): $$f\left(\sum_{i=1}^n\,t^ie_i\right)=\sum_{i=1}^n\,\sum_{j=1}^n\,f^j_i\left(t^i\right)\,e_j\,.$$

Hence, if $h$ need not be continuous, then we can take arbitrary solutions $f^j_i$ to Cauchy's functional equation to get a non-affine $h$. With continuity, each $h^j_i$ is continuous, so each of them is given by $f^j_i(t)=\alpha^j_i\,t$ for all $t\in\mathbb{R}$, where $\alpha^j_i$ is a fixed constant. Then, with $A=\left[\alpha^j_i\right]_{i,j=1,2,\ldots,n}$, we have $f(x)=Ax$ for all $x\in\mathbb{R}^n$, whence $$h(x)=f(x)+h(0)=Ax+h(0)$$ for all $x$.

To use the Jacobian approach under the continuity assumption, we note by some inductive arguments that $f(qx)=q\,f(x)$ for all $q\in\mathbb{Q}$, whence $f(tx)=t\,f(x)$ for all $t\in\mathbb{R}$ by continuity. Thus, $$\lim_{t\to0}\,\frac{f(x+ty)-f(x)}{t}=\lim_{t\to0}\,\frac{f(ty)}{t}=f(y)\,.$$ That is, $f$ is differentiable and the Jacobian matrix $J(x)$ of $f$ at $x$ satisfies $$J(x)\,y=f(y)\,,$$ for all $y$. This means $\big(J(x)-J(z)\big)\,y=f(y)-f(y)=0$ for all $x,y,z$. Since $y$ is arbitrary, $J(x)=J(z)$ for all $x$ and $z$, so for some constant matrix $A$, we have $J(x)=A$.

Answer 3

Let $g(x)=h(x)-h(0)$ so that $g(0)=0$. Then, \begin{align} g(x+y) & =h(x+y)-h(0) \\ & = h(x+y)+h(0)-2h(0) \\ & = h(x)+h(y)-2h(0) \\ & = g(x)+g(y). \end{align} Hence $0=g(0)=g(-x+x)=g(-x)+g(x)$, which gives $$ g(-x)=-g(x). $$ The function $g$ is linear over the rational scalars, but not necessarily over all reals. By induction, $$ g(nx) = ng(x),\;\;\ n=0,1,2,\cdots,\;\; x\in\mathbb{R}^n. $$ Therefore, if $q=\frac{m}{n}$ for integers $n,m$ with $n\ne 0$, $$ n g(qx) = g(nqx)=g(mx)=mg(x) \implies g(qx) = q g(x),\;\; q\in\mathbb{Q}. $$ This is not enough to imply linearity of $g$, unless, for example, you assume $g$ is continuous. To see why this is so, it is enough to find a function $h$ on $\mathbb{R}$ that is additive, but not linear. The Axiom of Choice is needed to construct $h : \mathbb{R}\rightarrow\mathbb{R}$ such that $h(x+y)=h(x)+h(y)$ for all $x,y\in\mathbb{R}$, but which is not linear. Using what has been done so far, you can see that $$ h(\frac{m}{n}r) = \frac{m}{n}h(r),\;\;\; m,n\in\mathbb{Z}, n \ne 0, r\in\mathbb{R}. $$ So $h$ is linear on $\{ qr : q \in \mathbb{Q} \}\subset\mathbb{R}$ with linear multiplier $h(r)$.

The problem is that you can have different multipliers on different sets of this type, $\{ qr : q \in\mathbb{Q} \}$. $\mathbb{Q}$ can be partitioned into disjoint subsets on which the $h$ must be linear. This is done by defining an equivalence relation on $\mathbb{R}\setminus\{0\}$ by $x \sim y$ iff $y/x \in \mathbb{Q}$ (or $x/y\in\mathbb{Q}$.) It's easy to check that this is an equivalence relation, meaning that it is reflexive, symmetric, and transitive. An equivalence class $[x]$ containing $x\in\mathbb{R}\setminus\{0\}$ has the form $[x]=\{ qx : q \in \mathbb{Q}, q\ne 0 \}$. The function $h$ is linear on $[x]$ with $h(qx)=qh(x)$, and this defines $h$ on $[x]$ because every $y\in [x]$ is a rational multiple of $x$. If $\{ [x_\alpha]\}_{\alpha\in\Lambda}$ are disjoint equivalence classes whose union is $\mathbb{R}\setminus\{0\}$, then $h$ is uniquely determined on $\mathbb{R}$ by the values of $h(x_\alpha)$, and these values may be assigned to be arbitrary real numbers. This describes all possible additive functions on $\mathbb{R}$. Such a function $h$ can be highly discontinuous, depending on the choice function $\alpha\mapsto h(x_\alpha)$. The different slopes are $m_{\alpha}=h(x_\alpha)/x_\alpha$. $h$ is continuous iff all these slopes $m_{\alpha}$ are the same. In fact, $h$ is Lebesgue measurable iff all the slopes are identical. This construction is intimately connected with the argument used to establish the existence of non-measurable subsets of $\mathbb{R}$, so much so that you show that there exists a discontinuous additive function $h$ on $\mathbb{R}$ iff there exists a non-measurable subset of $\mathbb{R}$; there are no such things without the Axiom of Choice.