Continuous $f$, $\sup_{x,y\in R^m}\|f(x+y)-f(x)-f(y)\|<\infty$, then exist only one real matrix $A$, such that $f(x)-Ax$ is bounded.

Question

Problem:

f is a continuous function, $f: \mathbb{R}^m\to \mathbb{R}^n$, $\sup_{x,y\in R^m}\|f(x+y)-f(x)-f(y)\|<\infty$, there exist only one real matrix $A$, such that $f(x)-Ax$ is bounded.

My thoughts 1:

$\sup_{x,y\in R^m}\|f(x+y)-f(x)-f(y)\|:=M<\infty$, then $\|f(2x)-2f(x)\|<M$, $\|f(3x)-f(2x)-f(x)\|<M$, so $\|f(3x)-3f(x)\|\leq\|f(3x)-f(2x)-f(x)+f(2x)-2f(x)\|\leq2M$, so by reduction, we have $\|f(nx)-nf(x)\|<(n-1)M$, then $\|\frac{f(nx)}{n}-f(x)\|<(1-\frac{1}{n})M$

I want to fix $x$ and let $n \to \infty$, but is hard to show $\lim_{n \to \infty}\frac{f(nx)}{n}$ exists.

My thoughts 2:

If we assert $\lim_{n \to \infty}\frac{f(nx)}{n}$ exists, define $G(x)=\lim_{n \to \infty}\frac{f(nx)}{n}$, I want to show $G(x)$ is linear.

If so, $G(x)=Ax$, then the $A$ is what we want.

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My thoughts: The limit may not exist.

Three steps: 1.$R^1$ case

2.each element in $R^n$ is actually the $R^1$ case

3.different elements are "independent"

Answer 1

Let us prove that $f_n(x) = \frac{f(2^nx)}{2^n}$ converges for all $x$ toward some $L(x)$. A powerful tool to obtain limits in metric spaces is completeness.

Let $n \in \mathbb{N}$. We have, $$ \|f_{n + 1}(x) - f_n(x)\| = \frac{1}{2^{n + 1}}\|f(2^{n + 1}x) - 2f(2^nx)\| \leqslant \frac{M}{2^{n + 1}}, $$ because $2^{n + 1}x = 2^nx + 2^nx$ and $2f(2^nx) = f(2^nx) + f(2^nx)$. It proves that the series of the $f_{n + 1} - f_n$ is normally convergent (the series of the $\frac{M}{2^{n + 1}}$ converges). By completeness of the space of bounded continuous functions with the sup norm, the series of the $f_{n + 1} - f_n$ converges, thus $f_n - f_0 \rightarrow f_\infty$ where $f_\infty$ is a bounded continuous function. Let $L = f_\infty + f_0 = f_\infty + f = \lim_{n \rightarrow +\infty} f_n$ (the limit is uniform). As $f_\infty$ is bounded, $f - L = -f_\infty$ is bounded too.

We only have left to show that $L$ is linear. Indeed, for all $x,y$ and all $n$, $$ \|f_n(x + y) - f_n(x) - f_n(y)\| = \frac{1}{2^n}\|f(2^nx + 2^ny) - f(2^nx) - f(2^ny)\| \leqslant \frac{M}{2^n} $$ When $n \rightarrow +\infty$, we deduce that $L(x + y) = L(x) + L(y)$. Knowing that $L$ is continuous, it is enough to conlude that it is linear. For this, begin by proving that for all $n \in \mathbb{N}$ and all $x$, $L(nx) = nL(x)$, then extend it to $n \in \mathbb{Z}$ and then prove that for all $q \in \mathbb{Q}$ and all $x$, $L(qx) = qL(x)$, and finally conlude using the density of $\mathbb{Q}$ in $\mathbb{R}$ and continuity of $L$.

Therefore, $L(x) = Ax$ for some matrix $A$ and $f(x) = L(x) + f_\infty(x) = Ax + \mathrm{O}(1)$ because $f_\infty$ is bounded.

Answer 2

A bit of a tedious proof. Assume wlog that $\sup_{x,y\in\mathbb{R}^m}\|f(x+y)-f(x)-f(y)\| = 1$.

Lemma 1

Let $x_1,x_2,\ldots,x_k\in\mathbb{R}^m, 1 < k$, then $\left\|f\left(\sum_{i=1}^k x_i\right) - \sum_{i=1}^kf(x_i)\right\| < k-1$

Proof: We use induction. The base case $k=2$ is just the given condition. Assume as induction hypothesis that $\left\|f\left(\sum_{i=1}^{k-1} x_i\right) - \sum_{i=1}^{k-1}f(x_i)\right\| < k-2$. Then $$ \left\|f\left(\sum_{i=1}^k x_i\right) - \sum_{i=1}^k f(x_i)\right\|= \left\|f\left(\sum_{i=1}^k x_i\right) - f\left(\sum_{i=1}^{k-1} x_i\right) +f\left(\sum_{i=1}^{k-1} x_i\right) - \sum_{i=1}^k f(x_i)\right\|\leq\\ \leq \left\|f\left(\sum_{i=1}^{k-1} x_i\right) - \sum_{i=1}^{k-1}f(x_i)\right\| + \left\|f\left(\sum_{i=1}^k x_i\right) - f\left(\sum_{i=1}^{k-1} x_i\right) - f(x_k)\right\| <k-2 + 1=k-1, $$ using the induction hypothesis and the base case, so the general case is proven by induction.

Lemma 2

For positive integer $p$ and a vector $x\in\mathbb{R}^m$, we have $\|f(-px)+pf(x)\| < p+1.$

Proof: Note that $\|f(0)\| = \|-f(0)-f(x)+f(x)\|\leq 1$, using the main condition. Thus using Lemma 1: $$ \|f(-px)+pf(x)\|\leq \|f(0)-f(-px)-pf(x)\|+\|f(0)\| < p+1. $$

Lemma 3

Let $t_1,t_2,\ldots,t_k\in\mathbb{R}$ be scalars and $x_1,x_2,\ldots,x_k\in\mathbb{R}^n$ vectors, then $\left\|f\left(\sum_{i=1}^kt_ix_i\right) - \sum_{i=1}^kt_if(x_i)\right\| < 2k - 1 + \sum_{i=1}^k |t_i|$.

Proof: First we prove the case $k=1$. Let $p \neq 0$ and $q > 0$ be integers. If $p > 0$, then using Lemma 1 we have that $$ \left\|\frac{p}{q}f(x)-f\left(\frac{p}{q}x\right)\right\|= \frac{1}{q}\left\|pf(x)-qf\left(\frac{p}{q}x\right)\right\|= \frac{1}{q}\left\|pf(x) - f(px)+f(px)-qf\left(\frac{p}{q}x\right)\right\|\leq\\ \leq \frac{1}{q}\left\|f(px) - pf(x) \right\| + \frac{1}{q}\left\|f(px) - qf\left(\frac{px}{q}\right)\right\| < \frac{p-1}{q} + \frac{q-1}{q} < p/q + 1. \tag{1}\label{ineq:1} $$ On the other hand if $p < 0$, then using Lemma 2 we have similarly as above that: $$ \left\|\frac{p}{q}f(x)-f\left(\frac{p}{q}x\right)\right\| < \left(\frac{|p|+1}{q} + \frac{q-1}{q} \right)M < \frac{|p| + 1}{q} + 1 \tag{2}\label{ineq:2} $$ Now let $\frac{p_1}{q_1},\frac{p_2}{q_2},\ldots$ be a rational sequence that converges to some nonzero $t\in\mathbb{R}$. Then by continuity of $f$ and of the norm, we have that $$ \left\|\frac{p_i}{q_i}f(x)-f\left(\frac{p_i}{q_i}x\right)\right\| \to \|tf(x)-f(tx)\|, $$ while the bounds $\eqref{ineq:1}$ and $\eqref{ineq:2}$ also converge to $|t|+1$, thus proving the case $k=1$.

Now we prove the general case $k>1$: $$ \left\|f\left(\sum_{i=1}^kt_ix_i\right) - \sum_{i=1}^kt_if(x_i)\right\|\leq \sum_{i=1}^k\| f(t_ix_i) - t_if(x_i) \| + \left\|f\left(\sum_{i=1}^kt_ix_i\right) - \sum_{i=1}^kf(t_ix_i)\right\| <\\< k-1+\sum_{i=1}^k |t_i|+1 = 2k + \sum_{i=1}^k |t_i|, $$ using Lemma 1 and the case $k=1$, which completes the proof.

Solution to the problem

First we prove that the sequence $\{f(tx)/t\}$ is convergent for any $x\in\mathbb{R}^m$, where $t$ is a positive integer. Indeed given any $\epsilon > 0$ and $N > 2\epsilon^{-1}$, we have for all positive integers $t,s > N$ that $$ \left\| \frac{f(tx)}{t} - \frac{f(sx)}{s} \right\| = \frac{1}{ts}\left\| sf(tx) - tf(sx)\right\| \leq \frac{1}{ts}\left\| f(tsx) - tf(sx)\right\| + \frac{1}{ts}\left\| f(tsx) - sf(tx)\right\| < \frac{t+s-2}{ts} < \epsilon $$ using Lemma 1, so $\{f(tx)/t\}$ is Cauchy, and hence convergent since $\mathbb{R}^n$ is complete.

Let $\{e_1,\ldots,e_m\}$ be the standard basis of $\mathbb{R}^m$, and let $A_t = \begin{pmatrix}f(te_1)/t&f(te_2)/t&\cdots&f(te_m)/t\end{pmatrix}$. If $x=\sum_{i=1}^m x_i e_i$, then by Lemma 3 we have $$ \|f(x)-A_tx\|=\left\|f\left(\sum_{i=1}^m x_i e_i \right)-\sum_{i=1}^m x_i\frac{f(te_i)}{t}\right\| < 2m - 1 + \frac{1}{t}\sum_{i=1}^m|x_i|. $$ Taking the limit, setting $A = \lim_{t\to\infty}A_t$ and using continuity gives us the result.