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Consider a map $T:X\to Y$ of the form $$T(u):=p(u)H(u),$$ where $H:X\to Y$ is a Lipschitz homogeneous map (here always meaning $H(\alpha u)=\alpha H(u)$ for all $u\in X$ and $\alpha\in\mathbb{K}$), and $p:X\to\mathbb{R}$ is a bounded function with the property that the image of any ball of fixed radius $R$ is the same. For instance, any bounded $R$-periodic function of the norm.

So for any $u\in X$ and $\alpha\in\mathbb{K}$ there is $v\in X$ such that $\| \alpha u -v\|\le R$ and $p(v)=p(u)$. Then $\alpha T(u)-T(v)=\alpha p(u) H(u)-p(v)H(v)=p(u)(H(\alpha u) -H(v)) $ so that $\|\alpha T(u)-T(v)\|\le\|p\|_\infty\operatorname{Lip}(H)R$. Thus $T$ satisfies your hypothesis with $N:=\max\big(1, \|p\|_\infty\operatorname{Lip}(H)\big)R.$

On the other hand, if a map $T$ satisfies your hypothesis, and also has the following mild property: for any $u\in X$
$$\sup_{\|\alpha u-w\|\le N} \|T(w)-T(\alpha u)\|=o(\alpha)\qquad (\text{for}\; |\alpha|\to+\infty),$$ (for instance, it is Lipschitz), then it is a homogeneous map. For, if $v$ is as in your notation, $$\alpha T(u)= T(\alpha u)+ \big( T(v)-T(\alpha u)\big)+ \big(\alpha T(u)-T(v)\big)= T(\alpha u)+o(\alpha)$$ so that $T(0)=0$, and for all $u\in X$ $$T(u)=\lim_{|\alpha|\to +\infty}{1\over\alpha}T(\alpha u)$$ which implies it is a homogeneous map.

1) "Other examples of $T$ aside from the class of homogeneous functions".

Consider a map $T:X\to Y$ of the form $$T(u):=p(u)H(u),$$ where $H:X\to Y$ is a Lipschitz homogeneous map (here always meaning $H(\alpha u)=\alpha H(u)$ for all $u\in X$ and $\alpha\in\mathbb{K}$), and $p:X\to\mathbb{R}$ is a bounded function that has the property that the whole image of $p$ is already reached on any ball of some fixed radius $R$. In formulas, $p(X)=p(B(x,R))$ for any $x\in X$. For instance, any bounded $R$-periodic function of the norm does.

So for any $u\in X$ and $\alpha\in\mathbb{K}$ there is $v\in X$ such that $\| \alpha u -v\|\le R$ and $p(v)=p(u)$. Then $\alpha T(u)-T(v)=\alpha p(u) H(u)-p(v)H(v)=p(u)(H(\alpha u) -H(v)) $ so that $\|\alpha T(u)-T(v)\|\le\|p\|_\infty\operatorname{Lip}(H)R$. Thus $T$ satisfies your hypothesis with $N:=\max\big(1, \|p\|_\infty\operatorname{Lip}(H)\big)R.$

2) "$T$ is almost homogeneous".

On the other hand, if a map $T$ satisfies your hypothesis, and also has the following mild property: for any $u\in X$
$$\sup_{\|\alpha u-w\|\le N} \|T(w)-T(\alpha u)\|=o(\alpha)\qquad (\text{for}\; |\alpha|\to+\infty),$$ (for instance, it is Lipschitz), then it is a homogeneous map. For, if $v$ is as in your notation, $$\alpha T(u)= T(\alpha u)+ \big( T(v)-T(\alpha u)\big)+ \big(\alpha T(u)-T(v)\big)= T(\alpha u)+o(\alpha)$$ so that $T(0)=0$, and for all $u\in X$ $$T(u)=\lim_{|\alpha|\to +\infty}{1\over\alpha}T(\alpha u)$$ which implies it is a homogeneous map.

Consider a map $T:X\to Y$ of the form $$T(u):=p(u)H(u),$$ where $H:X\to Y$ is a Lipschitz homogeneous map (here always meaning $H(\alpha u)=\alpha H(u)$ for all $u\in X$ and $\alpha\in\mathbb{K}$), and $p:X\to\mathbb{R}$ is a bounded function with the property that the image of any ball of fixed radius $R$ is the same. For instance, any bounded $R$-periodic function of the norm.

So for any $u\in X$ and $\alpha\in\mathbb{K}$ there is $v\in X$ such that $\| \alpha u -v\|\le R$ and $p(v)=p(u)$. Then $\alpha T(u)-T(v)=\alpha p(u) H(u)-p(v)H(v)=p(u)(H(\alpha u) -H(v)) $ so that $\|\alpha T(u)-T(v)\|\le\|p\|_\infty\operatorname{Lip}(H)R$. Thus $T$ satisfies your hypothesis with $N:=\max\big(1, \|p\|_\infty\operatorname{Lip}(H)\big)R.$

On the other hand, if a map $T$ satisfies your hypothesis, and also the following mild property: for any $u\in X$
$$\sup_{\|\alpha u-w\|\le N} \|T(w)-T(\alpha u)\|=o(\alpha)\qquad (\text{for}\; |\alpha|\to+\infty),$$ (for instance, it is Lipschitz), then it is homogeneous. For, if $v$ is as in your notation, $$\alpha T(u)= T(\alpha u)+ \big( T(v)-T(\alpha u)\big)+ \big(\alpha T(u)-T(v)\big)= T(\alpha u)+o(\alpha)$$ so that for all $u\in X$ $$T(u)=\lim_{|\alpha|\to +\infty}{1\over\alpha}T(\alpha u)$$ which implies it is a homogeneous map.

Consider a map $T:X\to Y$ of the form $$T(u):=p(u)H(u),$$ where $H:X\to Y$ is a Lipschitz homogeneous map (here always meaning $H(\alpha u)=\alpha H(u)$ for all $u\in X$ and $\alpha\in\mathbb{K}$), and $p:X\to\mathbb{R}$ is a bounded function with the property that the image of any ball of fixed radius $R$ is the same. For instance, any bounded $R$-periodic function of the norm.

So for any $u\in X$ and $\alpha\in\mathbb{K}$ there is $v\in X$ such that $\| \alpha u -v\|\le R$ and $p(v)=p(u)$. Then $\alpha T(u)-T(v)=\alpha p(u) H(u)-p(v)H(v)=p(u)(H(\alpha u) -H(v)) $ so that $\|\alpha T(u)-T(v)\|\le\|p\|_\infty\operatorname{Lip}(H)R$. Thus $T$ satisfies your hypothesis with $N:=\max\big(1, \|p\|_\infty\operatorname{Lip}(H)\big)R.$

On the other hand, if a map $T$ satisfies your hypothesis, and also has the following mild property: for any $u\in X$
$$\sup_{\|\alpha u-w\|\le N} \|T(w)-T(\alpha u)\|=o(\alpha)\qquad (\text{for}\; |\alpha|\to+\infty),$$ (for instance, it is Lipschitz), then it is a homogeneous map. For, if $v$ is as in your notation, $$\alpha T(u)= T(\alpha u)+ \big( T(v)-T(\alpha u)\big)+ \big(\alpha T(u)-T(v)\big)= T(\alpha u)+o(\alpha)$$ so that $T(0)=0$, and for all $u\in X$ $$T(u)=\lim_{|\alpha|\to +\infty}{1\over\alpha}T(\alpha u)$$ which implies it is a homogeneous map.

Consider a map $T:X\to Y$ of the form $$T(u):=p(u)H(u),$$ where $H:X\to Y$ is a Lipschitz homogeneous map (here always meaning $H(\alpha u)=\alpha H(u)$ for all $u\in X$ and $\alpha\in\mathbb{K}$), and $p:X\to\mathbb{R}$ is a bounded function with the property that the image of any ball of fixed radius $R$ is the same. For instance, any bounded $R$-periodic function of the norm.

So for any $u\in X$ and $\alpha\in\mathbb{K}$ there is $v\in X$ such that $\| \alpha u -v\|\le R$ and $p(v)=p(u)$. Then $\alpha T(u)-T(v)=\alpha p(u) H(u)-p(v)H(v)=p(u)(H(\alpha u) -H(v)) $ so that $\|\alpha T(u)-T(v)\|\le\|p\|_\infty\operatorname{Lip}(H)R$. Thus $T$ satisfies your hypothesis with $N:=\max\big(1, \|p\|_\infty\operatorname{Lip}(H)\big)R.$

On the other hand, if a map $T$ satisfies your hypothesis, and also the following mild property: for any $u\in X$
$$\sup_{\|\alpha u-w\|\le N} \|T(w)-T(\alpha u)\|=o(\alpha)\qquad (\text{for}\; |\alpha|\to+\infty),$$ (for instance, it is Lipschitz), then it is homogeneous. For, if $v$ is as in your notation, $$T(u)={1\over\alpha}T(\alpha u)+{1\over\alpha}\big(T(v)-T(\alpha u)\big)+O\big({1\over\alpha}\big)={1\over\alpha}T(\alpha u)+o(1)$$ so that for all $u\in X$ $$T(u)=\lim_{\alpha\to +\infty}{1\over\alpha}T(\alpha u)$$ which implies it is a homogeneous map.

Consider a map $T:X\to Y$ of the form $$T(u):=p(u)H(u),$$ where $H:X\to Y$ is a Lipschitz homogeneous map (here always meaning $H(\alpha u)=\alpha H(u)$ for all $u\in X$ and $\alpha\in\mathbb{K}$), and $p:X\to\mathbb{R}$ is a bounded function with the property that the image of any ball of fixed radius $R$ is the same. For instance, any bounded $R$-periodic function of the norm.

So for any $u\in X$ and $\alpha\in\mathbb{K}$ there is $v\in X$ such that $\| \alpha u -v\|\le R$ and $p(v)=p(u)$. Then $\alpha T(u)-T(v)=\alpha p(u) H(u)-p(v)H(v)=p(u)(H(\alpha u) -H(v)) $ so that $\|\alpha T(u)-T(v)\|\le\|p\|_\infty\operatorname{Lip}(H)R$. Thus $T$ satisfies your hypothesis with $N:=\max\big(1, \|p\|_\infty\operatorname{Lip}(H)\big)R.$

On the other hand, if a map $T$ satisfies your hypothesis, and also the following mild property: for any $u\in X$
$$\sup_{\|\alpha u-w\|\le N} \|T(w)-T(\alpha u)\|=o(\alpha)\qquad (\text{for}\; |\alpha|\to+\infty),$$ (for instance, it is Lipschitz), then it is homogeneous. For, if $v$ is as in your notation, $$\alpha T(u)= T(\alpha u)+ \big( T(v)-T(\alpha u)\big)+ \big(\alpha T(u)-T(v)\big)= T(\alpha u)+o(\alpha)$$ so that for all $u\in X$ $$T(u)=\lim_{|\alpha|\to +\infty}{1\over\alpha}T(\alpha u)$$ which implies it is a homogeneous map.