It may be a simple question to post it here, but I posted this question in the Math Stack Exchange forum and no one answered me.
Let $E$ be a (possibly infinite) alphabet and consider $X = E^{\mathbb{Z}^d}$. For each $\lambda = (\lambda_1, \dots, \lambda_d) \in \mathbb{Z}^d$, let $|\lambda| = \max_{1 \leq j \leq d}|\lambda_j|$ and consider $\Lambda_n = \{\lambda \in \mathbb{Z}^d \,:\, |\lambda| < n\}$ the open box of radius $n$ centered at the origin.
Given a continuous function $f\colon X \to \mathbb{R}$, define, for each $n \in \mathbb{N}$, $$\delta_n(f) = \sup\{|f(x) - f(y)| \,:\, x|_{\Lambda_n} = y|_{\Lambda_n}\}.$$
Now, let $g\colon X \to \mathbb{R}$ be a bounded uniformly continuous function and express it in the following way:
$$g(x) = g_1(x) + g_2(x) + g_3(x) + \dots,$$
where $g_1(x) = \inf g|_{ [x|_{\Lambda_1}]}$ and $g_n(x) = \inf g|_{[x|_{\Lambda_n}]} - \inf g|_{[ x|_{\Lambda_{n-1}}]},$ for $n \geq 2$.
$\textbf{Claim.}$ $||g_n||_{\infty} < \delta_{n-1}(g)$, $\forall n \geq 1$.
What I've done so far:
Since $g$ is bounded, we can suppose, without loss of generality, that $g$ is positive. Indeed, if there is $x \in X$ such that $g(x) < 0$, let $M = \inf_{x \in X} g(x)$. Then $ -\infty < M < 0$ and we can consider the new function $\tilde{g}(x) = g(x) + |M|$, which is positive and has the same properties as $g$.
Observe the following relation between the cylinder sets: $[x|_{\Lambda_{n}}] \subset [x|_{\Lambda_{n-1}}],$ since we have fewer restrictions in the second one. Therefore, we get
$$\inf g|_{[x|_{\Lambda_n}]} \geq \inf g|_{[x|_{\Lambda_{n-1}}]},$$ $$\sup g|_{[x|_{\Lambda_n}]} \leq \sup g|_{[x|_{\Lambda_{n-1}}]}.$$
I tried to use all these informations in the calculation of $||g_n||_{\infty}$ and the relations and properties of $\sup$ and $\inf$, but couldn't get to the claimed inequality. I guess I'm missing something very simple.
Could someone give me a hint on that?
Thank you!
