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Let $f:\mathbb{\,R}^{2}\longrightarrow\mathbb{R}\setminus\left\{ 0\right\} $ be a function such that $$\sup_{x,y\in\mathbb{R}:\,0<\left|x-y\right|\leq1}\left|f\left(x,y\right)\right|=M<+\infty.$$ I would like to prove that $$\sup_{x,y\in\mathbb{R}:\,0<\left|x-y\right|\leq1}\left|x-y\right|\left|f\left(x,y\right)\right|=\sup_{x,y\in\mathbb{R}:\,0<\left|x-y\right|\leq1}\left|x-y\right|\sup_{x,y\in\mathbb{R}:\,0<\left|x-y\right|\leq1}\left|f\left(x,y\right)\right|.\tag{1}$$ I write a proof but I don't know if it is correct.

My attempt: we know that if $A$ and $B$ are two set of positive real numbers and $\sup A<+\infty,$ $\sup B<+\infty,$ then $$\sup AB=\sup A\sup B.$$Now since$$\sup_{x,y\in\mathbb{R}:\,0<\left|x-y\right|\leq1}\left|x-y\right|=1$$ we can conlude that $(1)$ holds. Is my proof correct? Thank you.

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Your statement is not true: consider $f(x, y) = \mathrm{e}^{-|x-y|}$.

$t \mapsto \mathrm{e}^{-t}$ is strictly decreasing so the first supremum is attained for $|x-y| \to 0$, and it is $1$. However $|x-y|\mathrm{e}^{-|x-y|}$ attains its maximum for $|x-y| = 1$, as you can check by calculating the derivative. The maximum in this case is $\frac{1}{\mathrm{e}}$.

The problem with your proof is that the product of two sets $AB$ consists of all possible products of two elements $a \in A$ and $b \in B$. This is not the case with the product of two functions $fg$, as the range of the product consists only of products $f(t)g(t)$, for some $t$ from the domain, not of products $f(s)g(t)$ for any $s$ and $t$.

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    $\begingroup$ So the problem is that actually I consider only the "diagonal" elements. Thank you. $\endgroup$
    – user422009
    Aug 31 '17 at 19:31
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The equality $\sup AB=\sup A\sup B$ holds if the sets $A,B$ are subsets of $[0, +\infty)$.

So your idea of proving your statement is correct.

But you have to be more careful in general because the equality does not hold in cases where,fot instance, subsets of $(- \infty,0)$ get involded.

Take for instance $A=\{-1\}$ and $B=[0,1]$.

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