let $(X,d)$ be a COMPACT metric space. and $f$ is continuous function on $X$ that maps $X$ to real numbers, prove $f^2$ (pointwise product) is uniformly continuous.
I know by theorem, that $f$ is uniformly continuous on $X$.
I know is general this is not true that : if $f$ is uniformly continuous, then $f.f=f^2$ is uniformly continuous.

But can I say since $X$ is compact, so is $X\times X$. Hence $f^2$ is uniformly continuous on $X\times X$. ( and we know if $f$ is continuous on $X$, so is $f^2$ on $X\times X$).

Can this be considered as a proof for this question?

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    $\begingroup$ $f^2$ is a function on $X$, not on $X\times X$. $\endgroup$ – Eric Wofsey Mar 29 at 18:41

It seems you have the right general idea, but you got confused at the end: $f^2$ is still just a function on $X$, not on $X\times X$. It's defined by $f^2(x)=f(x)^2$.

But, as you mention, the key point is that on a compact metric space, continuity and uniform continuity are equivalent. So you actually only have to show that $f^2$ is continuous, and that is immediate since a product of two continuous functions is continuous. (Or more simply, you are composing $f$ with the continuous function $x\mapsto x^2$.)

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  • $\begingroup$ OOPs! right thank you $\endgroup$ – BesMath Mar 29 at 18:58

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