prove that if two real valued functions are uniformly continuous on some interval $I$ and each is bounded on $I$ then their product $f\cdot g=f(x)\cdot g(x)$ is also uniformly continuous on $I$ . Is boundedness of each function on $I$ is necessary for the product
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1$\begingroup$ What are your thoughts? $\endgroup$ – leo Aug 19 '13 at 0:32
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Hint: Let $x_0\in I$, using the definition of uniform continuity and the boundedness of each function we prove that $fg$ is uniformly continuous at $x_0$ (and then on $I$) by the inequality
$$|f(x)g(x)-f(x_0)g(x_0)|=|f(x)g(x)-f(x)g(x_0)+f(x)g(x_0)-f(x_0)g(x_0)|\\ \leq|f(x)||g(x)-g(x_0)|+|g(x_0)||f(x)-f(x_0)|\\\leq M|g(x)-g(x_0)|+M'|f(x)-f(x_0)|$$
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$\begingroup$ How did you put those two bounds instead of $f(x)$ and $g(x_0)$ ? $\endgroup$ – GinKin Dec 26 '13 at 14:03
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1$\begingroup$ $f$ and $g$ are bounded on $I$ means that there's $M,M'\ge0$ such that $$|f(x)|\le M\quad;\quad |g(x)|\le M'\;\forall x\in I$$ $\endgroup$ – user63181 Dec 26 '13 at 14:07
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1$\begingroup$ Let $\epsilon>0$ since $f$ is uniformly continuous then there's $\eta>0$ such that $|f(x)-f(x_0)|<\frac{\epsilon}{M}$ whatever $|x-x_0|<\eta$. Do the same thing for $g$ and we find $|g(x)-g(x_0)|<\frac{\epsilon}{M'}$ if $|x-x_0|<\eta'$. Now let $\delta=\min(\eta,\eta')$ and use the inequality of my answer to prove that $$|f(x)g(x)-f(x_0)g(x_0)|<\epsilon$$ if $|x-x_0|<\delta$. $\endgroup$ – user63181 Dec 26 '13 at 14:29
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1$\begingroup$ @GinKin Surely $\epsilon$ is arbitrary but I divide by $M$ to find the final result $<\epsilon$. Cauchy has a famous saying: "Just cut $\epsilon$ out of 2";-) $\endgroup$ – user63181 Dec 26 '13 at 14:50
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Yes. For example. $f(x)=x$ is uniformly continuous on $\mathbb{R}$, but not bounded. $f^{2}(x)=x^{2}$ is not uniformly continuous on $\mathbb{R}$
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$\begingroup$ i think you understand something wrong i want to ask if product is uniformly continuous then then it is not necessary that each function is bounded. $\endgroup$ – pagal Aug 19 '13 at 0:38
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Let $f$ be uniformly continuous on a real interval $I$ with endpoints $0$ and $1$.
For any $\epsilon>0$ there is a $\delta>0$ such that for all $x,y\in I$, $|x-y|<\delta\implies |f(x)-f(y)|<\epsilon$.
In particular, by the Archimedean property, there is an $n\in\Bbb Z_+$ such that for all $x,y\in I$, $|x-y|<\frac2n\implies|f(x)-f(y)|<1$. Then for any $x\in I$, $|f(x)-f(0)|<n$. Or something like that.
