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I am looking at the following exercise: $f,g:A \to R$ are uniformly continuous at $A$.If we suppose that $f,g$ are bounded,prove that $f \cdot g:A \to R$ is uniformly continuous at $A$. I have thought to do it like that: Let $\epsilon'>0$.Since $f$ is uniformly continuous at $A$, there is a $\delta_1$ such that $\forall x,y \in A$ with $$|x-y|<\delta_1 \Rightarrow |f(x)-f(y)|<\epsilon'$$ Since,g is uniformly continuous at $A$,there is a $\delta_2$ such that $\forall x,y \in A$ with $$|x-y|<\delta_2 \Rightarrow |g(x)-g(y)|<\epsilon'$$ We set $\delta=min\{\delta_1,\delta_2\}$. Therefore,we have :$\forall x,y \in A$ with $$|x-y|<\delta \Rightarrow |(fg)(x)-(fg)(y)|=|f(x)g(x)-f(y)g(y)|=|f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)|\leq|f(x)||g(x)-g(y)|+|g(y)||f(x)-f(y)|$$ $f$ is bounded,so there is a $M>0$ such that $\forall x,y \in A:|f(x)|\leq M$. Also,$g$ is bounded,so there is a $N>0$ such that $\forall x,y \in A:|g(x)|\leq N$. So,$|(fg)(x)-(fg)(y)|<\epsilon'(M+N)$.We set $\epsilon'=\frac{\epsilon}{M+N}$.So,we have that: $|(fg)(x)-(fg)(y)|<\epsilon$,so $f \cdot g$ is uniformly continuous.

BUT,in my textbook,they don't take $\delta_1,\delta_2$,when they use the definition that $f,g$ is uniformly continuous but the same $\delta$.Is the way that I did it wrong??

evinda
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1 Answers1

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You're not doing it wrong.

For each $\varepsilon$, knowing that $f$ is uniformly continuous gives you a $\delta$ that works for $f$, and knowing that $g$ is uniformly continuous gives you a possibly different $\delta$ that works for $g$.

As you observe, the minimum of these two $\delta$s actually works for both $f$ and $g$ (because of what exactly it means to "work" in this context).

If your textbooks jumps directly to having a single $\delta$ that works for both functions, it must be because it is implying exactly the reasoning you're writing out in full (and leaving it to the reader to figure out what goes on).

hmakholm left over Monica
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