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I was asked to prove that if $f\colon A \to \mathbb{R}$ is uniformly continuous and there exists $M \in \mathbb{R}$ such that $\left | f(x) \right | \ge M$, then $g(x)= 1/f(x)$ is uniformly continuous.

But taking $A = (0,\infty)$ and $f(x)=x$ then $g(x)=1/x$ is clearly not uniformly continuous (as the derivative is not bounded).

What am I missing?

Thank you

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You seem to be missing the condition $M \gt 0$. – t.b. Mar 30 '12 at 9:52
makes sense, thank you (sometimes even profs forget things) :D – scibuff Mar 30 '12 at 9:54
If the derivative of a derivable function is not bounded, we can't deduce that the function is not uniformly continuous, for example with $f(x)=x^{3/2}\sin 1/x$ if $x\in (0,1]$ and $f(0)=0$. – Davide Giraudo Mar 30 '12 at 11:05
up vote 8 down vote accepted

In fact, it was assumed that there exists a positive constant $M$ such that for all $x\in A$: $|f(x)|\geq M$.

A hint to show the result: write $$|g(x)-g(y)|=\left|\frac 1{f(x)}-\frac 1{f(y)} \right|=\frac{|f(x)-f(y)|}{|f(x)|\cdot|f(y)|}\leq \frac 1{M^2}|f(x)-f(y)|.$$

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