# When is convergence of uniformly bounded monotonically increasing functions uniform?

Let $(f_n)$ be a sequence of monotonically increasing (ie. nondecreasing) functions defined on $\mathbb{R}$ with $0\leq f_n\leq 1$ for all $n$. In addition, $f_n\to f$ pointwise for some continuous $f$.

Baby Rudin (Principles of Mathematical Analysis 3rd Ed.) 7.13.b asks us to prove that the convergence must be uniform, but I think I have a counterexample: put $I(x)=0$ on $x<0$, $I(x)=1$ on $x\geq0$ and $f_n(x)=I(x-n)$ so that $f_n\to0$ pointwise but not uniformly.

Under what extra conditions is the original statement true? For example, require that the bounds $0\leq f\leq1$ be tight.

Edit: clarified definition of monotonic (this is the definition that Rudin uses).

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Your functions $f_n$ are not monotonic. – JSchlather Dec 18 '12 at 1:53
@JacobSchlather: they are perhaps not strictly monotonic, but it wouldn't take much tweaking to make them so, and still satisfy the conditions. – Ben Millwood Dec 18 '12 at 1:58
@JacobSchlather: for example, use suitably scaled and translated versions of $I=\arctan$ or $I(x)=x/\sqrt(x^2+1)$. This shows that it is not sufficient to simply replace monotonic with strictly monotonic. – user39976 Dec 18 '12 at 2:25
In the first edition, the problem is to show uniform convergence on compact sets. – David Mitra Dec 18 '12 at 2:36