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## Homework Statement

Wondering if you guys could check my proof. This is my first problem with sequences of functions

Let [itex]a > 0 [/itex] and [itex]f_{n}(x) = \frac{nx}{1+nx}. [/itex]

a) Show that the sequence of functions [itex] (f_{n})[/itex] converges pointwise on [itex] [0,∞)[/itex]

b) Show that [itex](f_{n})[/itex] converges uniformly on [itex][a,∞) [/itex]

c) Show that [itex](f_{n})[/itex] does not converges uniformly on [itex][0,a) [/itex]

## Homework Equations

ummm, theorems for pointwise convergence and uniform convergence.

edit: definitions given here

http://www.math.psu.edu/wade/M401-notes1.pdf

## The Attempt at a Solution

OK so for part a) it seems that the function will converge uniformly on [itex] [0,∞)[/itex] if we let [itex] f(x) = 0 [/itex] for [itex] x=0[/itex] and [itex] f(x) = 1[/itex] for [itex] x>0[/itex]....I believe this is correct

For b), since [itex]a>0[/itex] this implies [itex]f(a) = 1[/itex]. So by the definition of uniform convergence, [itex]|\frac{na}{1+na}-1|< ε[/itex]. This can be simplified down to [itex]|\frac{1}{na}|< ε[/itex].

So now we can let [itex]ε>0[/itex] so that [itex]\frac{1}{aε}>0[/itex]. Then by Archimedes principle there exists an [itex]N(ε)[/itex] in the Reals such that [itex]N(ε)>\frac{1}{aε}[/itex]. Now if [itex]n>N(ε)[/itex] this implies that [itex]|f_{n}(x)-f(x)| = |\frac{1}{na}| = \frac{1}{na} < \frac{1}{N(ε)a}<ε[/itex]. And hence [itex]f_{n}

[/itex] converges uniformly on [itex][a,∞)[/itex].

I wasn't sure about leaving the "a" in there, but I think that might work?

For c)

I let [itex]ε=\frac{1}{2}[/itex], and I let [itex]x = \frac{1}{n}[/itex]. If you go through with this, you end up getting [itex]|f_{n}(x)-f(x)| = \frac{1}{2} = ε[/itex], which can't happen. Therefore it is not uniformly convergent on [itex][0,a)[/itex].

Did I go wrong anywhere? This is from a final exam practice for real analysis, which I have mine on monday next week and there are no answers given.