# Uniform conevergence

Uniform convergence

Hello!
I've got a short question to an example.
I should check the following sequence for uniform convergence on the whole of $\mathbb{R}$:
$f_n(x)=\frac{nx(7+sin(nx))}{4+n^2x^2}$
It says that the conevergence is nonuniform, because:
$sup_{\mathbb{R}}|\frac{nx(7+sin(nx))}{4+n^2x^2}|\ge{\frac{7+sin1}{5}$
Obviously they put $x=1/n$. I cannot see why this is true.
I tried to differentiate the given function sequence and see, wether the derivative can become zero or not, i.e. I look for maximum values, but the derivative term gets difficult to manage in the end.
How can I make it clear to myself that the above inequality holds?

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nazzard
Gold Member
Hello little Hilbert,

littleHilbert said:
evergence is nonuniform, because:
$sup_{\mathbb{R}}|\frac{nx(7+sin(nx))}{4+n^2x^2}|\ge{\frac{7+sin1}{5}$
Obviously they put $x=1/n$. I cannot see why this is true.

Are you sure that the inequality shouldn't read something like:

$sup_{\mathbb{R}}|\frac{nx(7+sin(nx))}{4+n^2x^2}|\leq{\frac{7+sin1}{5}$ ?

You need to find a supremum to show that the series of functions converges uniformly on R.

Obviously the sinus term can be 1 at max. And after canceling nx you will have the given numerator already. Can you go from there and find the denominator?

Regards,

nazzard

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No, the greater-equals-sign is correct.
Estimatation from above is:
$|\frac{nx(7+sin(nx))}{4+n^2x^2}|\le{\frac{8n|x|}{4+n^2x^2}$, because $|sin(nx)|\le{1}$...but it leads me to nowhere because I need a majorant independent of x.

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nazzard
Gold Member
My bad, I thought you'd have to show that it does converge uniformly, sorry. $|\frac{nx(7+sin(nx))}{4+n^2x^2}|\le{\frac{8n|x|}{4 +n^2x^2}\le{\frac{8n|x|}{n^2x^2}\le{\frac{8}{nx}$

and therefore f(x) is maximal for x = 1/n . You must now proove that

lim(n->infinity) [ sup (|fn(x)-f(x)|)] > 0

So : lim(n->infinity) [|8/(nx) - 8|] = 8 > 0

Now you need to write this properly but the main ideas are there. My experiences with analysis have teached me that you cannot be rigourous enough so if anyone sees a flaw in my reasoning, don't hesitate to rectify my monstrosities.

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shmoe
Homework Helper
The maximum of f(x) is not at x=1/n. It's probably at someplace gross. Fortunately we don't need to know where it is.

Every place you see an "n" in your function f_n(x), there's an "x" living next to it, you can think of it as a function of "nx". So if you set x=1/n the result will be entirely devoid of x and n, it's just a convenient place to evaluate f at that gives a constant. You could have taken x=2/n, or pi/n, etc.

So we know for any n,

$$f_n(1/n)=\frac{(7+sin(1))}{5}>1$$

Hence for any n we have

$$\sup_{\mathbb{R}}|f_n(x)|>1$$

since f_n(x) takes on a value larger than 1.

Taking x = 2/n or pi/n does not bound 8/(nx) maximaly but work just fine too given the fact that you must disprove the unif.conv. The use of x = 1/n seems to be forced by the " sup " part of the equation. Nevertheless, I seem to be using a slightly different theorem than you ( even if they are equiavlent) and I thank you for bringing to my attention an easier way to investigate such interesting matter.

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shmoe
Homework Helper
Gagle The Terrible said:
Taking x = 2/n or pi/n does not bound 8/(nx) maximaly but work just fine too given the fact that you must disprove the unif.conv.

Nor would x=1/n. But making 8/(nx) large won't help an upper bound.

All you've shown is there that f_n(1/n)<8. Bounding one point of the function from above does not tell you anything at all about the sup over the reals. (again I'll points out that the maximum of f_n(x) does not occur at x=1/n, and again that we don't even care)

You want to bound the sup from below in any case. For that it's enough to have a sequence of points where $$f_n(x_n)$$ is bounded away from zero, i.e. they are all >=k for some fixed k>0.