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# Functional analysis (uniform convergence) watch

1. So I've just proved the Weierstrass M-test; that is, if a series of functions is such that for all , where converges, then converges uniformly on

I'm given that , let and let for . I have to show that converges uniformly on for each .

Now I'm a bit confused here. If I'm supposed to use the Weierstrass M-test on this then each of the should be bounded; but is unbounded on any , for example; and indeed is unbounded on any with .

Showing uniform convergence of the series probably isn't too difficult just using my bare hands; I suppose I'm just being thrown off by the Weierstrass M-test in the first half of the question. So if someone could verify whether or not I'm going mad, I'd appreciate it
2. We can ignore the first R (or 2R if it makes life easier, which it might) terms of the series since that's a finite sum. The rest of the terms don't have the boundedness problem.
3. (Original post by DFranklin)
We can ignore the first R (or 2R if it makes life easier, which it might) terms of the series since that's a finite sum. The rest of the terms don't have the boundedness problem.
Duh, I'm a moron. Thanks
4. Just whilst I'm here, I might as well just check that I did the last part right.

I have to determine whether converges uniformly on . I've said it doesn't:

I've noted that for all and that each is unbounded for and bounded on . Suppose that and . Then I note that is bounded/unbounded on the same regions as .

The series converges uniformly on E iff uniformly on E iff as . If is bounded on for some then choosing any , we must have that is unbounded in a neighbourhood of and so the supremum is infinite. Similarly, if is unbounded on for all then since each is bounded on , the supremum is again infinite. In either case, the supremum certainly doesn't converge to zero, and so doesn't converge uniformly on .

Once again I'm fairly sure this is correct, but is it a bit clumsy? Is there a neater way of saying this?

Thanks to anyone for their help.
5. I think you can do it more painlessly: suppose the sum uniformly converges over the reals, take epsilon = 1. By assumption of unif conv we can find N s.t. n > N => |f_n(x) - f(x)| < 1 for all real x. But f_n+1(n+1.1) > f_n(n+1.1) + 100, and so f(n+1.1) > f_n(n+1.1) + 100. Contradiction.
6. (Original post by DFranklin)
I think you can do it more painlessly: suppose the sum uniformly converges over the reals, take epsilon = 1. By assumption of unif conv we can find N s.t. n > N => |f_n(x) - f(x)| < 1 for all real x. But f_n+1(n+1.1) > f_n(n+1.1) + 100, and so f(n+1.1) > f_n(n+1.1) + 100. Contradiction.
That's certainly less awkward; thanks.

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