Convergence pointwise and in the uniform (sup) metric

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Karen Bart - 29 Oct 2008 20:17 GMT

Dear All,

I would like to check and prove or disprove the convergence (firstly
pointwise and then under the uniform/sup metric) of the following two
sequences of functions in the space of bounded functions from [0, 2Pi] to
R=reals).

f_n (x) = (sin x)^(2n)
f_n (x) = sin(nx)

I have done some preliminary work with both of these but my argumentation is
flawed so I won't repeat it here.

Thank you.

Best.

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Brian M. Scott - 30 Oct 2008 00:19 GMT

On Wed, 29 Oct 2008 19:17:01 -0000, Karen Bart
<name@host.com> wrote in
<news:GE2Ok.115581$5A6.34079@newsfe09.ams2> in
alt.math.undergrad:

> I would like to check and prove or disprove the
> convergence (firstly pointwise and then under the
> uniform/sup metric) of the following two sequences of
> functions in the space of bounded functions from [0, 2Pi]
> to R=reals).

> f_n (x) = (sin x)^(2n)

It's not too hard to see what the pointwise limit is. If x
is neither pi/2 nor 3*pi/2, then -1 < sin x < 1, and hence
f_n (x) --> 0 as n --> oo. If x = pi/2 or x = 3*pi/2, on
the other hand, then f_n (x) = 1 for all n. In other words,
if f : [0, 2*pi] --> R is defined by f(pi/2) = 1 = f(3*pi/2)
and f(x) = 0 for x in [0, 2*pi]\{pi/2, 3*pi/2}, then the f_n
converge pointwise to f. Now all you have to do is decide
whether this convergence is uniform: either the f_n converge
uniformly to f, or they don't converge uniformly at all.

> f_n (x) = sin(nx)

Hint: show that the sequence <f_n (1) : n in N> doesn't
converge in R.

[...]

Brian

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Karen Bart - 30 Oct 2008 18:09 GMT

> On Wed, 29 Oct 2008 19:17:01 -0000, Karen Bart
> <name@host.com> wrote in
[quoted text clipped - 5 lines]
> Hint: show that the sequence <f_n (1) : n in N> doesn't
> converge in R.

Got the first bit. Thanks very much.
For the second, essentially we will show this is not pointwise and use the
fact that uniform conv. implies pointwise conv. and, thus, not pointwise
implies not uniform. Is this a suitable proof for showing it is not
pointwise?
f_n(1)=sin(x) and so if f_n(1) converges it must converge to f(x)=sin(x),
Thus, |f_n(x)-f(x)|=|sin(x)-sin(x)|=0 > epsilon.
Originally I thought to use x=1/n.
Then |f_n(x)-f(x)|=|sin(1)-sin(1)|=0>epsilon.
Is that okay? The assumption of f(x) seems dodgy to me.
Best wishes. and thanks again.

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Karen Bart - 30 Oct 2008 18:45 GMT

>> On Wed, 29 Oct 2008 19:17:01 -0000, Karen Bart
>> <name@host.com> wrote in
[quoted text clipped - 17 lines]
> Is that okay? The assumption of f(x) seems dodgy to me.
> Best wishes. and thanks again.

Oh dear, epsilon is >0 not <0!

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Brian M. Scott - 30 Oct 2008 19:21 GMT

On Thu, 30 Oct 2008 17:09:42 -0000, Karen Bart
<name@host.com> wrote in
<news:jTlOk.106297$WX2.67640@newsfe17.ams2> in
alt.math.undergrad:

>> On Wed, 29 Oct 2008 19:17:01 -0000, Karen Bart
>> <name@host.com> wrote in
>> <news:GE2Ok.115581$5A6.34079@newsfe09.ams2> in
>> alt.math.undergrad:

>>> f_n (x) = sin(nx)

>> Hint: show that the sequence <f_n (1) : n in N> doesn't
>> converge in R.

> Got the first bit. Thanks very much.

> For the second, essentially we will show this is not
> pointwise and use the fact that uniform conv. implies
> pointwise conv. and, thus, not pointwise implies not
> uniform. Is this a suitable proof for showing it is not
> pointwise?

> f_n(1)=sin(x)

You've got off on the wrong foot right at the start, I"m
afraid: f_n (1) = sin(n*1) = sin(n). But <sin(n) : n in N>
doesn't converge, so <f_n (1) : n in N> doesn't converge,
and therefore <f_n : n in N> has no pointwise limit.

I suspect from this mistake that you probably don't at this
point have a valid proof that <sin(n) : n in N> doesn't
converge in R; that's actually the hard part of the
argument.

[...]

Brian

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Karen Bart - 30 Oct 2008 20:16 GMT

> On Thu, 30 Oct 2008 17:09:42 -0000, Karen Bart
> <name@host.com> wrote in
[quoted text clipped - 30 lines]
> converge in R; that's actually the hard part of the
> argument.

Hi. Yes, you are right, sorry. If it was sin(n*Pi/2), n in N, then I would
simply show it has two subsquences converging to two distinct points,
e.g. -1 and 1, which would show it diverges but sin(n) by itself seems a
little tricky. Any suggestions?

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Karen Bart - 30 Oct 2008 21:05 GMT

> Hi. Yes, you are right, sorry. If it was sin(n*Pi/2), n in N, then I
> would simply show it has two subsquences converging to two distinct
> points, e.g. -1 and 1, which would show it diverges but sin(n) by itself
> seems a little tricky. Any suggestions?

Perhaps it's just sufficient to say as n grows without bounds sin(n)
continues to oscillate never settling to a single value or some such. It's
an argument and an argument is a proof...

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Brian M. Scott - 30 Oct 2008 21:25 GMT

On Thu, 30 Oct 2008 19:16:45 -0000, Karen Bart
<name@host.com> wrote in
<news:pKnOk.147491$dN6.56880@newsfe18.ams2> in
alt.math.undergrad:

>> On Thu, 30 Oct 2008 17:09:42 -0000, Karen Bart
>> <name@host.com> wrote in
>> <news:jTlOk.106297$WX2.67640@newsfe17.ams2> in
>> alt.math.undergrad:

>>>> On Wed, 29 Oct 2008 19:17:01 -0000, Karen Bart
>>>> <name@host.com> wrote in
>>>> <news:GE2Ok.115581$5A6.34079@newsfe09.ams2> in
>>>> alt.math.undergrad:

>>>>> f_n (x) = sin(nx)

>>>> Hint: show that the sequence <f_n (1) : n in N> doesn't
>>>> converge in R.

>>> Got the first bit. Thanks very much.

>>> For the second, essentially we will show this is not
>>> pointwise and use the fact that uniform conv. implies
>>> pointwise conv. and, thus, not pointwise implies not
>>> uniform. Is this a suitable proof for showing it is not
>>> pointwise?

>>> f_n(1)=sin(x)

>> You've got off on the wrong foot right at the start, I"m
>> afraid: f_n (1) = sin(n*1) = sin(n). But <sin(n) : n in N>
>> doesn't converge, so <f_n (1) : n in N> doesn't converge,
>> and therefore <f_n : n in N> has no pointwise limit.

>> I suspect from this mistake that you probably don't at this
>> point have a valid proof that <sin(n) : n in N> doesn't
>> converge in R; that's actually the hard part of the
>> argument.

> Hi. Yes, you are right, sorry. If it was sin(n*Pi/2), n
> in N, then I would simply show it has two subsquences
> converging to two distinct points, e.g. -1 and 1, which
> would show it diverges but sin(n) by itself seems a
> little tricky. Any suggestions?

Actually, I was making it too hard, simply because I know
that <sin n : n in N> is divergent. By all means look at
the sequence <f_n (pi/2) : n in N> instead; as you say,
that's <sin(n*pi/2) : n in N> = <(-1)^(n+1) : n in N>, which
clearly does not converge, and it follows that
<f_n : n in N> cannot converge pointwise.

Brian

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Paul Sperry - 30 Oct 2008 21:52 GMT

> On Thu, 30 Oct 2008 19:16:45 -0000, Karen Bart
> <name@host.com> wrote in
[quoted text clipped - 46 lines]
> the sequence <f_n (pi/2) : n in N> instead; as you say,
> that's <sin(n*pi/2) : n in N> = <(-1)^(n+1) : n in N>, which

Hmm. sin(2*(Pi/2)) = ? :-)

> clearly does not converge, and it follows that
> <f_n : n in N> cannot converge pointwise.
>
> Brian

Signature

Paul Sperry
Columbia, SC (USA)

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Brian M. Scott - 30 Oct 2008 22:01 GMT

On Thu, 30 Oct 2008 16:52:01 -0400, Paul Sperry
<plsperry@sc.rr.com> wrote in
<news:301020081652017150%plsperry@sc.rr.com> in
alt.math.undergrad:

>> On Thu, 30 Oct 2008 19:16:45 -0000, Karen Bart
>> <name@host.com> wrote in
>> <news:pKnOk.147491$dN6.56880@newsfe18.ams2> in
>> alt.math.undergrad:

>>>> On Thu, 30 Oct 2008 17:09:42 -0000, Karen Bart
>>>> <name@host.com> wrote in
>>>> <news:jTlOk.106297$WX2.67640@newsfe17.ams2> in
>>>> alt.math.undergrad:

>>>>>> On Wed, 29 Oct 2008 19:17:01 -0000, Karen Bart
>>>>>> <name@host.com> wrote in
>>>>>> <news:GE2Ok.115581$5A6.34079@newsfe09.ams2> in
>>>>>> alt.math.undergrad:

>>>>>>> f_n (x) = sin(nx)

>>>>>> Hint: show that the sequence <f_n (1) : n in N> doesn't
>>>>>> converge in R.

>>>>> Got the first bit. Thanks very much.

>>>>> For the second, essentially we will show this is not
>>>>> pointwise and use the fact that uniform conv. implies
>>>>> pointwise conv. and, thus, not pointwise implies not
>>>>> uniform. Is this a suitable proof for showing it is not
>>>>> pointwise?

>>>>> f_n(1)=sin(x)

>>>> You've got off on the wrong foot right at the start, I"m
>>>> afraid: f_n (1) = sin(n*1) = sin(n). But <sin(n) : n in N>
>>>> doesn't converge, so <f_n (1) : n in N> doesn't converge,
>>>> and therefore <f_n : n in N> has no pointwise limit.

>>>> I suspect from this mistake that you probably don't at this
>>>> point have a valid proof that <sin(n) : n in N> doesn't
>>>> converge in R; that's actually the hard part of the
>>>> argument.

>>> Hi. Yes, you are right, sorry. If it was sin(n*Pi/2), n
>>> in N, then I would simply show it has two subsquences
>>> converging to two distinct points, e.g. -1 and 1, which
>>> would show it diverges but sin(n) by itself seems a
>>> little tricky. Any suggestions?

>> Actually, I was making it too hard, simply because I know
>> that <sin n : n in N> is divergent. By all means look at
>> the sequence <f_n (pi/2) : n in N> instead; as you say,
>> that's <sin(n*pi/2) : n in N> = <(-1)^(n+1) : n in N>, which

> Hmm. sin(2*(Pi/2)) = ? :-)

Oops; for some reason I was thinking of powers rather than
multiples. Of course <f_n (pi/2) : n in N> =
<sin(n*pi/2) : n in N> = <0, 1, 0, -1, 0, 1, 0, -1, ...>.

>> clearly does not converge, and it follows that
>> <f_n : n in N> cannot converge pointwise.

Brian

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Karen Bart - 31 Oct 2008 20:26 GMT

>>> Actually, I was making it too hard, simply because I know
>>> that <sin n : n in N> is divergent. By all means look at
[quoted text clipped - 6 lines]
> multiples. Of course <f_n (pi/2) : n in N> =
> <sin(n*pi/2) : n in N> = <0, 1, 0, -1, 0, 1, 0, -1, ...>.

Thanks very much! I got that too myself. It does the trick nicely :)

Thanks for all your guidance.

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Paul Sperry - 30 Oct 2008 19:33 GMT

> > On Wed, 29 Oct 2008 19:17:01 -0000, Karen Bart
> > <name@host.com> wrote in
[quoted text clipped - 11 lines]
> implies not uniform. Is this a suitable proof for showing it is not
> pointwise?

Calculate f_1(Pi/2), f_2(Pi/2), f_3(Pi/2), f_4(Pi/2) etc. and you'll
see what's going on.

Signature

Paul Sperry
Columbia, SC (USA)

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Math Forum / Mathematics / Undergraduate Math / October 2008

Convergence pointwise and in the uniform (sup) metric