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Topic: Proof that mixed partials commute.
Replies: 20   Last Post: Nov 22, 2013 11:57 PM

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Hetware

Posts: 148
Registered: 4/13/13
Re: Proof that mixed partials commute.
Posted: Nov 20, 2013 7:27 PM
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On 11/20/2013 11:07 AM, dullrich@sprynet.com wrote:
> On Wed, 20 Nov 2013 06:55:30 -0500, Hetware <hattons@speakyeasy.net>
> wrote:
>

>> On 11/20/2013 3:22 AM, Robin Chapman wrote:
>>> On 19/11/2013 23:59, Hetware wrote:
>>>>>
>>>>> That's very bad notation. It's not one limit, it's the limit of
>>>>> a limit. Should be
>>>>>
>>>>> Limt(Limt(...)[x->c][y->c].
>>>>>
>>>>> And now the big question is why
>>>>>
>>>>> Limt(Limt(...)[x->c][y->c] = Limt(Limt(...)[y->c][x->c]

>>>>
>>>> I guess I should have included the intermediate steps. I had intended
>>>> that the order of taking limits should be ambiguous.

>>>
>>> That's the nub of the matter. Iterated limits need not commute.
>>> One has to show that in this case they do. Putting in deliberate
>>> ambiguities in your notation sounds a really bad idea.
>>>
>>> Of course there are examples where mixed partials are different,
>>> so your original argument can't have been valid, since it didn't
>>> use the necessary hypotheses about continuity of partials etc.
>>>

>>
>> But I added my reason for assuming the limits commute. I expressed a
>> function of two independent variables as the function of a single
>> variable and appealed to the limit rules for a function of a single
>> variable to the result.
>>
>> The question is whether that reasoning is valid.

>
> It can't be valid, since it "proves" something false!
> Mixed partials are the same _under_ certain hypotheses.
> Your proof, if valid, would show that they commute
> _wiithout_ those hypotheses. And that's not true.
>
> Many theorems in analysis amount to showing
> that some particular two limits commuute.
>
>


The starting point for the proof, as stated in the OP:
"Theorem. If the function w=f(x,y) together with the partial derivatives
f_x, f_y, f_xy and f_yx are continuous, then f_xy = f_yx."





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