Uniform Convergence and integration

[angryfaceofdr]

What can I conclude using the following theorem?

Let the functions u_n (x) be continuous on the closed interval a \le x \le b and let them converge uniformly on this interval to the limit function u(x). Then

<br /> \int_a^b u (x) \, dx=\lim_{n \to \infty} \int_a^b u_n (x) \, dx<br />

Can I conclude that if \int_a^b u (x) \, dx \neq \lim_{n \to \infty} \int_a^b u_n (x) \, dx, then the sequence of functions u_n (x) is NOT uniformly convergent on the interval?

-Also what is the contrapositive of the above theorem? (I am confused on how to negate P=" Let the functions u_n (x) be continuous on the closed interval a \le x \le b and let them converge uniformly on this interval to the limit function u(x)." )

 

[jostpuur]

If you get confused by logic, then try to assign symbols for different statements, and reduce the problem to a simpler form. If you know

<br /> A\quad\textrm{and}\quad B\implies C<br />

then you also know

<br /> \textrm{not}\; C\implies (\textrm{not}\;A)\quad\textrm{or}\quad(\textrm{not}\;B)<br />

Your example is slightly more confusing than it would need to be, because continuity of u_n is not highly essential. If u_n and u are integrable, and u_n\to u uniformly, then

<br /> \lim_{n\to\infty} \int\limits_a^b u_n(x)dx = \int\limits_a^b u(x) dx.<br />

Substituting A to be the knowledge that u_n are all continuous, B to be the knowledge that u_n\to u uniformly, and C to be the knowledge that the limit and integral commute, might bring some clarity.

I would also set D to be the knowledge that u_n and u are integrable. Then

<br /> B\quad\textrm{and}\quad D\implies C<br />

and

<br /> A\quad\textrm{and}\quad B\implies D<br />

(additional comment: I just realized that if u_n are integrable and if u_n\to u uniformly, then probably u is integrable too, but I'm not 100% sure of this right now. It could be that some parts of my response are not the most sense making, but I think there is nothing incorrect there anyway.)

 

[angryfaceofdr]

How would you get \neg A? Do you negate closed and the interval as a&gt;x&gt;b?

Or in other words,

" Assume the functions u_n (x) are not continuous on the not closed interval a &gt; x &gt; b"?

 

[jostpuur]

How would you get \neg A? Do you negate closed and the interval as a&gt;x&gt;b?

I would keep the assumption that the functions are of form u_n:[a,b]\to\mathbb{C} untouched, and only let A concern the continuity.

It should be recognized that there is true ambiguity in questions like this. If I declare that u_n will always have the domain [a,b], and then state that A means that u_n are all continuous, than \neg A will simply mean that u_n are not all continuous. This is what I meant originally.

Alternatively one could state that A means that u_n all have the domain [a,b] and that u_n are all continuous. Then \neg A would mean that u_n don't all have the domain [a,b] or that u_n are not all continuous.

You must know yourself what you mean by A, before asking what \neg A is.

 

[angryfaceofdr]

I would keep the assumption that the functions are of form u_n:[a,b]\to\mathbb{C} untouched, and only let A concern the continuity.

How does one know when to do this?

 

[angryfaceofdr]

You must know yourself what you mean by A, before asking what \neg A is.

My fear is what if I "choose" the wrong statement A like I did in post #3? How would I know what the "right" statement is?

 

[jostpuur]

There are no right or wrong choices for A. What matters is that you know what you have chosen.

Once A has been fixed, there can be right or wrong deductions, however.

(There was a movie where somebody said something like "there's no right or wrong decisions. What matters is that you dare to make a decision". It could be this is slightly off topic, though.)