Uniform Convergence and integration

1. Aug 13, 2009

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

$$\int_a^b u (x) \, dx=\lim_{n \to \infty} \int_a^b u_n (x) \, dx$$

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 $u(x)$." )

2. Aug 13, 2009

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

$$A\quad\textrm{and}\quad B\implies C$$

then you also know

$$\textrm{not}\; C\implies (\textrm{not}\;A)\quad\textrm{or}\quad(\textrm{not}\;B)$$

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

$$\lim_{n\to\infty} \int\limits_a^b u_n(x)dx = \int\limits_a^b u(x) dx.$$

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

$$B\quad\textrm{and}\quad D\implies C$$

and

$$A\quad\textrm{and}\quad B\implies D$$

(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.)

Last edited: Aug 13, 2009
3. Aug 13, 2009

angryfaceofdr

How would you get $$\neg A$$? Do you negate closed and the interval as $a>x>b$?

Or in other words,

" Assume the functions $$u_n (x)$$ are not continuous on the not closed interval $$a > x > b$$"?

Last edited: Aug 13, 2009
4. Aug 13, 2009

jostpuur

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.

Last edited: Aug 13, 2009
5. Aug 13, 2009

angryfaceofdr

How does one know when to do this?

6. Aug 13, 2009

angryfaceofdr

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?

7. Aug 14, 2009

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.)