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Let $(x_n)$ be any function sequence such that

$ \int_0^1x_n(t)dt=1 \qquad \forall n $

$ \lim_{n\to\infty}x_n = x $

I'm trying to prove that the limit $x$ also has the property $\int_0^1x(t)dt=1$. I don't think I could construct a "bounding" function to use the dominated convergence theorem. Could I have a hint?

  • 3
    Depending on the type of convergence it may or may not be true2012-11-07

3 Answers 3

4

You can't prove it because it's not true. Take $x_n=n\chi_{(0,1/n]}$ and $x(t)=0$.

3

Assuming that you're talking about a pointwise limit, this doesn't work. Consider $x_n(t)=\begin{cases}-2n^2t+2n & 0 These converge pointwise to the zero function.

Now, if we happen to know that the functions $x_n$ converge uniformly, then we can let the limit "pass through" the integral, and get $1=\lim_{n\to\infty} 1=\lim_{n\to\infty}\int_0^1x_n(t)\,dt=\int_0^1\left(\lim_{n\to\infty}x_n(t)\right)\,dt=\int_0^1 x(t)\,dt.$

2

This is not necessarily true. Consider the function

$f_n(x)=nxe^{-nx^2}$

Note that $\int_0^1 {{f_n}} (x)dx = \int_0^1 {nx{e^{ - n{x^2}}}dx} = \left( { - \frac{1}{2}} \right)\left. {{e^{ - n{x^2}}}} \right|_0^1 = - \frac{1}{2}{e^{ - n}} + \frac{1}{2}$

Then $\lim \int_0^1 {{f_n}} (x)dx=\frac 1 2$

But

$f_n\to 0$

so $\int_0^1 f(x) dx =0$

A sufficient condition for this to be true is that the $f_n$ are all integrable and $f_n$ converges uniformly to $f$ over $[a,b]$.