# Thread: counterexample of convergence of integral again

1. ## counterexample of convergence of integral again

fn, are continuous on [0,1] and converges pointwisely to f,

and,

$\displaystyle \int_{0}^{1}{f_{n}(x)}\rightarrow\int_{0}^{1}{f(x) }$ again,

must f be continuous ?

give counterexample

2. Originally Posted by silversand
fn, are continuous on [0,1] and converges pointwisely to f,

and, $\displaystyle \int_{0}^{1}{f_{n}(x)}\rightarrow\int_{0}^{1}{f(x) }$ again, must f be continuous ?

give counterexample
here's a counter-example: $\displaystyle f_n(x)=\sin^n(\pi x).$

3. Originally Posted by silversand
fn, are continuous on [0,1] and converges pointwisely to f,

and,

$\displaystyle \int_{0}^{1}{f_{n}(x)}\rightarrow\int_{0}^{1}{f(x) }$ again,

must f be continuous ?

give counterexample
Here is an easier conterexample.

Let $\displaystyle f_n(x) = x^n$ then each one is continous and converges to $\displaystyle f(x)$ which is $\displaystyle 0$ for $\displaystyle 0\leq x < 1$ with $\displaystyle f(1) = 1$.

Now $\displaystyle \int_0^1 x^n dx = \frac{1}{n+1} \to 0 = \int_0^1 f(x) dx$

But $\displaystyle f$ is not continous.

4. Originally Posted by ThePerfectHacker
Here is an easier conterexample.

Let $\displaystyle f_n(x) = x^n$ then each one is continous and converges to $\displaystyle f(x)$ which is $\displaystyle 0$ for $\displaystyle 0\leq x < 1$ with $\displaystyle f(1) = 1$.

Now $\displaystyle \int_0^1 x^n dx = \frac{1}{n+1} \to 0 = \int_0^1 f(x) dx$

But $\displaystyle f$ is not continous.
that's certainly an easier example but i like mine better! haha ... let's see, how would you prove that $\displaystyle \lim_{n\to\infty} \int_0^1 \sin^n(\pi x) \ dx = 0$?

5. Originally Posted by NonCommAlg
that's certainly an easier example but i like mine better! haha ... let's see, how would you prove that $\displaystyle \lim_{n\to\infty} \int_0^1 \sin^n(\pi x) \ dx = 0$?
Reduction formula: $\displaystyle \int \sin^n(\pi x)\,dx = -\frac{1}{n}\cos(\pi x)sin^{n-1}(\pi x) + \frac{n-1}{n}\int sin^{n-2}(\pi x)\,dx$

There are some $\displaystyle \pi$'s and $\displaystyle \frac{1}{\pi}$'s that should be in there somewhere, but I'm too lazy to find out where, and they shouldn't make any difference anyway. As $\displaystyle n\to\infty$, the non-integral part clearly goes to 0, and then you might be able to prove by induction that everything is 0. Or something.

6. Originally Posted by NonCommAlg
that's certainly an easier example but i like mine better! haha ... let's see, how would you prove that $\displaystyle \lim_{n\to\infty} \int_0^1 \sin^n(\pi x) \ dx = 0$?
If $\displaystyle t=\pi x$ then we get $\displaystyle \frac{1}{\pi}\int_0^{\pi} \sin^n t dt$.
The function $\displaystyle \sin^n t$ is symmetric about $\displaystyle t=\frac{\pi}{2}$.
Thus, $\displaystyle \frac1{\pi}\int_0^{\pi/2}\sin^n t = \frac{2}{\pi}\int_0^{\pi/2}\sin^n t dt = {{2n}\choose n}\cdot \frac{1}{2^{2n}}$

It remains to show that expression goes to zero (such as using Stirling's approximation).
----

Notice something interesting, there is a power series:
$\displaystyle \sum_{n=0}^{\infty} {{2n}\choose n}z^n = \frac{1}{\sqrt{1-4z}}$ for $\displaystyle |z|<\tfrac{1}{4}$
Therefore, $\displaystyle {{2n}\choose n}z^n \to 0$ if $\displaystyle |z| < \tfrac{1}{4}$.
But in our case $\displaystyle z=\tfrac{1}{4}$ to get $\displaystyle {{2n}\choose n}\cdot \frac{1}{2^{2n}}$.
However, $\displaystyle \sum_{n=0}^{\infty} {{2n}\choose n}z^n$ cannot be analytically continued at $\displaystyle z=\tfrac{1}{4}$ since $\displaystyle \frac1{\sqrt{1-4z}}$ explodes that at point.
Somethings seems to be wrong.