counterexample of convergence of integral again

**silversand** · May 1st 2009, 02:32 PM

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

and,

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

must f be continuous ?

give counterexample

**NonCommAlg** · May 1st 2009, 02:59 PM

Originally Posted by silversand

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

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

give counterexample

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

**ThePerfectHacker** · May 1st 2009, 03:25 PM

Originally Posted by silversand

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

and,

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

must f be continuous ?

give counterexample

Here is an easier conterexample.

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

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

But $f$ is not continous.

**NonCommAlg** · May 1st 2009, 03:50 PM

Originally Posted by ThePerfectHacker

Here is an easier conterexample.

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

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

But $f$ is not continous.

that's certainly an easier example but i like mine better! haha ... let's see, how would you prove that $\lim_{n\to\infty} \int_0^1 \sin^n(\pi x) \ dx = 0$ ?

**redsoxfan325** · May 1st 2009, 04:11 PM

Originally Posted by NonCommAlg

that's certainly an easier example but i like mine better! haha ... let's see, how would you prove that $\lim_{n\to\infty} \int_0^1 \sin^n(\pi x) \ dx = 0$ ?

Reduction formula: $\int \sin^n(\pi x)\,dx = -\frac{1}{n}\cos(\pi x)sin^{n-1}(\pi x) + \frac{n-1}{n}\int sin^{n-2}(\pi x)\,dx$

There are some $\pi$ 's and $\frac{1}{\pi}$ '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 $n\to\infty$ , the non-integral part clearly goes to 0, and then you might be able to prove by induction that everything is 0. Or something.

**ThePerfectHacker** · May 1st 2009, 06:36 PM

Originally Posted by NonCommAlg

that's certainly an easier example but i like mine better! haha ... let's see, how would you prove that $\lim_{n\to\infty} \int_0^1 \sin^n(\pi x) \ dx = 0$ ?

If $t=\pi x$ then we get $\frac{1}{\pi}\int_0^{\pi} \sin^n t dt$ .
The function $\sin^n t$ is symmetric about $t=\frac{\pi}{2}$ .
Thus, $\frac1{\pi}\int_0^{\pi/2}\sin^n t = \frac{2}{\pi}\int_0^{\pi/2}\sin^n t dt = {{2n}\choose n}\cdot \frac{1}{2^{2n}}$

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

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

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