Help: Minor detail in proof of uniform convergence

**paupsers** · April 11th 2010, 02:22 PM

The problem states:

Prove that the following limit exists and find the limit.

$lim_{n\rightarrow\infty}\int_0^{\pi/2}1-\frac{\sqrt{x\sin {nx}}}{n}dx$

I've already shown that the pointwise limit is 1, but when I try to show that it converges uniformly using the definition that

If $|f_n(x)-f(x)|<\epsilon$

then it's uniformly convergent, I run into a slight problem... I get to a point where I have

$|\frac{\sqrt x\sqrt{\sin nx}}{n}|$ and I want to show (after a few more inequalities) that this is less than epsilon.

However, how can I account for the fact that sometimes the $\sqrt{\sin nx}$ is negative, thus giving me a nonreal value?

What this all boils down to, is can I say that

$|\frac{\sqrt x\sqrt{\sin nx}}{n}|\leq|\frac{\sqrt x}{n}|$

**Plato** · April 11th 2010, 02:40 PM

Originally Posted by paupsers

The problem states:

Prove that the following limit exists and find the limit.

$lim_{n\rightarrow\infty}\int_0^{\pi/2}1-\frac{\sqrt{x\sin {nx}}}{n}dx$

I've already shown that the pointwise limit is 1, but when I try to show that it converges uniformly using the definition that

If $|f_n(x)-f(x)|<\epsilon$

then it's uniformly convergent, I run into a slight problem... I get to a point where I have

$|\frac{\sqrt x\sqrt{\sin nx}}{n}|$ and I want to show (after a few more inequalities) that this is less than epsilon.

However, how can I account for the fact that sometimes the $\sqrt{\sin nx}$ is negative, thus giving me a nonreal value?

What this all boils down to, is can I say that

$|\frac{\sqrt x\sqrt{\sin nx}}{n}|\leq|\frac{\sqrt x}{n}|$

I am not sure that I follow all of that.
But $\left|\frac{\sqrt x\sqrt{\sin nx}}{n}\right|\leq\left|\frac{\sqrt x}{n}\right|$ is correct.
But have you been careful about domain issues? i.e. about $\sqrt{\sin nx}$

**paupsers** · April 11th 2010, 02:45 PM

Well, that's another issue. I'm trying to use a theorem that states I can interchange the order of the limit and integral, but only if ${f_n}$ is a sequence of CONTINUOUS functions... Are they not continuous due to this "glitch" with the complex answers?

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