How to prove uniform convergence

Question

I have the following sequence of functions before me:

$f_n(x)=0, \text{if} x<\dfrac {1}{n+1}$

$=\sin(\dfrac {π}{x}),\text{if} \dfrac{1}{n+1}\leq{x} \leq{\dfrac{1}{n}}$

$=0,\text{if} x>\dfrac{1}{n}$.

I have to prove that this sequence of functions converges to a continuous function but not uniformly.

The sequence converges to $f(x)=0$ in a pointwise manner. To show that it does not converge uniformly, my idea to is to integrate this sequence of functions and if I am able to get a non-zero value of the integral then I am done.

The integral is :

$\int_\dfrac{1}{n+1}^\dfrac{1}{n} \sin\dfrac{π}{x} dx$

I performed a change of variable using $\dfrac{1}{x}=t$ which gives me the integral

$\int_{n}^{n+1}\sin{\dfrac{πt}{t^2}} dt$

How can I now show that this integral has a non-zero value? Or am I being completely off track in solving this problem? Please suggest.

Answer 1

Your approach does not work. You have $$ \left | \int_{\Bbb R} f_n(x) \, dx \right| \le \int_{1/(n+1)}^{1/n} 1 \, dx = \frac{1}{n(n+1)} $$ so that $\int_{\Bbb R} f_n(x) \, dx\to 0$ for $n \to \infty$.

That does not disprove the uniform convergence. As pointed out in the comments and in the other answer you can consider $\sup f_n(x)$ instead.

Answer 2

Since $(f_n)_{n\in\Bbb N}$ converges pointwise to the null function, if it converged uniformly, it could only converge uniformly to the null function. Bu it doesn't, since$$(\forall n\in\Bbb N):\left|f_n\left(\frac1{n+1/2}\right)\right|=1.$$