# Study convergence of $f_n(x)=\frac{\sin(nx)}{\sqrt{n}}$

$$f_n(x)=\frac{\sin(nx)}{\sqrt{n}}$$

Pointwise convergence:

$$\lim_{n \rightarrow \infty } \ f_n (x)=0$$

It converges to the function $$f(x)\equiv0$$

Uniform convergence

$$f'_n(x)=n \frac{\cos(nx)}{\sqrt{n}}$$ $$f''_n(x)=-n^2 \frac{\sin(nx)}{\sqrt{n}}$$

I have to find the points $$x_0$$ that: $$f'_n(x_0)=0$$ and $$f''_n(x_0)<0$$ to apply the definition of uniform convergence.

So: $$\cos(nx_0)=0 \Longleftrightarrow n x_0=\frac{\pi}{2}+k \pi$$ $$x_0=\frac{\pi}{2n}+\frac{k \pi}{n}$$

To respect $$f''_n(x_0)<0$$:

$$x_0=\frac{\pi}{2n}+\frac{2k \pi}{n}$$

So:

$$f_n(x_0)=\frac{1}{\sqrt{n}}$$

and

$$\lim_{n \rightarrow \infty } \left| \frac{1}{\sqrt{n}}-0 \right|=0$$

It respects uniform convergence

Is it correct?

Much easier: $$|f_n(x)-0| = \left|\frac{\sin(nx)}{\sqrt{n}}\right|\le\frac1{\sqrt{n}}.$$