# Proving almost uniform convergence of $n\sin(\frac{x}{n})$

I'm learning about uniform convergence. For example, consider a function sequence

$$f_n: \mathbb{R} \rightarrow \mathbb{R}, f_n(x) = n\sin(\frac{x}{n})$$

This sequence converges pointwise to $$f(x) = x$$, but on $$\mathbb{R}$$ the convergence is not uniform (one can look at $$f(\frac{\pi}{2n}) = n$$ ).

I'm supposed to prove that on any closed interval $$[a,b] \subset \mathbb{R}$$ the convergence is uniform.

To prove that I need to show that the supremum of

$$| x - n\sin(\frac{x}{n}) |$$

converges to $$0$$ as $$n \rightarrow \infty$$.

My take is this: if $$x \in [a,b] \subset \mathbb{R}$$, then by Weierstrass theorem a continuous function on closed interval attains its maximum value, so there exists $$x_1 \in [a,b]$$ that for every $$x$$

$$| x - n\sin(\frac{x}{n}) | \leq | x_1 - n\sin(\frac{x_1}{n}) |$$

and then it's easy to see that as $$n \rightarrow \infty$$ the RHS also converges to $$0$$. But RHS is the supremum which converges to $$0$$, so the convergence is uniform on any $$[a,b] \subset \mathbb{R}$$.

Is this method/trick correct? It seems like if it is, then proving almost uniform convergence can sometimes be easy just by invoking the Weierstrass theorem: suspiciously easy, that's why I'm asking.

• The value of $x_1$ depends on $n$.
– TSF
Apr 21 '19 at 11:57

Actually, you don't have $$f_n\left(\frac\pi{2n}\right)=n$$. You can prove that the convergence is not uniform on $$\mathbb R$$ usinge the fact that $$f_n\left(\frac{\pi n}2\right)=n$$ for each natural $$n$$.
And your proof is wrong, since $$x_1$$ depends upon $$n$$.
I suggest that you define $$g_n(x)=f_n(x)-x=n\sin\left(\frac xn\right)-x$$ and then you prove that it converges uniformly to the null function on $$[a,b]$$. Use the fact that $$g_n'(x)=\cos\left(\frac xn\right)-1$$.