Uniform convergence of $\sin(nx) \cdot\frac{x}{n} + x$

Question

I've shown that $f_n(x) = \sin(nx)\cdot \frac{x}{n} + x$ converges pointwise to $f(x) = x$. Now if I consider $|f_n(x) - f(x)| = |\frac{x}{n}\sin(nx)|$ it looks that it is dependent over $x$ so I think that the convergence isn't uniform over $\mathbb{R}$. I am trying to find some $x$ such that $|f_n(x) - f(x)| \geq c > 0$.

My idea was to try and make the argument inside the sine function equal to $1$. I've tried with $x = \frac{n\pi}{2}$ then $|f_n(x) - f(x)| = |\frac{\pi}{2}\sin(n^2\pi/2)|$ but that $n^2$ term is ruining it. Any suggestions on how to prove or disprove my claim that $f_n$ isn't uniformly convergent?

Answer 1

The sequence is not uniformly convergent. To see this, fix some $\epsilon > 0$. Then, as you noticed, we have $$\vert f_n(x) - f(x) \vert = \vert \sin(nx)\frac{x}{n}\vert.$$ Let $x$ be an odd multiple of $\pi/2$ so that $\vert f_n(x) - f(x) \vert = \frac{x}{n}$. For any $n$, we can find $x$ large enough so that $\frac{x}{n} > \epsilon$. Hence, the convergence is not uniform.

Answer 2

Let $\epsilon=1$

$$\forall N, \exists n=2N,m=2N+1, x=\frac{(2N+1)\pi}2, $$

such that

$$\begin{align}|f_n(x)&-f_m(x)| =\\ \\ &= \left|\sin\left(2N\cdot \frac{(2N+1)\pi}2\right) \frac{(2N+1)\pi}{4N} -\sin\left((2N+1)\cdot \frac{(2N+1)\pi}2\right) \frac{(2N+1)\pi}{2(2N+1)} \right|\\ \\ &=\frac\pi2>1=\epsilon\end{align}$$