Uniform convergence of a sequence of functions on real line.

Question

Let the sequence $\{f_n(x)\}$ be given by $$f_n(x)=\cos\frac{(n+1)x}{n}.$$ Then, it is easy to see that $$\lim_{n\to\infty}f_n(x)=\cos x:=f(x).$$ Is the above convergence uniform over $\mathbb{R}$. I tried to obtain an upper bound on $|f_n(x)-f(x)|$ which is $$|f_n(x)-f(x)|=\bigg|\cos\frac{(n+1)x}{n}-\cos x\bigg|$$ $$=2\bigg|\sin \frac{(n+1)x+nx}{2n}\bigg|\ \bigg|\sin \frac{x}{2n}\bigg|.$$ How to proceed further? Please help.

Answer 1

The convergence is not uniform. For example consider the sequence $x_n=n\pi$. Then, \begin{align} \bigl|f_n(x_n)-f(x_n)|&=\bigl|\cos(n+1)\pi-\cos n\pi\bigr|\\ &=\bigl|\cos n \pi\cos \pi-\cos n\pi\bigr|\\ &=2|\cos n \pi|=2 \end{align} If the convergence was uniform, then for $\epsilon=1$ we would be able to find $N$ such that $$\sup_{x\in \mathbb{R}}\bigl|f_n(x)-f(x)|\leq 1$$ for every $n\geq N$. But this contradicts the fact that $|f_n(x_n)-f(x_n)|=2>1$ for all $n$.