I am trying to prove that the sequence of functions $f_{k}:\mathbb{R}\to\mathbb{R}$ defined by:
$$f_{k}(x):= \begin{cases}|x-k|-1,&\text{if }x\in[k-1,k+1]\\ 0,&\text{if }x \notin [k-1,k+1] \end{cases}$$
Does not converge uniformly to a function in $\mathbb{R}$. First, I would like to show that it does pointwise converges to a function $f:\mathbb{R}\to\mathbb{R}$. Namely, the constant function $f\equiv0$. To do this, we choose an arbitrary $x\in\mathbb{R}$ and then, an appropriate $k\in\mathbb{N^{+}}$ such that our inverted triangle is far from $x$. Is this correct?
Second, if the function converged uniformly to a function $f$, it should necessarily converge to $f\equiv0$, since uniform convergence to a function implies pointwise convergence to the function. But that cannot happen, because for $\epsilon:=\frac{1}{2}$ and for any natural $n_{0}$ we can find an $f_{k}$ such that $|f_{k}(x)-0|\geq\epsilon$. Is there any error in this reasoning?
At first I thought there was not even pointwise convergence and even though I think my first argument holds I have the sense that something is wrong with it. Thank you for your comments.
Your argument that $f_k$ converges to $0$ pointwise is correct. It is also correct that if $f_k$ converges uniformly to $f$ then $f$ must be $\equiv 0$.
For the last part, while your idea is correct, I would like to be a bit more precise.
Suppose $f_k$ converges uniformly to $0$. Then, for any $\varepsilon >0$, there exists $N$ such that $|f_k(x)-0| \leq \varepsilon$ for any $x \in \mathbb{R}$ and $k\geq N$.
However, if $\varepsilon = 1/2$, then for any integer $N\geq 1$, $|f_N(N)|=1 > \varepsilon$ (actually $|f_k(k)|=1$ for any $k$ in this case). This is a contradiction.