I'd like to know if following sequence converges uniformly: $$f_n (x)=f(x)\chi_{[-n,n]}(x)$$ Where $f\in\mathcal S(\mathbb R)$ and $\chi_{[-n,n]}$ is indicator function of interval [-n,n].
Now we know that $f_n=f$ on interval [-n,n] and outside that interval $f(x)<C/x^2$ where C is specific for a given schwartz function. We can now put: $$\epsilon=C/n^2$$ and therefore for every $\epsilon>0$ there exists $n$ so that $|f_n(x)-f(x)|<\epsilon$ for every $x\in \mathbb R$. We have proven uniform convergence on $\mathbb R$.
In the book that I am reading they are using it in a proof but the indicator function is first convoluted with mollifier. They use the properties of the mollified function in other parts of the proof. I want to know if my proof is correct even though the sequence does not consist of continuous functions.
According to the comments the proof is correct so indeed the product of Schwartz function and indicator function can converge uniformly to the Schwartz function.