I have to find example of 2 sequences of functions $ f_{n},g_{n} $ and 2 function $ f_{0},g_{0} $
such that $ f_n $ converge uniformly to $ f_0 $ and $ g_n $ converge uniformly to $ g_0 $ but
$ f_{n}g_{n} $ does not converge uniformly to $ f_{0}g_{0}$. I have try something, first I'll not that Im not allowed to use any theorem in order to prove covergence/disconvergence. Only by definitions.
I'll be glad if someone can tell if im abusing the definitions or if my attempt is successful.
So what I tried is:
Consider $ f_{n}\left(x\right)=g_{n}\left(x\right)=\frac{1}{n}+\frac{1}{x} $. and my domain will be $ I=(0,1] $
I want to argue that those sequences of functions converge uniformly to $ f_{0}=g_{0}=\frac{1}{x} $ I think this is the easy part so I'll not show my proof for this (I guess its intuitive any its not the part that I want to ask about).
Now I struggle to show that $ f_{n}g_{n}=\frac{1}{n^{2}}+\frac{2}{nx}+\frac{1}{x^{2}} $ does not converge uniformly to $ f_{0}g_{0}=\frac{1}{x^{2}} $.
So by definition of the uniform convergence, I'll have to find $ \varepsilon>0 $ such that for any $ N \in \mathbb{N} $ there exists $ n>N $ and $ x \in I $ such that $ |f_{n}g_{n}\left(x\right)-f_{0}g_{0}\left(x\right)|\geq\varepsilon $.
So if we consider $ \varepsilon=1 $, and let $ N $ be some natural number, we can take $ n=N+1 $ and $ x $ to be close enough to $ 0 $ such that $ (N+1)x < \frac{1}{4} $ and then we'll get that $ \frac{1}{\left(N+1\right)x}>4 $. So for those $ \varepsilon $ , $ n $ and $x $ that we chose :
$ |f_{N+1}g_{N+1}\left(x\right)-f_{0}g_{0}\left(x\right)|=|\frac{1}{\left(N+1\right)^{2}}+\frac{1}{\left(N+1\right)x}+\frac{1}{x^{2}}-\frac{1}{x^{2}}|=\frac{1}{\left(N+1\right)^{2}}+\frac{2}{\left(N+1\right)x}>\frac{1}{\left(N+1\right)^{2}}+8>1=\varepsilon $
But I feel like Im abusing the definitions. I'd be glad to hear what you think about this proof, is it legit?
Thanks in advance