# Limit of a special function

I'm posting this question just to be sure if my solution is correct or not.

I have a sequence $\{f_{n}\}$, where each $f_{n}:\mathbb{R}\to\mathbb{R}$ is continuous on $\mathbb{R}$ and $|f_{n}(x)|< a_{n}$ for all $n=1,2,3,\dots$ , for some sequence $a_{n}$ converges to 0. I'm trying to prove that $\frac{|f_{n}(x)|}{a_{n}}\to 0$ as $n\to\infty$.

I started with: since $|f_{n}(x)|\to 0$ as $n\to\infty$ (because $a_{n}\to 0$) for all $x\in\mathbb{R}$, this means that:

for any $\epsilon>0$ there exists $N_{\epsilon}>0$ such that $|f_{n}(x)|< \epsilon$ for all $n> N_{\epsilon}$ and all $x\in \mathbb{R}$, (uniform convergence)

Also, we can say: for any $j\in \mathbb{N}$ (by taking $\epsilon=a_{j}^{2}$) there exists $N_{j}>0$ such that $|f_{n}(x)|< a_{j}^{2}$ for all $n> N_{j}$ and all $x\in \mathbb{R}$, therefore: $$\frac{|f_{n}(x)|}{a_{n}}<\frac{a_{j}^{2}}{a_{n}} \to 0,\; as \;j\to \infty$$ for all $n>N_{j}$ and all $x$.

So, for large $n$ we have $\frac{|f_{n}(x)|}{a_{n}}\to 0$ by taking the limit as $j\to \infty$.

I think there is something wrong somewhere in my argument, if so please correct me!

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Just a note: lets forget about constant functions like $f_{n}(x)=1/(n+1)$ (since we will get $1/(n+1)< 1/n$ and the limit will be 1 not 0). –  Dimon Jun 19 '12 at 1:34
The argument that went wrong in your proof is that when letting $j \rightarrow \infty$ then in the inequality your $n$ is $n(j)$ it depends on $j$ so as far as we know $n$ could be must larger than $j$ thus much nearer to $0$ and that means $\frac{a_{j}^2}{a_n}$could be big –  clark Jun 19 '12 at 1:44
We usually can get one of the indices to tend to $\infty$ when they are independent. For example if you have something like that $\forall m,n \geq n_0$ we have some inequality then we can make $m$ or $n$ tend to $\infty$ –  clark Jun 19 '12 at 1:58
@Dimon: You should accept the answer below, as it completely answers your question. –  Vobo Jun 21 '12 at 13:16

The stated result is not true, so there cannot be a correct argument for it. For a counterexample, let $a_n=\frac{1}{n}$ and let $f_n(x)=\frac{1}{2n}e^{-x^2}$.
Then $|f_n(x)|\lt a_n$ for all $n$, but $\frac{f_n(x)}{a_n}=\frac{1}{2}e^{-x^2}$.
I cannot think of anything close. When we divide by the $a_n$ that witnesses the smallness of $f_n(x)$, we are spoiling that smallness. –  André Nicolas Jun 19 '12 at 1:43