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If $\{h_{k}\}_{k=1}^{\infty}$ is a sequence of real-valued continuous functions on the real line, and $0<h_{k}(x)\leq 1,\;\forall x\in\mathbb{R}$ and all $k$. Assume that $h_{k}(x)\to 0$ uniformly on $\mathbb{R}$. Is it true that $h_{k}(x+x_{k})\to 0$ as $k\to \infty$, for any sequence of points $\{x_{k}\}\subset \mathbb{R} $ and any $x\in \mathbb{R}$? If so how to prove it?

Thanks for help in advance!

Ok, what I have tried: I'm trying to use induction on $x_{k}$: Fix any $x\in\mathbb{R}, $we have

for $x_{1}$: $h_{k}(x+x_{1})\to 0$

for $x_{2}$: $h_{k}(x+x_{2})\to 0$

for $x_{3}$: $h_{k}(x+x_{3})\to 0$

.

.

.

for $x_{m}$: $h_{k}(x+x_{m})\to 0$

so, its true for all $m$, hence for $k$. Not sure!

Edit: I had a typo, sorry. My previous question was about $h_{k}(x_{k})\to 0$, but the correct question is about $h_{k}(x+x_{k})\to 0$

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1  
What have you tried? –  Jonas Meyer Jun 18 '12 at 19:56
    
Let $a_k=h_k(x_k)$. The question is whether this sequence of numbers $a_1,a_2,a_3,\ldots$ converges to $0$. It might help to write out exactly what that would mean, to know what goal you are trying to achieve. Because the sequence $(x_k)$ is completely arbitrary, the important hypothesis is that $(h_k)$ converges to $0$ uniformly. I might help to write out exactly what that means, to know how you can use it to show that $(a_k)$ converges to $0$. I don't understand your work. –  Jonas Meyer Jun 18 '12 at 20:07
    
Do you know what it means for $h_k(x)$ to converge uniformly to $0$? If you write down the definition, it will help a lot. –  N.U. Jun 18 '12 at 20:09
    
@Jonas Meyer: Oh ok. Let $k\geq 1$, then the point $x_{k}$ is the point where the $\sup$ of $h_{k}(x)$ is attained, i.e. $\sup_{x\in\mathbb{R}}|h_{k}(x)|=|h_{k}(x_{k})|$, (of course we could have more than one point where the sup is attained but we can take the first occurrence). –  Zack M Jun 18 '12 at 20:11
    
@ZackM: The $\sup$ might not be attained. A counter-example is $h_n(x)=\frac1n\left(1-\frac1{2+x^2}\right)$. –  Dejan Govc Jun 18 '12 at 20:16

2 Answers 2

This should be fairly direct.

The statement that $h_k(x) \to 0$ uniformly on $\mathbb{R}$ means that for any $\epsilon > 0$, there exists some $K_\epsilon$ such that $k>K_\epsilon$ implies that $|h_k(x)|<\epsilon$ for all $x \in \mathbb{R}$. If $\{x_k\}$ is some sequence in $\mathbb{R}$, then we certainly have that $|h_k(x_k)| < \epsilon$ for all $k>K_\epsilon$, and so $h_k(x_k) \to 0$.

Uniform convergence is a very powerful property.

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He changed the question, but your proof is still correct (with $x+x_k$ instead of $x_k$). That is how nice uniform convergence is =) –  N.U. Jun 18 '12 at 20:26

The idea Greg Zitelli points out in his answer should work with the new version of the problem too.

I'd just like to add that your idea of proof (if I understand it correctly) won't work. Here's an example that should demonstrate what the problem is. Let $$f_n(x)=\begin{cases}0;&x<n\\1;&\text{otherwise}\end{cases}$$ and let $x_n=n$. Then for every term $x_i$ of this sequence you have: $f_n(x_i)\to0$, but $f_n(x_n)\to1$. Of course here the convergence is not uniform and the functions are not continuous, but it still shows that the convergence of $f_n(x_n)$ does not follow from that of $f_n(x_i)$.

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