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Continuity in a Metric Space

dray

Member
Feb 5, 2012
37
Let $a=(a_k)$ be a fixed element of the metric space $l_p$, where $l_p$ is equipped with its usual metric. I need to prove that the function $f:{l_p} \rightarrow {l_p}$ defined by $f:{(x_k)}\mapsto({5x_k}+{a_k})$ is continuous. Is is uniformly continuous?

Can anyone help?
 
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girdav

Member
Feb 1, 2012
96
Let $x:=(x_k)$ and $y:=(y_k)$ two sequences in $\ell_p$. What is $||f(x)-f(y)||_p$?
 

dray

Member
Feb 5, 2012
37
Let $x:=(x_k)$ and $y:=(y_k)$ two sequences in $\ell_p$. What is $||f(x)-f(y)||_p$?
I'm quite happy using the epsilon-delta definition of continuity in a metric space to prove continuity, but I'm not sure what $||f(x)-f(y)||$ is. this is my sticking point. How do I handle the fixed element $(a_k)$?
 

girdav

Member
Feb 1, 2012
96
If $p\geq 1$ then $||u||_p=\left(\sum_{n=0}^{+\infty}|u_n|^p\right)^{\frac 1p}$. Do you have to deal with the case $p\in (0,1)$?
 

dray

Member
Feb 5, 2012
37
If $p\geq 1$ then $||u||_p=\left(\sum_{n=0}^{+\infty}|u_n|^p\right)^{\frac 1p}$. Do you have to deal with the case $p\in (0,1)$?
No. I'm just considering the case $p\geq 1$.

---------- Post added at 07:06 AM ---------- Previous post was at 05:22 AM ----------

I'm quite happy using the epsilon-delta definition of continuity in a metric space to prove continuity, but I'm not sure what $||f(x)-f(y)||$ is. this is my sticking point. How do I handle the fixed element $(a_k)$?
$||f(x)-f(y)||_p= ||{5x+a_k}-{5y+a_k}||_p= ||5x-5y||_p= 5||x-y||_p$.

Given $\epsilon>0$, let $\delta=\frac{\epsilon}{5}>0$. If $||x-y||_p<\delta$, $||f(x)-f(y)||_p<\epsilon$. Thus $f$ is continuous at $y\in{l_p}$.

Any good?
 

girdav

Member
Feb 1, 2012
96
It's good (except a typo at the first line, it shoul be $5x+a-(5y+a)$ instead of $5x+a-5y+a$. Now what can you conclude about uniform continuity?
 

dray

Member
Feb 5, 2012
37
Since $\delta(\epsilon)$ depends only on $\epsilon$ and not on any particular $x\in{l_p}$ the function $f$ is uniformly continuous?
 

girdav

Member
Feb 1, 2012
96
That's right.