Uniform continuity of $f(x)=sum_{n=1}^infty f_n(x)$ on $mathbb R$
$begingroup$
Suppose for $nin mathbb N$, $f_n(x) =
begin{cases}
n(x-n+frac{1}{n}) & text{if $xin [n-frac{1}{n},n]$} \
n(n+frac{1}{n}-x) & text{if $xin [n,n+frac{1}{n}]$} & \ 0 &text{otherwise} end{cases}$.
Let $f(x)=sum_{n=1}^infty f_n(x)$. Check whether the given function $f(x)$ uniformly continuous or not on $mathbb R$.
My attempt:-
I draw the graph of $f_1, f_2, f_3,..$
From the graph, I can deduce that $f$ is not uniformly continuous on $mathbb R$. For analytical proof. I want to prove there exists an $epsilon_oin mathbb R^+$ such that $ forall delta in mathbb R^+$. $|x-y|<delta implies |f(x)-f(y)|ge epsilon_o.$ As we move farther from the origin, In the $epsilon-delta$ definition of continuity. delta is decreasing for a given $epsilon$. I am not able to prove formally.
real-analysis continuity uniform-continuity
asked Dec 23 '18 at 7:03
Unknown xUnknown x
2,54211028
$endgroup$
add a comment |
$begingroup$
Suppose for $nin mathbb N$, $f_n(x) =
begin{cases}
n(x-n+frac{1}{n}) & text{if $xin [n-frac{1}{n},n]$} \
n(n+frac{1}{n}-x) & text{if $xin [n,n+frac{1}{n}]$} & \ 0 &text{otherwise} end{cases}$.
Let $f(x)=sum_{n=1}^infty f_n(x)$. Check whether the given function $f(x)$ uniformly continuous or not on $mathbb R$.
My attempt:-
I draw the graph of $f_1, f_2, f_3,..$
From the graph, I can deduce that $f$ is not uniformly continuous on $mathbb R$. For analytical proof. I want to prove there exists an $epsilon_oin mathbb R^+$ such that $ forall delta in mathbb R^+$. $|x-y|<delta implies |f(x)-f(y)|ge epsilon_o.$ As we move farther from the origin, In the $epsilon-delta$ definition of continuity. delta is decreasing for a given $epsilon$. I am not able to prove formally.
real-analysis continuity uniform-continuity
asked Dec 23 '18 at 7:03
Unknown xUnknown x
2,54211028
$endgroup$
add a comment |
$begingroup$
Suppose for $nin mathbb N$, $f_n(x) =
begin{cases}
n(x-n+frac{1}{n}) & text{if $xin [n-frac{1}{n},n]$} \
n(n+frac{1}{n}-x) & text{if $xin [n,n+frac{1}{n}]$} & \ 0 &text{otherwise} end{cases}$.
Let $f(x)=sum_{n=1}^infty f_n(x)$. Check whether the given function $f(x)$ uniformly continuous or not on $mathbb R$.
My attempt:-
I draw the graph of $f_1, f_2, f_3,..$
From the graph, I can deduce that $f$ is not uniformly continuous on $mathbb R$. For analytical proof. I want to prove there exists an $epsilon_oin mathbb R^+$ such that $ forall delta in mathbb R^+$. $|x-y|<delta implies |f(x)-f(y)|ge epsilon_o.$ As we move farther from the origin, In the $epsilon-delta$ definition of continuity. delta is decreasing for a given $epsilon$. I am not able to prove formally.
real-analysis continuity uniform-continuity
asked Dec 23 '18 at 7:03
Unknown xUnknown x
2,54211028
$endgroup$
Suppose for $nin mathbb N$, $f_n(x) =
begin{cases}
n(x-n+frac{1}{n}) & text{if $xin [n-frac{1}{n},n]$} \
n(n+frac{1}{n}-x) & text{if $xin [n,n+frac{1}{n}]$} & \ 0 &text{otherwise} end{cases}$.
Let $f(x)=sum_{n=1}^infty f_n(x)$. Check whether the given function $f(x)$ uniformly continuous or not on $mathbb R$.
My attempt:-
I draw the graph of $f_1, f_2, f_3,..$
From the graph, I can deduce that $f$ is not uniformly continuous on $mathbb R$. For analytical proof. I want to prove there exists an $epsilon_oin mathbb R^+$ such that $ forall delta in mathbb R^+$. $|x-y|<delta implies |f(x)-f(y)|ge epsilon_o.$ As we move farther from the origin, In the $epsilon-delta$ definition of continuity. delta is decreasing for a given $epsilon$. I am not able to prove formally.
real-analysis continuity uniform-continuity
real-analysis continuity uniform-continuity
asked Dec 23 '18 at 7:03
Unknown xUnknown x
2,54211028
asked Dec 23 '18 at 7:03
Unknown xUnknown x
2,54211028
asked Dec 23 '18 at 7:03
Unknown xUnknown x
2,54211028
asked Dec 23 '18 at 7:03
Unknown xUnknown x
2,54211028
asked Dec 23 '18 at 7:03
Unknown xUnknown x
2,54211028
2,54211028
add a comment |
add a comment |
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