Mixing
Approximating a continuous function by a sawtooth function
I need to understand a proof in the book "Lecture notes on topology and geometry" by Singer and Thorpe, from 2.5 Applications
Theorem 1: There exists a continuous real-valued function $fin C([0,1])$ such that $f$ has a derivative in no point of $[0,1]$.
Proof: Define for all $nin mathbb{N}$ the set
$$
C_n
=left{fin mathcal{C}([0,1]): Big|dfrac{f(t+h)-f(t)}{h}Big|leqslant n,text{for some $t$ and all $h$ with $t+h$ $in [0,1]$} right}
$$
Show that $C_n$ is nowhere dense for each $n$, then since $C([0,1])$ is a complete metric space we have
$$
mathcal{C}([0,1])neq bigcup_{nin mathbb{N}} C_n
$$
it must exists a function $fin mathcal{C}([0,1])$ such that $fnotin C_n$ for all $n$ and this means it has no derivative.
- I understand why this last remark is true because if $fnotin C_n$ for all $n$ then
$$
Big|dfrac{f(t+h)-f(t)}{h}Big|> n
$$ and then
$$limsup_{hrightarrow0}Big|dfrac{f(t+h)-f(t)}{h}Big|=infty$$ for all $tin [0,1]$ thus the derivative does not exists.
- I understand how to show that $overline C_n=C_n$ because is well explained in the book, I must considere a sequence ${f_k}$ in $C_n$ which converges to $f$ and show that $fin C_n$.
I have a problem showing that $C_n$ is nowhere dense, the book says that given any $epsilon >0$ and any $gin C_n$ there exists $f in C([0,1])$ such that $||f-g||<epsilon$ and $fnotin C_n$, and suggest i should do that by approximating $g$ (using the fact that $g$ is uniformly continuous) by a piecewise linear function $g_1$ and then use a "sawtooth" function for each linear piece of $g_1$
and then "to patch". Also note that the norm is $||f||=sup|f(x)|$ for all $xin [0,1]$
I ask for an explanation on this last remark, a good reference to understand it or any other help that you can provide.
general-topology continuity metric-spaces proof-explanation uniform-continuity
edited Dec 6 '18 at 5:40
Alex Ravsky
39.4k32181
asked Nov 21 '18 at 19:59
AlfdavAlfdav
436
add a comment |
I need to understand a proof in the book "Lecture notes on topology and geometry" by Singer and Thorpe, from 2.5 Applications
Theorem 1: There exists a continuous real-valued function $fin C([0,1])$ such that $f$ has a derivative in no point of $[0,1]$.
Proof: Define for all $nin mathbb{N}$ the set
$$
C_n
=left{fin mathcal{C}([0,1]): Big|dfrac{f(t+h)-f(t)}{h}Big|leqslant n,text{for some $t$ and all $h$ with $t+h$ $in [0,1]$} right}
$$
Show that $C_n$ is nowhere dense for each $n$, then since $C([0,1])$ is a complete metric space we have
$$
mathcal{C}([0,1])neq bigcup_{nin mathbb{N}} C_n
$$
it must exists a function $fin mathcal{C}([0,1])$ such that $fnotin C_n$ for all $n$ and this means it has no derivative.
- I understand why this last remark is true because if $fnotin C_n$ for all $n$ then
$$
Big|dfrac{f(t+h)-f(t)}{h}Big|> n
$$ and then
$$limsup_{hrightarrow0}Big|dfrac{f(t+h)-f(t)}{h}Big|=infty$$ for all $tin [0,1]$ thus the derivative does not exists.
- I understand how to show that $overline C_n=C_n$ because is well explained in the book, I must considere a sequence ${f_k}$ in $C_n$ which converges to $f$ and show that $fin C_n$.
I have a problem showing that $C_n$ is nowhere dense, the book says that given any $epsilon >0$ and any $gin C_n$ there exists $f in C([0,1])$ such that $||f-g||<epsilon$ and $fnotin C_n$, and suggest i should do that by approximating $g$ (using the fact that $g$ is uniformly continuous) by a piecewise linear function $g_1$ and then use a "sawtooth" function for each linear piece of $g_1$
and then "to patch". Also note that the norm is $||f||=sup|f(x)|$ for all $xin [0,1]$
I ask for an explanation on this last remark, a good reference to understand it or any other help that you can provide.
general-topology continuity metric-spaces proof-explanation uniform-continuity
edited Dec 6 '18 at 5:40
Alex Ravsky
39.4k32181
asked Nov 21 '18 at 19:59
AlfdavAlfdav
436
This is just a visualization: function $f$ must satisfy $||f-g||<epsilon$. Function $g_1$ is a good try, since it is linear, but it can have small derivatives at some points. Notice that a piecewise linear $f notin C_n$ needs to have angular coeficient with modulus greater than $n$, everywhere. The idea is then to build $f$ from $g_1$ by replacing each linear piece by a new piecewise linear with same starting and ending, but always having high angular coeficient. For a similar construction, check Munkres, page 304.
– Daniel
Nov 21 '18 at 21:00
add a comment |
I need to understand a proof in the book "Lecture notes on topology and geometry" by Singer and Thorpe, from 2.5 Applications
Theorem 1: There exists a continuous real-valued function $fin C([0,1])$ such that $f$ has a derivative in no point of $[0,1]$.
Proof: Define for all $nin mathbb{N}$ the set
$$
C_n
=left{fin mathcal{C}([0,1]): Big|dfrac{f(t+h)-f(t)}{h}Big|leqslant n,text{for some $t$ and all $h$ with $t+h$ $in [0,1]$} right}
$$
Show that $C_n$ is nowhere dense for each $n$, then since $C([0,1])$ is a complete metric space we have
$$
mathcal{C}([0,1])neq bigcup_{nin mathbb{N}} C_n
$$
it must exists a function $fin mathcal{C}([0,1])$ such that $fnotin C_n$ for all $n$ and this means it has no derivative.
- I understand why this last remark is true because if $fnotin C_n$ for all $n$ then
$$
Big|dfrac{f(t+h)-f(t)}{h}Big|> n
$$ and then
$$limsup_{hrightarrow0}Big|dfrac{f(t+h)-f(t)}{h}Big|=infty$$ for all $tin [0,1]$ thus the derivative does not exists.
- I understand how to show that $overline C_n=C_n$ because is well explained in the book, I must considere a sequence ${f_k}$ in $C_n$ which converges to $f$ and show that $fin C_n$.
I have a problem showing that $C_n$ is nowhere dense, the book says that given any $epsilon >0$ and any $gin C_n$ there exists $f in C([0,1])$ such that $||f-g||<epsilon$ and $fnotin C_n$, and suggest i should do that by approximating $g$ (using the fact that $g$ is uniformly continuous) by a piecewise linear function $g_1$ and then use a "sawtooth" function for each linear piece of $g_1$
and then "to patch". Also note that the norm is $||f||=sup|f(x)|$ for all $xin [0,1]$
I ask for an explanation on this last remark, a good reference to understand it or any other help that you can provide.
general-topology continuity metric-spaces proof-explanation uniform-continuity
edited Dec 6 '18 at 5:40
Alex Ravsky
39.4k32181
asked Nov 21 '18 at 19:59
AlfdavAlfdav
436
I need to understand a proof in the book "Lecture notes on topology and geometry" by Singer and Thorpe, from 2.5 Applications
Theorem 1: There exists a continuous real-valued function $fin C([0,1])$ such that $f$ has a derivative in no point of $[0,1]$.
Proof: Define for all $nin mathbb{N}$ the set
$$
C_n
=left{fin mathcal{C}([0,1]): Big|dfrac{f(t+h)-f(t)}{h}Big|leqslant n,text{for some $t$ and all $h$ with $t+h$ $in [0,1]$} right}
$$
Show that $C_n$ is nowhere dense for each $n$, then since $C([0,1])$ is a complete metric space we have
$$
mathcal{C}([0,1])neq bigcup_{nin mathbb{N}} C_n
$$
it must exists a function $fin mathcal{C}([0,1])$ such that $fnotin C_n$ for all $n$ and this means it has no derivative.
- I understand why this last remark is true because if $fnotin C_n$ for all $n$ then
$$
Big|dfrac{f(t+h)-f(t)}{h}Big|> n
$$ and then
$$limsup_{hrightarrow0}Big|dfrac{f(t+h)-f(t)}{h}Big|=infty$$ for all $tin [0,1]$ thus the derivative does not exists.
- I understand how to show that $overline C_n=C_n$ because is well explained in the book, I must considere a sequence ${f_k}$ in $C_n$ which converges to $f$ and show that $fin C_n$.
I have a problem showing that $C_n$ is nowhere dense, the book says that given any $epsilon >0$ and any $gin C_n$ there exists $f in C([0,1])$ such that $||f-g||<epsilon$ and $fnotin C_n$, and suggest i should do that by approximating $g$ (using the fact that $g$ is uniformly continuous) by a piecewise linear function $g_1$ and then use a "sawtooth" function for each linear piece of $g_1$
and then "to patch". Also note that the norm is $||f||=sup|f(x)|$ for all $xin [0,1]$
I ask for an explanation on this last remark, a good reference to understand it or any other help that you can provide.
general-topology continuity metric-spaces proof-explanation uniform-continuity
general-topology continuity metric-spaces proof-explanation uniform-continuity
edited Dec 6 '18 at 5:40
Alex Ravsky
39.4k32181
asked Nov 21 '18 at 19:59
AlfdavAlfdav
436
edited Dec 6 '18 at 5:40
Alex Ravsky
39.4k32181
asked Nov 21 '18 at 19:59
AlfdavAlfdav
436
edited Dec 6 '18 at 5:40
Alex Ravsky
39.4k32181
edited Dec 6 '18 at 5:40
Alex Ravsky
39.4k32181
edited Dec 6 '18 at 5:40
Alex Ravsky
39.4k32181
39.4k32181
asked Nov 21 '18 at 19:59
AlfdavAlfdav
436
asked Nov 21 '18 at 19:59
AlfdavAlfdav
436
asked Nov 21 '18 at 19:59
AlfdavAlfdav
436
436
This is just a visualization: function $f$ must satisfy $||f-g||<epsilon$. Function $g_1$ is a good try, since it is linear, but it can have small derivatives at some points. Notice that a piecewise linear $f notin C_n$ needs to have angular coeficient with modulus greater than $n$, everywhere. The idea is then to build $f$ from $g_1$ by replacing each linear piece by a new piecewise linear with same starting and ending, but always having high angular coeficient. For a similar construction, check Munkres, page 304.
– Daniel
Nov 21 '18 at 21:00
add a comment |
This is just a visualization: function $f$ must satisfy $||f-g||<epsilon$. Function $g_1$ is a good try, since it is linear, but it can have small derivatives at some points. Notice that a piecewise linear $f notin C_n$ needs to have angular coeficient with modulus greater than $n$, everywhere. The idea is then to build $f$ from $g_1$ by replacing each linear piece by a new piecewise linear with same starting and ending, but always having high angular coeficient. For a similar construction, check Munkres, page 304.
– Daniel
Nov 21 '18 at 21:00
This is just a visualization: function $f$ must satisfy $||f-g||<epsilon$. Function $g_1$ is a good try, since it is linear, but it can have small derivatives at some points. Notice that a piecewise linear $f notin C_n$ needs to have angular coeficient with modulus greater than $n$, everywhere. The idea is then to build $f$ from $g_1$ by replacing each linear piece by a new piecewise linear with same starting and ending, but always having high angular coeficient. For a similar construction, check Munkres, page 304.
– Daniel
Nov 21 '18 at 21:00
This is just a visualization: function $f$ must satisfy $||f-g||<epsilon$. Function $g_1$ is a good try, since it is linear, but it can have small derivatives at some points. Notice that a piecewise linear $f notin C_n$ needs to have angular coeficient with modulus greater than $n$, everywhere. The idea is then to build $f$ from $g_1$ by replacing each linear piece by a new piecewise linear with same starting and ending, but always having high angular coeficient. For a similar construction, check Munkres, page 304.
– Daniel
Nov 21 '18 at 21:00
add a comment |
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This is just a visualization: function $f$ must satisfy $||f-g||<epsilon$. Function $g_1$ is a good try, since it is linear, but it can have small derivatives at some points. Notice that a piecewise linear $f notin C_n$ needs to have angular coeficient with modulus greater than $n$, everywhere. The idea is then to build $f$ from $g_1$ by replacing each linear piece by a new piecewise linear with same starting and ending, but always having high angular coeficient. For a similar construction, check Munkres, page 304.
– Daniel
Nov 21 '18 at 21:00