The space of piecewise continuous sawtooth functions with arbitrarily large slope is dense in...
$begingroup$
Fix $L>0$. Show that the set
$$ A={fin mathcal{C}([0,1]): exists tin [0,1] s.t. |f(t)-f(s)|leq L|t-s| forall sin [0,1] } $$
has empty interior in $mathcal{C}([0,1])$.
Hint: Show that the complement of $A$ is dense by approximating an arbitrary continuous function by "sawtooth" functions with arbitrarily large slope.
Anyone have a hint for how to construct sawtooth functions the way the hint mentioned? I tried, but it ended looking really ugly. Is that inevitable?
real-analysis general-topology functional-analysis
asked Dec 1 '18 at 7:57
Joe Man AnalysisJoe Man Analysis
33419
$endgroup$
add a comment |
$begingroup$
Fix $L>0$. Show that the set
$$ A={fin mathcal{C}([0,1]): exists tin [0,1] s.t. |f(t)-f(s)|leq L|t-s| forall sin [0,1] } $$
has empty interior in $mathcal{C}([0,1])$.
Hint: Show that the complement of $A$ is dense by approximating an arbitrary continuous function by "sawtooth" functions with arbitrarily large slope.
Anyone have a hint for how to construct sawtooth functions the way the hint mentioned? I tried, but it ended looking really ugly. Is that inevitable?
real-analysis general-topology functional-analysis
asked Dec 1 '18 at 7:57
Joe Man AnalysisJoe Man Analysis
33419
$endgroup$
add a comment |
$begingroup$
Fix $L>0$. Show that the set
$$ A={fin mathcal{C}([0,1]): exists tin [0,1] s.t. |f(t)-f(s)|leq L|t-s| forall sin [0,1] } $$
has empty interior in $mathcal{C}([0,1])$.
Hint: Show that the complement of $A$ is dense by approximating an arbitrary continuous function by "sawtooth" functions with arbitrarily large slope.
Anyone have a hint for how to construct sawtooth functions the way the hint mentioned? I tried, but it ended looking really ugly. Is that inevitable?
real-analysis general-topology functional-analysis
asked Dec 1 '18 at 7:57
Joe Man AnalysisJoe Man Analysis
33419
$endgroup$
Fix $L>0$. Show that the set
$$ A={fin mathcal{C}([0,1]): exists tin [0,1] s.t. |f(t)-f(s)|leq L|t-s| forall sin [0,1] } $$
has empty interior in $mathcal{C}([0,1])$.
Hint: Show that the complement of $A$ is dense by approximating an arbitrary continuous function by "sawtooth" functions with arbitrarily large slope.
Anyone have a hint for how to construct sawtooth functions the way the hint mentioned? I tried, but it ended looking really ugly. Is that inevitable?
real-analysis general-topology functional-analysis
real-analysis general-topology functional-analysis
asked Dec 1 '18 at 7:57
Joe Man AnalysisJoe Man Analysis
33419
asked Dec 1 '18 at 7:57
Joe Man AnalysisJoe Man Analysis
33419
asked Dec 1 '18 at 7:57
Joe Man AnalysisJoe Man Analysis
33419
asked Dec 1 '18 at 7:57
Joe Man AnalysisJoe Man Analysis
33419
asked Dec 1 '18 at 7:57
Joe Man AnalysisJoe Man Analysis
33419
33419
add a comment |
add a comment |
1 Answer
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$begingroup$
Just define the triangle function
$$g(x) = begin{cases} x & text{ if } x in [0,1/2] \ 1-x & text{ if } x in (1/2,1].end{cases}$$
and set $g(x) =0$ for $x notin [0,1]$. We combine this functions in order to get appropiate sawtooth functions as follows:
$$g_n(x) := sum_{k=1}^{n} g(nx-k).$$ In any point $x in (k/n,(k+1/2)/n)$ we have $g'(x) = n$. Thus $$|g(x)-g(y)| = n |x-y| quad text{for} quad x,y in (k/n,(k+1/2)/n).$$
For an $f in mathcal{A}$, we can define $h_varepsilon := f+ varepsilon g_n$, where $n$ is taken so large that $varepsilon n ge 3L$. Then $$|f-h_varepsilon| = varepsilon |g_n| le varepsilon,$$
but for $x in (k/n,(k+1/2)/n)$
$$|h_varepsilon(x)-h_varepsilon(y)| ge varepsilon n |x-y| - |f(x)-f(y)| > L|x-y|,$$
i.e. $h_varepsilon notin mathcal{A}$. We conclude that $mathcal{A}$ has empty interior and thus is nowhere dense.
Note that $mathcal{A}$ is closed: Take a sequence $(f_n)_{n in mathbb{N}} subset mathcal{A}$ with $f_n rightarrow f$ uniformly. Then there exists a $t_n in [0,1]$ with $$tag{1}|f_n(t_n)-f_n(s)| le L|t_n-s| quad text{for all} quad s in [0,1].$$ By passing to a subsequence we can suppose that $(t_n)_n$ is convergent with limes $t in [0,1]$ - here we have just used Bolzano-Weierstraß. Taking $n rightarrow infty$ in (1) shows that $f in mathcal{A}$.
answered Dec 1 '18 at 9:18
p4schp4sch
4,945217
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Just define the triangle function
$$g(x) = begin{cases} x & text{ if } x in [0,1/2] \ 1-x & text{ if } x in (1/2,1].end{cases}$$
and set $g(x) =0$ for $x notin [0,1]$. We combine this functions in order to get appropiate sawtooth functions as follows:
$$g_n(x) := sum_{k=1}^{n} g(nx-k).$$ In any point $x in (k/n,(k+1/2)/n)$ we have $g'(x) = n$. Thus $$|g(x)-g(y)| = n |x-y| quad text{for} quad x,y in (k/n,(k+1/2)/n).$$
For an $f in mathcal{A}$, we can define $h_varepsilon := f+ varepsilon g_n$, where $n$ is taken so large that $varepsilon n ge 3L$. Then $$|f-h_varepsilon| = varepsilon |g_n| le varepsilon,$$
but for $x in (k/n,(k+1/2)/n)$
$$|h_varepsilon(x)-h_varepsilon(y)| ge varepsilon n |x-y| - |f(x)-f(y)| > L|x-y|,$$
i.e. $h_varepsilon notin mathcal{A}$. We conclude that $mathcal{A}$ has empty interior and thus is nowhere dense.
Note that $mathcal{A}$ is closed: Take a sequence $(f_n)_{n in mathbb{N}} subset mathcal{A}$ with $f_n rightarrow f$ uniformly. Then there exists a $t_n in [0,1]$ with $$tag{1}|f_n(t_n)-f_n(s)| le L|t_n-s| quad text{for all} quad s in [0,1].$$ By passing to a subsequence we can suppose that $(t_n)_n$ is convergent with limes $t in [0,1]$ - here we have just used Bolzano-Weierstraß. Taking $n rightarrow infty$ in (1) shows that $f in mathcal{A}$.
answered Dec 1 '18 at 9:18
p4schp4sch
4,945217
$endgroup$
add a comment |
$begingroup$
Just define the triangle function
$$g(x) = begin{cases} x & text{ if } x in [0,1/2] \ 1-x & text{ if } x in (1/2,1].end{cases}$$
and set $g(x) =0$ for $x notin [0,1]$. We combine this functions in order to get appropiate sawtooth functions as follows:
$$g_n(x) := sum_{k=1}^{n} g(nx-k).$$ In any point $x in (k/n,(k+1/2)/n)$ we have $g'(x) = n$. Thus $$|g(x)-g(y)| = n |x-y| quad text{for} quad x,y in (k/n,(k+1/2)/n).$$
For an $f in mathcal{A}$, we can define $h_varepsilon := f+ varepsilon g_n$, where $n$ is taken so large that $varepsilon n ge 3L$. Then $$|f-h_varepsilon| = varepsilon |g_n| le varepsilon,$$
but for $x in (k/n,(k+1/2)/n)$
$$|h_varepsilon(x)-h_varepsilon(y)| ge varepsilon n |x-y| - |f(x)-f(y)| > L|x-y|,$$
i.e. $h_varepsilon notin mathcal{A}$. We conclude that $mathcal{A}$ has empty interior and thus is nowhere dense.
Note that $mathcal{A}$ is closed: Take a sequence $(f_n)_{n in mathbb{N}} subset mathcal{A}$ with $f_n rightarrow f$ uniformly. Then there exists a $t_n in [0,1]$ with $$tag{1}|f_n(t_n)-f_n(s)| le L|t_n-s| quad text{for all} quad s in [0,1].$$ By passing to a subsequence we can suppose that $(t_n)_n$ is convergent with limes $t in [0,1]$ - here we have just used Bolzano-Weierstraß. Taking $n rightarrow infty$ in (1) shows that $f in mathcal{A}$.
answered Dec 1 '18 at 9:18
p4schp4sch
4,945217
$endgroup$
add a comment |
$begingroup$
Just define the triangle function
$$g(x) = begin{cases} x & text{ if } x in [0,1/2] \ 1-x & text{ if } x in (1/2,1].end{cases}$$
and set $g(x) =0$ for $x notin [0,1]$. We combine this functions in order to get appropiate sawtooth functions as follows:
$$g_n(x) := sum_{k=1}^{n} g(nx-k).$$ In any point $x in (k/n,(k+1/2)/n)$ we have $g'(x) = n$. Thus $$|g(x)-g(y)| = n |x-y| quad text{for} quad x,y in (k/n,(k+1/2)/n).$$
For an $f in mathcal{A}$, we can define $h_varepsilon := f+ varepsilon g_n$, where $n$ is taken so large that $varepsilon n ge 3L$. Then $$|f-h_varepsilon| = varepsilon |g_n| le varepsilon,$$
but for $x in (k/n,(k+1/2)/n)$
$$|h_varepsilon(x)-h_varepsilon(y)| ge varepsilon n |x-y| - |f(x)-f(y)| > L|x-y|,$$
i.e. $h_varepsilon notin mathcal{A}$. We conclude that $mathcal{A}$ has empty interior and thus is nowhere dense.
Note that $mathcal{A}$ is closed: Take a sequence $(f_n)_{n in mathbb{N}} subset mathcal{A}$ with $f_n rightarrow f$ uniformly. Then there exists a $t_n in [0,1]$ with $$tag{1}|f_n(t_n)-f_n(s)| le L|t_n-s| quad text{for all} quad s in [0,1].$$ By passing to a subsequence we can suppose that $(t_n)_n$ is convergent with limes $t in [0,1]$ - here we have just used Bolzano-Weierstraß. Taking $n rightarrow infty$ in (1) shows that $f in mathcal{A}$.
answered Dec 1 '18 at 9:18
p4schp4sch
4,945217
$endgroup$
Just define the triangle function
$$g(x) = begin{cases} x & text{ if } x in [0,1/2] \ 1-x & text{ if } x in (1/2,1].end{cases}$$
and set $g(x) =0$ for $x notin [0,1]$. We combine this functions in order to get appropiate sawtooth functions as follows:
$$g_n(x) := sum_{k=1}^{n} g(nx-k).$$ In any point $x in (k/n,(k+1/2)/n)$ we have $g'(x) = n$. Thus $$|g(x)-g(y)| = n |x-y| quad text{for} quad x,y in (k/n,(k+1/2)/n).$$
For an $f in mathcal{A}$, we can define $h_varepsilon := f+ varepsilon g_n$, where $n$ is taken so large that $varepsilon n ge 3L$. Then $$|f-h_varepsilon| = varepsilon |g_n| le varepsilon,$$
but for $x in (k/n,(k+1/2)/n)$
$$|h_varepsilon(x)-h_varepsilon(y)| ge varepsilon n |x-y| - |f(x)-f(y)| > L|x-y|,$$
i.e. $h_varepsilon notin mathcal{A}$. We conclude that $mathcal{A}$ has empty interior and thus is nowhere dense.
Note that $mathcal{A}$ is closed: Take a sequence $(f_n)_{n in mathbb{N}} subset mathcal{A}$ with $f_n rightarrow f$ uniformly. Then there exists a $t_n in [0,1]$ with $$tag{1}|f_n(t_n)-f_n(s)| le L|t_n-s| quad text{for all} quad s in [0,1].$$ By passing to a subsequence we can suppose that $(t_n)_n$ is convergent with limes $t in [0,1]$ - here we have just used Bolzano-Weierstraß. Taking $n rightarrow infty$ in (1) shows that $f in mathcal{A}$.
answered Dec 1 '18 at 9:18
p4schp4sch
4,945217
edited Dec 3 '18 at 11:43
answered Dec 1 '18 at 9:18
p4schp4sch
4,945217
answered Dec 1 '18 at 9:18
p4schp4sch
4,945217
answered Dec 1 '18 at 9:18
p4schp4sch
4,945217
4,945217
add a comment |
add a comment |
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