Does $f_t(x) to mbox{sgn}(x)$ as $ t to 0$ imply $|f_t(x) - mbox{sgn}(x)|< t$?












0












$begingroup$



Define a function $f: mathbb{R} to mathbb{R}$ as
$$begin{equation}
f_t(x) =
begin{cases}
-1 & text{if}; ,x<-t \
sin left(frac{pi x}{2 t}right) & text{if};, |x| le t\
1 & text{if};, x>t
end{cases}
end{equation} $$
Then $f(x)$ converges pointwise to $mbox{sgn}(x)$ as $t to 0$.
I'm wondering if something stronger can be proven, can we say $|f_t(x) - mbox{sgn}(x)|< t$ ?







real-analysis analysis







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$endgroup$








  • 1




    $begingroup$
    Do you mean begin{equation} f_t(x) = begin{cases} -1 & text{if}; ,x<-t \ sin left(frac{pi x}{2 t}right) & text{if};, |x| le t\ 1 & text{if};, x>t end{cases}? end{equation} I.e., $x<-t$ in the first case.
    $endgroup$
    – Jens Schwaiger
    Dec 25 '18 at 8:14












  • $begingroup$
    @JensSchwaiger Yes, thank you !
    $endgroup$
    – Goal123
    Dec 25 '18 at 10:28
















0












$begingroup$



Define a function $f: mathbb{R} to mathbb{R}$ as
$$begin{equation}
f_t(x) =
begin{cases}
-1 & text{if}; ,x<-t \
sin left(frac{pi x}{2 t}right) & text{if};, |x| le t\
1 & text{if};, x>t
end{cases}
end{equation} $$
Then $f(x)$ converges pointwise to $mbox{sgn}(x)$ as $t to 0$.
I'm wondering if something stronger can be proven, can we say $|f_t(x) - mbox{sgn}(x)|< t$ ?







real-analysis analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Do you mean begin{equation} f_t(x) = begin{cases} -1 & text{if}; ,x<-t \ sin left(frac{pi x}{2 t}right) & text{if};, |x| le t\ 1 & text{if};, x>t end{cases}? end{equation} I.e., $x<-t$ in the first case.
    $endgroup$
    – Jens Schwaiger
    Dec 25 '18 at 8:14












  • $begingroup$
    @JensSchwaiger Yes, thank you !
    $endgroup$
    – Goal123
    Dec 25 '18 at 10:28














0












0








0





$begingroup$



Define a function $f: mathbb{R} to mathbb{R}$ as
$$begin{equation}
f_t(x) =
begin{cases}
-1 & text{if}; ,x<-t \
sin left(frac{pi x}{2 t}right) & text{if};, |x| le t\
1 & text{if};, x>t
end{cases}
end{equation} $$
Then $f(x)$ converges pointwise to $mbox{sgn}(x)$ as $t to 0$.
I'm wondering if something stronger can be proven, can we say $|f_t(x) - mbox{sgn}(x)|< t$ ?







real-analysis analysis







share|cite|improve this question











$endgroup$





Define a function $f: mathbb{R} to mathbb{R}$ as
$$begin{equation}
f_t(x) =
begin{cases}
-1 & text{if}; ,x<-t \
sin left(frac{pi x}{2 t}right) & text{if};, |x| le t\
1 & text{if};, x>t
end{cases}
end{equation} $$
Then $f(x)$ converges pointwise to $mbox{sgn}(x)$ as $t to 0$.
I'm wondering if something stronger can be proven, can we say $|f_t(x) - mbox{sgn}(x)|< t$ ?







real-analysis analysis




real-analysis analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 25 '18 at 10:27







Goal123

















asked Dec 25 '18 at 7:15









Goal123Goal123

505212




505212








  • 1




    $begingroup$
    Do you mean begin{equation} f_t(x) = begin{cases} -1 & text{if}; ,x<-t \ sin left(frac{pi x}{2 t}right) & text{if};, |x| le t\ 1 & text{if};, x>t end{cases}? end{equation} I.e., $x<-t$ in the first case.
    $endgroup$
    – Jens Schwaiger
    Dec 25 '18 at 8:14












  • $begingroup$
    @JensSchwaiger Yes, thank you !
    $endgroup$
    – Goal123
    Dec 25 '18 at 10:28














  • 1




    $begingroup$
    Do you mean begin{equation} f_t(x) = begin{cases} -1 & text{if}; ,x<-t \ sin left(frac{pi x}{2 t}right) & text{if};, |x| le t\ 1 & text{if};, x>t end{cases}? end{equation} I.e., $x<-t$ in the first case.
    $endgroup$
    – Jens Schwaiger
    Dec 25 '18 at 8:14












  • $begingroup$
    @JensSchwaiger Yes, thank you !
    $endgroup$
    – Goal123
    Dec 25 '18 at 10:28








1




1




$begingroup$
Do you mean begin{equation} f_t(x) = begin{cases} -1 & text{if}; ,x<-t \ sin left(frac{pi x}{2 t}right) & text{if};, |x| le t\ 1 & text{if};, x>t end{cases}? end{equation} I.e., $x<-t$ in the first case.
$endgroup$
– Jens Schwaiger
Dec 25 '18 at 8:14






$begingroup$
Do you mean begin{equation} f_t(x) = begin{cases} -1 & text{if}; ,x<-t \ sin left(frac{pi x}{2 t}right) & text{if};, |x| le t\ 1 & text{if};, x>t end{cases}? end{equation} I.e., $x<-t$ in the first case.
$endgroup$
– Jens Schwaiger
Dec 25 '18 at 8:14














$begingroup$
@JensSchwaiger Yes, thank you !
$endgroup$
– Goal123
Dec 25 '18 at 10:28




$begingroup$
@JensSchwaiger Yes, thank you !
$endgroup$
– Goal123
Dec 25 '18 at 10:28










2 Answers
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Such an estimate cannot hold. It would imply that $f_t(x) to operatorname{sgn}(x)$ uniformly in $x$, and – since each $f_t$ is a continuous function – that the limit function is continuous as well (which it isn't).






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    The inequality you want is not true. If $|f_t(x)-sgn(x)|<t$ then, taking $x=frac t n$ we get $|sin(frac {pi} {2n})-1|<t$ for all $n$ which implies $1leq t$.






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      2 Answers
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      2 Answers
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      $begingroup$

      Such an estimate cannot hold. It would imply that $f_t(x) to operatorname{sgn}(x)$ uniformly in $x$, and – since each $f_t$ is a continuous function – that the limit function is continuous as well (which it isn't).






      share|cite|improve this answer









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        1












        $begingroup$

        Such an estimate cannot hold. It would imply that $f_t(x) to operatorname{sgn}(x)$ uniformly in $x$, and – since each $f_t$ is a continuous function – that the limit function is continuous as well (which it isn't).






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Such an estimate cannot hold. It would imply that $f_t(x) to operatorname{sgn}(x)$ uniformly in $x$, and – since each $f_t$ is a continuous function – that the limit function is continuous as well (which it isn't).






          share|cite|improve this answer









          $endgroup$



          Such an estimate cannot hold. It would imply that $f_t(x) to operatorname{sgn}(x)$ uniformly in $x$, and – since each $f_t$ is a continuous function – that the limit function is continuous as well (which it isn't).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '18 at 12:19









          Martin RMartin R

          29.8k33558




          29.8k33558























              1












              $begingroup$

              The inequality you want is not true. If $|f_t(x)-sgn(x)|<t$ then, taking $x=frac t n$ we get $|sin(frac {pi} {2n})-1|<t$ for all $n$ which implies $1leq t$.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                The inequality you want is not true. If $|f_t(x)-sgn(x)|<t$ then, taking $x=frac t n$ we get $|sin(frac {pi} {2n})-1|<t$ for all $n$ which implies $1leq t$.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The inequality you want is not true. If $|f_t(x)-sgn(x)|<t$ then, taking $x=frac t n$ we get $|sin(frac {pi} {2n})-1|<t$ for all $n$ which implies $1leq t$.






                  share|cite|improve this answer











                  $endgroup$



                  The inequality you want is not true. If $|f_t(x)-sgn(x)|<t$ then, taking $x=frac t n$ we get $|sin(frac {pi} {2n})-1|<t$ for all $n$ which implies $1leq t$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 25 '18 at 7:26

























                  answered Dec 25 '18 at 7:19









                  Kavi Rama MurthyKavi Rama Murthy

                  66.9k53067




                  66.9k53067






























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