Integral going to zero

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The question is:



Let
beginalign*
varphi(x) =
begincases
1, & 0 < x < 1/2,\
0, & 1/2 < x < 1,
endcases
endalign*
be a $1$-periodic function, and define $varphi_n(x) = varphi(nx)$. Show that
beginalign*
int_a^b left[varphi_n(x) - 1/2right]dx rightarrow 0, quad text as quad n rightarrow infty,
endalign*
for any interval $(a, b)$.



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real-analysis analysis







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  • What if $x=frac12$?
    – cansomeonehelpmeout
    38 mins ago










  • The value of at $x=frac12$ does not matter for the integral.
    – GurrVasa
    34 mins ago














up vote
3
down vote

favorite












The question is:



Let
beginalign*
varphi(x) =
begincases
1, & 0 < x < 1/2,\
0, & 1/2 < x < 1,
endcases
endalign*
be a $1$-periodic function, and define $varphi_n(x) = varphi(nx)$. Show that
beginalign*
int_a^b left[varphi_n(x) - 1/2right]dx rightarrow 0, quad text as quad n rightarrow infty,
endalign*
for any interval $(a, b)$.



I'm having a hard time showing this. Any suggestions?






real-analysis analysis







share|cite|improve this question























  • What if $x=frac12$?
    – cansomeonehelpmeout
    38 mins ago










  • The value of at $x=frac12$ does not matter for the integral.
    – GurrVasa
    34 mins ago












up vote
3
down vote

favorite









up vote
3
down vote

favorite











The question is:



Let
beginalign*
varphi(x) =
begincases
1, & 0 < x < 1/2,\
0, & 1/2 < x < 1,
endcases
endalign*
be a $1$-periodic function, and define $varphi_n(x) = varphi(nx)$. Show that
beginalign*
int_a^b left[varphi_n(x) - 1/2right]dx rightarrow 0, quad text as quad n rightarrow infty,
endalign*
for any interval $(a, b)$.



I'm having a hard time showing this. Any suggestions?






real-analysis analysis







share|cite|improve this question















The question is:



Let
beginalign*
varphi(x) =
begincases
1, & 0 < x < 1/2,\
0, & 1/2 < x < 1,
endcases
endalign*
be a $1$-periodic function, and define $varphi_n(x) = varphi(nx)$. Show that
beginalign*
int_a^b left[varphi_n(x) - 1/2right]dx rightarrow 0, quad text as quad n rightarrow infty,
endalign*
for any interval $(a, b)$.



I'm having a hard time showing this. Any suggestions?






real-analysis analysis




real-analysis analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 44 mins ago

























asked 49 mins ago









GurrVasa

795




795











  • What if $x=frac12$?
    – cansomeonehelpmeout
    38 mins ago










  • The value of at $x=frac12$ does not matter for the integral.
    – GurrVasa
    34 mins ago
















  • What if $x=frac12$?
    – cansomeonehelpmeout
    38 mins ago










  • The value of at $x=frac12$ does not matter for the integral.
    – GurrVasa
    34 mins ago















What if $x=frac12$?
– cansomeonehelpmeout
38 mins ago




What if $x=frac12$?
– cansomeonehelpmeout
38 mins ago












The value of at $x=frac12$ does not matter for the integral.
– GurrVasa
34 mins ago




The value of at $x=frac12$ does not matter for the integral.
– GurrVasa
34 mins ago










1 Answer
1






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4
down vote



accepted










Hint. Note that for any real $a$
$$int_a^a+1 left[varphi(x) - 1/2right] dx=0.$$
Therefore, for $n>0$, after letting $t=nx$, we have that
$$left|int_a^b left[varphi_n(x) - 1/2right] dxright|=frac1nleft|int_na^nb left[varphi(t) - 1/2right]dtright|leq
frac1nint_0^1 left|varphi(t) - 1/2right|dt.$$






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    4
    down vote



    accepted










    Hint. Note that for any real $a$
    $$int_a^a+1 left[varphi(x) - 1/2right] dx=0.$$
    Therefore, for $n>0$, after letting $t=nx$, we have that
    $$left|int_a^b left[varphi_n(x) - 1/2right] dxright|=frac1nleft|int_na^nb left[varphi(t) - 1/2right]dtright|leq
    frac1nint_0^1 left|varphi(t) - 1/2right|dt.$$






    share|cite|improve this answer


























      up vote
      4
      down vote



      accepted










      Hint. Note that for any real $a$
      $$int_a^a+1 left[varphi(x) - 1/2right] dx=0.$$
      Therefore, for $n>0$, after letting $t=nx$, we have that
      $$left|int_a^b left[varphi_n(x) - 1/2right] dxright|=frac1nleft|int_na^nb left[varphi(t) - 1/2right]dtright|leq
      frac1nint_0^1 left|varphi(t) - 1/2right|dt.$$






      share|cite|improve this answer
























        up vote
        4
        down vote



        accepted







        up vote
        4
        down vote



        accepted






        Hint. Note that for any real $a$
        $$int_a^a+1 left[varphi(x) - 1/2right] dx=0.$$
        Therefore, for $n>0$, after letting $t=nx$, we have that
        $$left|int_a^b left[varphi_n(x) - 1/2right] dxright|=frac1nleft|int_na^nb left[varphi(t) - 1/2right]dtright|leq
        frac1nint_0^1 left|varphi(t) - 1/2right|dt.$$






        share|cite|improve this answer














        Hint. Note that for any real $a$
        $$int_a^a+1 left[varphi(x) - 1/2right] dx=0.$$
        Therefore, for $n>0$, after letting $t=nx$, we have that
        $$left|int_a^b left[varphi_n(x) - 1/2right] dxright|=frac1nleft|int_na^nb left[varphi(t) - 1/2right]dtright|leq
        frac1nint_0^1 left|varphi(t) - 1/2right|dt.$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 23 mins ago

























        answered 32 mins ago









        Robert Z

        88.1k1056127




        88.1k1056127



























             

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