Show that $phi_x(t) = int_{0}^{t}x(s)ds$ is continuous on $C[0,1]$












-2














Let $X=Y=C[0,1]$ with the uniform metric $d(x,y)=$sup$_{tin [0,1]};{|x(t) - y(t)|}$



Define $$phi:X to Y$$



$$phi_x(t) = int_{0}^{t}x(s)ds$$ Show that $phi$ is continuous on $X$



$d(phi_x,phi_y)=$sup$_{tin [0,1]};{|int_{0}^{t}x(s)ds - int_{0}^{t}y(s)ds|}$=sup$_{tin [0,1]};{|int_{0}^{t}(x(s)-y(s))ds|}$



$|int_{0}^{t}(x(s)-y(s))ds|leq int_{0}^{t}|x(s)-y(s)|ds$



How can I use this to prove it?






general-topology limits continuity metric-spaces







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  • 1




    you assume $|x(s)-y(s)| le delta$ for each $s$, which gives a bound of $int_0^t delta ds = delta t le delta$. Therefore, $d(x,y) < epsilon implies d(phi_x,phi_y) < epsilon$.
    – mathworker21
    Nov 25 at 5:19
















-2














Let $X=Y=C[0,1]$ with the uniform metric $d(x,y)=$sup$_{tin [0,1]};{|x(t) - y(t)|}$



Define $$phi:X to Y$$



$$phi_x(t) = int_{0}^{t}x(s)ds$$ Show that $phi$ is continuous on $X$



$d(phi_x,phi_y)=$sup$_{tin [0,1]};{|int_{0}^{t}x(s)ds - int_{0}^{t}y(s)ds|}$=sup$_{tin [0,1]};{|int_{0}^{t}(x(s)-y(s))ds|}$



$|int_{0}^{t}(x(s)-y(s))ds|leq int_{0}^{t}|x(s)-y(s)|ds$



How can I use this to prove it?






general-topology limits continuity metric-spaces







share|cite|improve this question


















  • 1




    you assume $|x(s)-y(s)| le delta$ for each $s$, which gives a bound of $int_0^t delta ds = delta t le delta$. Therefore, $d(x,y) < epsilon implies d(phi_x,phi_y) < epsilon$.
    – mathworker21
    Nov 25 at 5:19














-2












-2








-2







Let $X=Y=C[0,1]$ with the uniform metric $d(x,y)=$sup$_{tin [0,1]};{|x(t) - y(t)|}$



Define $$phi:X to Y$$



$$phi_x(t) = int_{0}^{t}x(s)ds$$ Show that $phi$ is continuous on $X$



$d(phi_x,phi_y)=$sup$_{tin [0,1]};{|int_{0}^{t}x(s)ds - int_{0}^{t}y(s)ds|}$=sup$_{tin [0,1]};{|int_{0}^{t}(x(s)-y(s))ds|}$



$|int_{0}^{t}(x(s)-y(s))ds|leq int_{0}^{t}|x(s)-y(s)|ds$



How can I use this to prove it?






general-topology limits continuity metric-spaces







share|cite|improve this question













Let $X=Y=C[0,1]$ with the uniform metric $d(x,y)=$sup$_{tin [0,1]};{|x(t) - y(t)|}$



Define $$phi:X to Y$$



$$phi_x(t) = int_{0}^{t}x(s)ds$$ Show that $phi$ is continuous on $X$



$d(phi_x,phi_y)=$sup$_{tin [0,1]};{|int_{0}^{t}x(s)ds - int_{0}^{t}y(s)ds|}$=sup$_{tin [0,1]};{|int_{0}^{t}(x(s)-y(s))ds|}$



$|int_{0}^{t}(x(s)-y(s))ds|leq int_{0}^{t}|x(s)-y(s)|ds$



How can I use this to prove it?






general-topology limits continuity metric-spaces




general-topology limits continuity metric-spaces






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asked Nov 25 at 5:11









So Lo

63719




63719








  • 1




    you assume $|x(s)-y(s)| le delta$ for each $s$, which gives a bound of $int_0^t delta ds = delta t le delta$. Therefore, $d(x,y) < epsilon implies d(phi_x,phi_y) < epsilon$.
    – mathworker21
    Nov 25 at 5:19














  • 1




    you assume $|x(s)-y(s)| le delta$ for each $s$, which gives a bound of $int_0^t delta ds = delta t le delta$. Therefore, $d(x,y) < epsilon implies d(phi_x,phi_y) < epsilon$.
    – mathworker21
    Nov 25 at 5:19








1




1




you assume $|x(s)-y(s)| le delta$ for each $s$, which gives a bound of $int_0^t delta ds = delta t le delta$. Therefore, $d(x,y) < epsilon implies d(phi_x,phi_y) < epsilon$.
– mathworker21
Nov 25 at 5:19




you assume $|x(s)-y(s)| le delta$ for each $s$, which gives a bound of $int_0^t delta ds = delta t le delta$. Therefore, $d(x,y) < epsilon implies d(phi_x,phi_y) < epsilon$.
– mathworker21
Nov 25 at 5:19










2 Answers
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$int_0^{t}|x(s)-y(s)|ds leq d(x,y)$ so $d(phi_x,phi_y) leq d(x,y)$.






share|cite|improve this answer





























    0














    I'd write $phi(x)(t) = int_0^t x(s)ds$.
    Then for any $x,y in X$ and any $t in [0,1]$:



    $$|phi(x)(t) - phi(y)(t)| = |int_0^t (x(s)-y(s))ds| le int_0^t |x(s)-y(s)| ds$$



    and as for all $s$, $|x(s) -y(s)|le d(x,y)$ we can estimate the final term above
    by $t d(x,y) le d(x,y)$ (as $t le 1$).



    So know that for all $t$, $$|phi(x)(t) - phi(y)(t)| le d(x,y)$$



    and taking the sup on the left:



    $$d(phi(x), phi(y)) le d(x,y)$$ showing that $phi$ is distance-shrinking and is even uniformly continuous: taking $delta=varepsilon$ will work at any point.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      1














      $int_0^{t}|x(s)-y(s)|ds leq d(x,y)$ so $d(phi_x,phi_y) leq d(x,y)$.






      share|cite|improve this answer


























        1














        $int_0^{t}|x(s)-y(s)|ds leq d(x,y)$ so $d(phi_x,phi_y) leq d(x,y)$.






        share|cite|improve this answer
























          1












          1








          1






          $int_0^{t}|x(s)-y(s)|ds leq d(x,y)$ so $d(phi_x,phi_y) leq d(x,y)$.






          share|cite|improve this answer












          $int_0^{t}|x(s)-y(s)|ds leq d(x,y)$ so $d(phi_x,phi_y) leq d(x,y)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 at 5:19









          Kavi Rama Murthy

          48.7k31854




          48.7k31854























              0














              I'd write $phi(x)(t) = int_0^t x(s)ds$.
              Then for any $x,y in X$ and any $t in [0,1]$:



              $$|phi(x)(t) - phi(y)(t)| = |int_0^t (x(s)-y(s))ds| le int_0^t |x(s)-y(s)| ds$$



              and as for all $s$, $|x(s) -y(s)|le d(x,y)$ we can estimate the final term above
              by $t d(x,y) le d(x,y)$ (as $t le 1$).



              So know that for all $t$, $$|phi(x)(t) - phi(y)(t)| le d(x,y)$$



              and taking the sup on the left:



              $$d(phi(x), phi(y)) le d(x,y)$$ showing that $phi$ is distance-shrinking and is even uniformly continuous: taking $delta=varepsilon$ will work at any point.






              share|cite|improve this answer


























                0














                I'd write $phi(x)(t) = int_0^t x(s)ds$.
                Then for any $x,y in X$ and any $t in [0,1]$:



                $$|phi(x)(t) - phi(y)(t)| = |int_0^t (x(s)-y(s))ds| le int_0^t |x(s)-y(s)| ds$$



                and as for all $s$, $|x(s) -y(s)|le d(x,y)$ we can estimate the final term above
                by $t d(x,y) le d(x,y)$ (as $t le 1$).



                So know that for all $t$, $$|phi(x)(t) - phi(y)(t)| le d(x,y)$$



                and taking the sup on the left:



                $$d(phi(x), phi(y)) le d(x,y)$$ showing that $phi$ is distance-shrinking and is even uniformly continuous: taking $delta=varepsilon$ will work at any point.






                share|cite|improve this answer
























                  0












                  0








                  0






                  I'd write $phi(x)(t) = int_0^t x(s)ds$.
                  Then for any $x,y in X$ and any $t in [0,1]$:



                  $$|phi(x)(t) - phi(y)(t)| = |int_0^t (x(s)-y(s))ds| le int_0^t |x(s)-y(s)| ds$$



                  and as for all $s$, $|x(s) -y(s)|le d(x,y)$ we can estimate the final term above
                  by $t d(x,y) le d(x,y)$ (as $t le 1$).



                  So know that for all $t$, $$|phi(x)(t) - phi(y)(t)| le d(x,y)$$



                  and taking the sup on the left:



                  $$d(phi(x), phi(y)) le d(x,y)$$ showing that $phi$ is distance-shrinking and is even uniformly continuous: taking $delta=varepsilon$ will work at any point.






                  share|cite|improve this answer












                  I'd write $phi(x)(t) = int_0^t x(s)ds$.
                  Then for any $x,y in X$ and any $t in [0,1]$:



                  $$|phi(x)(t) - phi(y)(t)| = |int_0^t (x(s)-y(s))ds| le int_0^t |x(s)-y(s)| ds$$



                  and as for all $s$, $|x(s) -y(s)|le d(x,y)$ we can estimate the final term above
                  by $t d(x,y) le d(x,y)$ (as $t le 1$).



                  So know that for all $t$, $$|phi(x)(t) - phi(y)(t)| le d(x,y)$$



                  and taking the sup on the left:



                  $$d(phi(x), phi(y)) le d(x,y)$$ showing that $phi$ is distance-shrinking and is even uniformly continuous: taking $delta=varepsilon$ will work at any point.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 25 at 6:59









                  Henno Brandsma

                  104k346113




                  104k346113






























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