Showing an inequality in the Banach space $C^1([a,b])$?

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Given $f in C^1([a,b]),$ how can it be shown that $$|f|_infty^2 le dfrac_2^2b-a + 2 |f|_2 |f'|_2?$$



I suppose I could use Cauchy-Schwarz, but how to handle the derivative? And I don't think Holder's applies...



This is a problem from an old Preliminary exam.




banach-spaces norm cauchy-schwarz-inequality




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    Given $f in C^1([a,b]),$ how can it be shown that $$|f|_infty^2 le dfrac_2^2b-a + 2 |f|_2 |f'|_2?$$



    I suppose I could use Cauchy-Schwarz, but how to handle the derivative? And I don't think Holder's applies...



    This is a problem from an old Preliminary exam.




    banach-spaces norm cauchy-schwarz-inequality




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Given $f in C^1([a,b]),$ how can it be shown that $$|f|_infty^2 le dfrac_2^2b-a + 2 |f|_2 |f'|_2?$$



      I suppose I could use Cauchy-Schwarz, but how to handle the derivative? And I don't think Holder's applies...



      This is a problem from an old Preliminary exam.




      banach-spaces norm cauchy-schwarz-inequality




      share|cite|improve this question











      Given $f in C^1([a,b]),$ how can it be shown that $$|f|_infty^2 le dfrac_2^2b-a + 2 |f|_2 |f'|_2?$$



      I suppose I could use Cauchy-Schwarz, but how to handle the derivative? And I don't think Holder's applies...



      This is a problem from an old Preliminary exam.





      banach-spaces norm cauchy-schwarz-inequality





      share|cite|improve this question










      share|cite|improve this question




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          One idea is to let $xinarg min |f^2(x)|$, fix $yin [a,b]$, and apply the FTC to $f^2$:
          $$
          f(y)^2-f(x)^2=2int_x^yf(t)f'(t)dt,
          $$
          so that by Holder,
          $$
          sup_yin[a,b]f(y)^2le f(x)^2+2|f|_2|f'|_2.
          $$
          To deal with the $f(x)^2$ term,
          $$
          f(x)^2=frac1b-aint_a^bf(x)^2dtle frac1b-aint_a^bf(t)^2dt
          $$
          since $f(x)^2le f(t)^2$ for all $t$.






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            up vote
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            down vote



            accepted










            One idea is to let $xinarg min |f^2(x)|$, fix $yin [a,b]$, and apply the FTC to $f^2$:
            $$
            f(y)^2-f(x)^2=2int_x^yf(t)f'(t)dt,
            $$
            so that by Holder,
            $$
            sup_yin[a,b]f(y)^2le f(x)^2+2|f|_2|f'|_2.
            $$
            To deal with the $f(x)^2$ term,
            $$
            f(x)^2=frac1b-aint_a^bf(x)^2dtle frac1b-aint_a^bf(t)^2dt
            $$
            since $f(x)^2le f(t)^2$ for all $t$.






            share|cite|improve this answer

























              up vote
              2
              down vote



              accepted










              One idea is to let $xinarg min |f^2(x)|$, fix $yin [a,b]$, and apply the FTC to $f^2$:
              $$
              f(y)^2-f(x)^2=2int_x^yf(t)f'(t)dt,
              $$
              so that by Holder,
              $$
              sup_yin[a,b]f(y)^2le f(x)^2+2|f|_2|f'|_2.
              $$
              To deal with the $f(x)^2$ term,
              $$
              f(x)^2=frac1b-aint_a^bf(x)^2dtle frac1b-aint_a^bf(t)^2dt
              $$
              since $f(x)^2le f(t)^2$ for all $t$.






              share|cite|improve this answer























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                One idea is to let $xinarg min |f^2(x)|$, fix $yin [a,b]$, and apply the FTC to $f^2$:
                $$
                f(y)^2-f(x)^2=2int_x^yf(t)f'(t)dt,
                $$
                so that by Holder,
                $$
                sup_yin[a,b]f(y)^2le f(x)^2+2|f|_2|f'|_2.
                $$
                To deal with the $f(x)^2$ term,
                $$
                f(x)^2=frac1b-aint_a^bf(x)^2dtle frac1b-aint_a^bf(t)^2dt
                $$
                since $f(x)^2le f(t)^2$ for all $t$.






                share|cite|improve this answer













                One idea is to let $xinarg min |f^2(x)|$, fix $yin [a,b]$, and apply the FTC to $f^2$:
                $$
                f(y)^2-f(x)^2=2int_x^yf(t)f'(t)dt,
                $$
                so that by Holder,
                $$
                sup_yin[a,b]f(y)^2le f(x)^2+2|f|_2|f'|_2.
                $$
                To deal with the $f(x)^2$ term,
                $$
                f(x)^2=frac1b-aint_a^bf(x)^2dtle frac1b-aint_a^bf(t)^2dt
                $$
                since $f(x)^2le f(t)^2$ for all $t$.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered 2 days ago









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