Continuously differentiable composition/convolution.












1












$begingroup$


My undergrad analysis is super rusty and I am getting ready for GRE subject and I am completely stuck, I usually have an attempt but I am stuck. However a hint will suffice, I don't need the whole answer. (apparently only 24% get this correct on GRE subject)



Let $f$ and $g$ be continuous functions over the reals s.t.



$g(x) = int_0^x f(y)(y-x) dy$ $forall x$



and $g$ is three times continuously differentiable,



what is the greatest integer $n$ s.t. $f$ is $n$ times continuously differentiable?



my guess is using some theorem or lemma or corollary regarding convolutions? is $g$ not a convolution of $f$?



(feel free to edit the title to best fit my question)










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    My undergrad analysis is super rusty and I am getting ready for GRE subject and I am completely stuck, I usually have an attempt but I am stuck. However a hint will suffice, I don't need the whole answer. (apparently only 24% get this correct on GRE subject)



    Let $f$ and $g$ be continuous functions over the reals s.t.



    $g(x) = int_0^x f(y)(y-x) dy$ $forall x$



    and $g$ is three times continuously differentiable,



    what is the greatest integer $n$ s.t. $f$ is $n$ times continuously differentiable?



    my guess is using some theorem or lemma or corollary regarding convolutions? is $g$ not a convolution of $f$?



    (feel free to edit the title to best fit my question)










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      My undergrad analysis is super rusty and I am getting ready for GRE subject and I am completely stuck, I usually have an attempt but I am stuck. However a hint will suffice, I don't need the whole answer. (apparently only 24% get this correct on GRE subject)



      Let $f$ and $g$ be continuous functions over the reals s.t.



      $g(x) = int_0^x f(y)(y-x) dy$ $forall x$



      and $g$ is three times continuously differentiable,



      what is the greatest integer $n$ s.t. $f$ is $n$ times continuously differentiable?



      my guess is using some theorem or lemma or corollary regarding convolutions? is $g$ not a convolution of $f$?



      (feel free to edit the title to best fit my question)










      share|cite|improve this question









      $endgroup$




      My undergrad analysis is super rusty and I am getting ready for GRE subject and I am completely stuck, I usually have an attempt but I am stuck. However a hint will suffice, I don't need the whole answer. (apparently only 24% get this correct on GRE subject)



      Let $f$ and $g$ be continuous functions over the reals s.t.



      $g(x) = int_0^x f(y)(y-x) dy$ $forall x$



      and $g$ is three times continuously differentiable,



      what is the greatest integer $n$ s.t. $f$ is $n$ times continuously differentiable?



      my guess is using some theorem or lemma or corollary regarding convolutions? is $g$ not a convolution of $f$?



      (feel free to edit the title to best fit my question)







      real-analysis calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 7 at 22:30









      Hossien SahebjameHossien Sahebjame

      1169




      1169






















          1 Answer
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          $begingroup$

          We rewrite $g(x)$ in terms of integral where $x$ appears only in the upper bound:
          $$g(x)=int_0^xf(y)ydy-xint_0^xf(y)dy$$
          Taking the derivative with respect to $x$, and using $$frac {d}{dx}int_0^x w(t)dt=w(x)$$ we have $$frac{dg(x)}{dx}=f(x)x-xf(x)-int_0^xf(y)dy=-int_0^xf(y)dy$$
          Taking the second derivative you get $g''(x)=-f(x)$. You should be able to finish.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            got thanks so much, the line with $w$ helped me a whole lot, together with splitting up the integral. so since $g$ is 3 times and they are "off by 2" integers of degree of differentiability if you will, so $f$ can only be 1 time differentiable? so $n = 1$?
            $endgroup$
            – Hossien Sahebjame
            Jan 8 at 21:45








          • 1




            $begingroup$
            You are correct.
            $endgroup$
            – Andrei
            Jan 8 at 21:47












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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          We rewrite $g(x)$ in terms of integral where $x$ appears only in the upper bound:
          $$g(x)=int_0^xf(y)ydy-xint_0^xf(y)dy$$
          Taking the derivative with respect to $x$, and using $$frac {d}{dx}int_0^x w(t)dt=w(x)$$ we have $$frac{dg(x)}{dx}=f(x)x-xf(x)-int_0^xf(y)dy=-int_0^xf(y)dy$$
          Taking the second derivative you get $g''(x)=-f(x)$. You should be able to finish.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            got thanks so much, the line with $w$ helped me a whole lot, together with splitting up the integral. so since $g$ is 3 times and they are "off by 2" integers of degree of differentiability if you will, so $f$ can only be 1 time differentiable? so $n = 1$?
            $endgroup$
            – Hossien Sahebjame
            Jan 8 at 21:45








          • 1




            $begingroup$
            You are correct.
            $endgroup$
            – Andrei
            Jan 8 at 21:47
















          2












          $begingroup$

          We rewrite $g(x)$ in terms of integral where $x$ appears only in the upper bound:
          $$g(x)=int_0^xf(y)ydy-xint_0^xf(y)dy$$
          Taking the derivative with respect to $x$, and using $$frac {d}{dx}int_0^x w(t)dt=w(x)$$ we have $$frac{dg(x)}{dx}=f(x)x-xf(x)-int_0^xf(y)dy=-int_0^xf(y)dy$$
          Taking the second derivative you get $g''(x)=-f(x)$. You should be able to finish.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            got thanks so much, the line with $w$ helped me a whole lot, together with splitting up the integral. so since $g$ is 3 times and they are "off by 2" integers of degree of differentiability if you will, so $f$ can only be 1 time differentiable? so $n = 1$?
            $endgroup$
            – Hossien Sahebjame
            Jan 8 at 21:45








          • 1




            $begingroup$
            You are correct.
            $endgroup$
            – Andrei
            Jan 8 at 21:47














          2












          2








          2





          $begingroup$

          We rewrite $g(x)$ in terms of integral where $x$ appears only in the upper bound:
          $$g(x)=int_0^xf(y)ydy-xint_0^xf(y)dy$$
          Taking the derivative with respect to $x$, and using $$frac {d}{dx}int_0^x w(t)dt=w(x)$$ we have $$frac{dg(x)}{dx}=f(x)x-xf(x)-int_0^xf(y)dy=-int_0^xf(y)dy$$
          Taking the second derivative you get $g''(x)=-f(x)$. You should be able to finish.






          share|cite|improve this answer









          $endgroup$



          We rewrite $g(x)$ in terms of integral where $x$ appears only in the upper bound:
          $$g(x)=int_0^xf(y)ydy-xint_0^xf(y)dy$$
          Taking the derivative with respect to $x$, and using $$frac {d}{dx}int_0^x w(t)dt=w(x)$$ we have $$frac{dg(x)}{dx}=f(x)x-xf(x)-int_0^xf(y)dy=-int_0^xf(y)dy$$
          Taking the second derivative you get $g''(x)=-f(x)$. You should be able to finish.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 22:44









          AndreiAndrei

          13.2k21230




          13.2k21230












          • $begingroup$
            got thanks so much, the line with $w$ helped me a whole lot, together with splitting up the integral. so since $g$ is 3 times and they are "off by 2" integers of degree of differentiability if you will, so $f$ can only be 1 time differentiable? so $n = 1$?
            $endgroup$
            – Hossien Sahebjame
            Jan 8 at 21:45








          • 1




            $begingroup$
            You are correct.
            $endgroup$
            – Andrei
            Jan 8 at 21:47


















          • $begingroup$
            got thanks so much, the line with $w$ helped me a whole lot, together with splitting up the integral. so since $g$ is 3 times and they are "off by 2" integers of degree of differentiability if you will, so $f$ can only be 1 time differentiable? so $n = 1$?
            $endgroup$
            – Hossien Sahebjame
            Jan 8 at 21:45








          • 1




            $begingroup$
            You are correct.
            $endgroup$
            – Andrei
            Jan 8 at 21:47
















          $begingroup$
          got thanks so much, the line with $w$ helped me a whole lot, together with splitting up the integral. so since $g$ is 3 times and they are "off by 2" integers of degree of differentiability if you will, so $f$ can only be 1 time differentiable? so $n = 1$?
          $endgroup$
          – Hossien Sahebjame
          Jan 8 at 21:45






          $begingroup$
          got thanks so much, the line with $w$ helped me a whole lot, together with splitting up the integral. so since $g$ is 3 times and they are "off by 2" integers of degree of differentiability if you will, so $f$ can only be 1 time differentiable? so $n = 1$?
          $endgroup$
          – Hossien Sahebjame
          Jan 8 at 21:45






          1




          1




          $begingroup$
          You are correct.
          $endgroup$
          – Andrei
          Jan 8 at 21:47




          $begingroup$
          You are correct.
          $endgroup$
          – Andrei
          Jan 8 at 21:47


















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