Local minimum of the difference of two functions












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Consider the function $$u(x)=begin{cases}x qquad qquad xin[0,frac{1}{2}]\-x+1 qquad x in[frac{1}{2},1]end{cases}.$$



Certainly $u$ is continuous in $[0,1]$, derivable at each point except at $x=frac{1}{2}$.



enter image description here



I wonder if there is a formal way to prove that there is no $phiin C^1((0,1))$ such that $phi(frac{1}{2})=u(frac{1}{2})$ and its graph is locally under the graph of $u$, ie there exist $epsilon>0$ such that $phi(y)<u(y)$ for all $y in B_epsilon(frac{1}{2})={yin (0,1): |y-frac{1}{2}|<epsilon}$, $yneq frac{1}{2}$.










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    $begingroup$


    Consider the function $$u(x)=begin{cases}x qquad qquad xin[0,frac{1}{2}]\-x+1 qquad x in[frac{1}{2},1]end{cases}.$$



    Certainly $u$ is continuous in $[0,1]$, derivable at each point except at $x=frac{1}{2}$.



    enter image description here



    I wonder if there is a formal way to prove that there is no $phiin C^1((0,1))$ such that $phi(frac{1}{2})=u(frac{1}{2})$ and its graph is locally under the graph of $u$, ie there exist $epsilon>0$ such that $phi(y)<u(y)$ for all $y in B_epsilon(frac{1}{2})={yin (0,1): |y-frac{1}{2}|<epsilon}$, $yneq frac{1}{2}$.










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      $begingroup$


      Consider the function $$u(x)=begin{cases}x qquad qquad xin[0,frac{1}{2}]\-x+1 qquad x in[frac{1}{2},1]end{cases}.$$



      Certainly $u$ is continuous in $[0,1]$, derivable at each point except at $x=frac{1}{2}$.



      enter image description here



      I wonder if there is a formal way to prove that there is no $phiin C^1((0,1))$ such that $phi(frac{1}{2})=u(frac{1}{2})$ and its graph is locally under the graph of $u$, ie there exist $epsilon>0$ such that $phi(y)<u(y)$ for all $y in B_epsilon(frac{1}{2})={yin (0,1): |y-frac{1}{2}|<epsilon}$, $yneq frac{1}{2}$.










      share|cite|improve this question











      $endgroup$




      Consider the function $$u(x)=begin{cases}x qquad qquad xin[0,frac{1}{2}]\-x+1 qquad x in[frac{1}{2},1]end{cases}.$$



      Certainly $u$ is continuous in $[0,1]$, derivable at each point except at $x=frac{1}{2}$.



      enter image description here



      I wonder if there is a formal way to prove that there is no $phiin C^1((0,1))$ such that $phi(frac{1}{2})=u(frac{1}{2})$ and its graph is locally under the graph of $u$, ie there exist $epsilon>0$ such that $phi(y)<u(y)$ for all $y in B_epsilon(frac{1}{2})={yin (0,1): |y-frac{1}{2}|<epsilon}$, $yneq frac{1}{2}$.







      real-analysis derivatives extremal-graph-theory






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      edited Jan 13 at 18:03







      eleguitar

















      asked Jan 13 at 17:52









      eleguitareleguitar

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          Assume by contrary that $epsilon>0$ exists such that $$left|x-{1over 2}right|<epsilonto phi(x)le u(x)$$since $phi(0.5)=u(0.5)$ we can write$$phi(x)-phi(0.5)le u(x)-u(0.5)$$for $0.5<x<0.5+epsilon$ we obtain $${phi(x)-phi(0.5)over x-0.5}le {u(x)-u(0.5)over x-0.5}$$ and by tending $xto 0.5^+$ since $phiin C^1[(0,1)]$ we have $$phi'(0.5)le u'^+(0.5)=-1$$where $u'^+(0.5)$ is the right derivative of $u$ in $0.5$. Similarly $0.5-epsilon<x<0.5$ yields to $${phi(x)-phi(0.5)over x-0.5}ge {u(x)-u(0.5)over x-0.5}$$and $$phi'(0.5)ge u'^-(0.5)=1$$which leads to the impossible inequality $1le-1$. Therefore such $epsilon$ does not exist and the proof is complete.






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            $begingroup$

            Assume by contrary that $epsilon>0$ exists such that $$left|x-{1over 2}right|<epsilonto phi(x)le u(x)$$since $phi(0.5)=u(0.5)$ we can write$$phi(x)-phi(0.5)le u(x)-u(0.5)$$for $0.5<x<0.5+epsilon$ we obtain $${phi(x)-phi(0.5)over x-0.5}le {u(x)-u(0.5)over x-0.5}$$ and by tending $xto 0.5^+$ since $phiin C^1[(0,1)]$ we have $$phi'(0.5)le u'^+(0.5)=-1$$where $u'^+(0.5)$ is the right derivative of $u$ in $0.5$. Similarly $0.5-epsilon<x<0.5$ yields to $${phi(x)-phi(0.5)over x-0.5}ge {u(x)-u(0.5)over x-0.5}$$and $$phi'(0.5)ge u'^-(0.5)=1$$which leads to the impossible inequality $1le-1$. Therefore such $epsilon$ does not exist and the proof is complete.






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              2












              $begingroup$

              Assume by contrary that $epsilon>0$ exists such that $$left|x-{1over 2}right|<epsilonto phi(x)le u(x)$$since $phi(0.5)=u(0.5)$ we can write$$phi(x)-phi(0.5)le u(x)-u(0.5)$$for $0.5<x<0.5+epsilon$ we obtain $${phi(x)-phi(0.5)over x-0.5}le {u(x)-u(0.5)over x-0.5}$$ and by tending $xto 0.5^+$ since $phiin C^1[(0,1)]$ we have $$phi'(0.5)le u'^+(0.5)=-1$$where $u'^+(0.5)$ is the right derivative of $u$ in $0.5$. Similarly $0.5-epsilon<x<0.5$ yields to $${phi(x)-phi(0.5)over x-0.5}ge {u(x)-u(0.5)over x-0.5}$$and $$phi'(0.5)ge u'^-(0.5)=1$$which leads to the impossible inequality $1le-1$. Therefore such $epsilon$ does not exist and the proof is complete.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Assume by contrary that $epsilon>0$ exists such that $$left|x-{1over 2}right|<epsilonto phi(x)le u(x)$$since $phi(0.5)=u(0.5)$ we can write$$phi(x)-phi(0.5)le u(x)-u(0.5)$$for $0.5<x<0.5+epsilon$ we obtain $${phi(x)-phi(0.5)over x-0.5}le {u(x)-u(0.5)over x-0.5}$$ and by tending $xto 0.5^+$ since $phiin C^1[(0,1)]$ we have $$phi'(0.5)le u'^+(0.5)=-1$$where $u'^+(0.5)$ is the right derivative of $u$ in $0.5$. Similarly $0.5-epsilon<x<0.5$ yields to $${phi(x)-phi(0.5)over x-0.5}ge {u(x)-u(0.5)over x-0.5}$$and $$phi'(0.5)ge u'^-(0.5)=1$$which leads to the impossible inequality $1le-1$. Therefore such $epsilon$ does not exist and the proof is complete.






                share|cite|improve this answer









                $endgroup$



                Assume by contrary that $epsilon>0$ exists such that $$left|x-{1over 2}right|<epsilonto phi(x)le u(x)$$since $phi(0.5)=u(0.5)$ we can write$$phi(x)-phi(0.5)le u(x)-u(0.5)$$for $0.5<x<0.5+epsilon$ we obtain $${phi(x)-phi(0.5)over x-0.5}le {u(x)-u(0.5)over x-0.5}$$ and by tending $xto 0.5^+$ since $phiin C^1[(0,1)]$ we have $$phi'(0.5)le u'^+(0.5)=-1$$where $u'^+(0.5)$ is the right derivative of $u$ in $0.5$. Similarly $0.5-epsilon<x<0.5$ yields to $${phi(x)-phi(0.5)over x-0.5}ge {u(x)-u(0.5)over x-0.5}$$and $$phi'(0.5)ge u'^-(0.5)=1$$which leads to the impossible inequality $1le-1$. Therefore such $epsilon$ does not exist and the proof is complete.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 13 at 18:25









                Mostafa AyazMostafa Ayaz

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                18.1k31040






























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