Local minimum of the difference of two functions
$begingroup$
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 $x=frac{1}{2}$.
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
asked Jan 13 at 17:52
eleguitareleguitar
140114
$endgroup$
add a comment |
$begingroup$
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 $x=frac{1}{2}$.
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
asked Jan 13 at 17:52
eleguitareleguitar
140114
$endgroup$
add a comment |
$begingroup$
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 $x=frac{1}{2}$.
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
asked Jan 13 at 17:52
eleguitareleguitar
140114
$endgroup$
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 $x=frac{1}{2}$.
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
real-analysis derivatives extremal-graph-theory
asked Jan 13 at 17:52
eleguitareleguitar
140114
asked Jan 13 at 17:52
eleguitareleguitar
140114
edited Jan 13 at 18:03
eleguitar
asked Jan 13 at 17:52
eleguitareleguitar
140114
asked Jan 13 at 17:52
eleguitareleguitar
140114
asked Jan 13 at 17:52
eleguitareleguitar
140114
140114
add a comment |
add a comment |
1 Answer
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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.
answered Jan 13 at 18:25
Mostafa AyazMostafa Ayaz
18.1k31040
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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.
answered Jan 13 at 18:25
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 13 at 18:25
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 13 at 18:25
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
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.
answered Jan 13 at 18:25
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 13 at 18:25
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 13 at 18:25
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 13 at 18:25
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
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