Exercise 9.10

Question

Exercise 10: 
    If $f$ is a real function defined in a convex open set $E\subset\R n$, such that $(D_1f)(\v x)=0$ for every $\v x\in E$, prove that $f(\v x)$ depends only on $x_2,\ldots,x_n$.

Show that the convexity of $E$ can be replaced by a weaker condition, but that some condition is required. For example, if $n=2$ and $E$ is shaped like a horseshoe, the statement may be false.

analambanomenos · Accepted Answer

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Let $E$ satisfy the weaker condition: if $(x,x_2,\ldots,x_n)\in E$ and $(y,x_2,\ldots,x_n)\in E$, then for all $x<z<y$ we have $(z,x_2,\ldots,x_n)\in E$. For such points, define $f_{x_2,\ldots,x_n}
    (z)=f(z,x_2,\ldots,x_n)$, then $f_{x_2,\ldots,x_n}'(z)=(D_1f)(z,x_2,\ldots,x_n)=0$, so by the Mean Value Theorem $f_{x_2,\ldots,x_n}$ must be constant on $[x,y]$. Hence $f(z,x_2,\ldots,x_n)$
    must equal this constant value for all $z$ such that $(z,x_2,\ldots,x_n)\in E$, so that $f(\v x)$ depends only on $x_2,\ldots,x_n$.

To see a counterexample, let $E\subset\R2$ be the square $(x,y)$, $-1<x<1$, $-1<y<1$ less the positive $y$-axis, $(0,y)$, $0<y<1$. Define
    $$
        f(x,y)=
        \begin{cases}
            0 & -1<x<1,\;-1<y<0 \
            -y & -1<x<0,\;0\le y<1 \
            y & 0<x<1,\;0\le y<1
        \end{cases}
    $$
    Then $f$ is continuous, $(D_1f)(x,y)=0$ for all $(x,y)\in E$, but the value of $f$ does not depend only on $y$.

Exercise 9.10

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Exercise 9.10

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