$g(x) = f(x,0)$ e $h(y) = f(0,y)$. If $x = 0$ is local minimum of $g$ and $y = 0$ is local minimum of $h$, then $(0,0)$ is local minimum of $f$.

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Prove or disprove. Let $f: mathbbR^2 to mathbbR$, $g(x) = f(x,0)$ e $h(y) = f(0,y)$. If $x = 0$ is local minimum point of $g$ and $y = 0$ is local minimum point of $h$, then $(0,0)$ is local minimum point of $f$.




I couldn't prove it to be true. I don't know what I can use, since there are not many assumptions about $f$. So I'm trying to get a counterexample, but I was not successful.



First, I tried to get a differentiable function that had no local minimum and after, to get $g$ and $h$. But in all cases, at least one of the two don't satisfy the hypotheses.




real-analysis




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    Prove or disprove. Let $f: mathbbR^2 to mathbbR$, $g(x) = f(x,0)$ e $h(y) = f(0,y)$. If $x = 0$ is local minimum point of $g$ and $y = 0$ is local minimum point of $h$, then $(0,0)$ is local minimum point of $f$.




    I couldn't prove it to be true. I don't know what I can use, since there are not many assumptions about $f$. So I'm trying to get a counterexample, but I was not successful.



    First, I tried to get a differentiable function that had no local minimum and after, to get $g$ and $h$. But in all cases, at least one of the two don't satisfy the hypotheses.




    real-analysis




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      up vote
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      down vote

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      up vote
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      down vote

      favorite












      Prove or disprove. Let $f: mathbbR^2 to mathbbR$, $g(x) = f(x,0)$ e $h(y) = f(0,y)$. If $x = 0$ is local minimum point of $g$ and $y = 0$ is local minimum point of $h$, then $(0,0)$ is local minimum point of $f$.




      I couldn't prove it to be true. I don't know what I can use, since there are not many assumptions about $f$. So I'm trying to get a counterexample, but I was not successful.



      First, I tried to get a differentiable function that had no local minimum and after, to get $g$ and $h$. But in all cases, at least one of the two don't satisfy the hypotheses.




      real-analysis




      share|cite|improve this question












      Prove or disprove. Let $f: mathbbR^2 to mathbbR$, $g(x) = f(x,0)$ e $h(y) = f(0,y)$. If $x = 0$ is local minimum point of $g$ and $y = 0$ is local minimum point of $h$, then $(0,0)$ is local minimum point of $f$.




      I couldn't prove it to be true. I don't know what I can use, since there are not many assumptions about $f$. So I'm trying to get a counterexample, but I was not successful.



      First, I tried to get a differentiable function that had no local minimum and after, to get $g$ and $h$. But in all cases, at least one of the two don't satisfy the hypotheses.





      real-analysis





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      share|cite|improve this question




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      asked 7 hours ago









      Lucas Corrêa

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          Let us consider $f:mathbbR^2 to mathbbR,~ f(x,y):=-xy$.
          Then $h=gequiv 0$, but $f$ has no local minimum in $(0,0)$.






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            Let us consider $f:mathbbR^2 to mathbbR,~ f(x,y):=-xy$.
            Then $h=gequiv 0$, but $f$ has no local minimum in $(0,0)$.






            share|cite|improve this answer

























              up vote
              3
              down vote













              Let us consider $f:mathbbR^2 to mathbbR,~ f(x,y):=-xy$.
              Then $h=gequiv 0$, but $f$ has no local minimum in $(0,0)$.






              share|cite|improve this answer























                up vote
                3
                down vote










                up vote
                3
                down vote









                Let us consider $f:mathbbR^2 to mathbbR,~ f(x,y):=-xy$.
                Then $h=gequiv 0$, but $f$ has no local minimum in $(0,0)$.






                share|cite|improve this answer













                Let us consider $f:mathbbR^2 to mathbbR,~ f(x,y):=-xy$.
                Then $h=gequiv 0$, but $f$ has no local minimum in $(0,0)$.







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered 6 hours ago









                Jonas Lenz

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