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Consider a smooth function $f : \mathbb{R}^2 \to \mathbb{R}$, I wonder that any contour (curve) in $\mathbb{R}^2$ where every point of it is a local maxima of $f$, need be a smooth curve?

Edit : $f$ need to be smooth.

Edit 2 : By contour I mean curve of nonzero arc length.

Elaboration (after comments by Will and copper.hat)

Let the function be $f(x,y)$. I want the contour to have at every point on it, the $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial^2 f}{\partial x^2} < 0$. Is any such contour which is not smooth possible?

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    What sort of smoothness do you want on $f$? You could be really obnoxious and let $f$ be the characteristic function of a nonsmooth curve if you make no assumptions.2012-10-24
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    @Zach L. : Sorry, I forgot to mention the actual thing. $f$ is smooth that is $\mathcal{C}^{\infty}$2012-10-24
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    What do you mean by 'where every point of it is a local maxima of $f$'? The function $f(x)=-(x_1^2+x_2^2)$ has just one (local) maximum at $(0,0)$, does this constitute a smooth curve?2012-10-24
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    @copper.hat : Thanks for the comment. The curve should be of non zero arc length.2012-10-24

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$$ f(x,y) = - x^2 y^2 {}{}{}{}{} $$

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    What is the contour of maxima which is not smooth for this function?2012-10-24
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    @RajeshD, the $x$ axis and the $y$ axis. Not smooth at the origin.2012-10-24
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    But it doesn't have a local maximum at $(0,0)$2012-10-24
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    @copper.hat [The partial derivative](http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427eqjj0nf926t) with respect to $x$ is $0$at $(0,0)$ and the double partial derivative with respect to $x$ is also $0$ at the origin. Hence it does not have a maximum along the x-axis at the origin.2012-10-24
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    @RajeshD, how about if you tell us the definition of local maximum? Meanwhile, how about $$ g(x,y) = -(x^2 - y^2)^2 \; \; ? $$2012-10-24
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    A local maximum is a maximum restricted to some ball. Since $0$ is the maximum (global and local) value, and the function attains the value on the axes, it has global & local maxima on the axes.2012-10-24
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    @copper.hat : Thanks for the explanation and clarification. Now your answer does seem to fit the question. But my intention is a bit different and not realizing the complexity I think I have not asked my question properly. I need contour to have at every point, a local maxima along one of the co-ordinate axis and let it be $x-axis$. I want partial derivative with repect to $x$ be zero and the partial second derivative with respect to $x$ to be negative. I don't want the kind of maxima you have mentioned.2012-10-24
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    Perhaps you should elaborate the question a little?2012-10-24
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    @Rajesh D: Take the function $g(x,y) = -(x^2-y^2)^2$ that Will gave in a comment above. (This is just the old function $f$ under a change of coordinates.) The union of two lines $x=\pm y$ satisfies all your new conditions.2012-10-24
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    @RahulNarain : Thanks for the clarification. I guess its a rotation of axis by $45^o$ to form the new function.2012-10-24