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Let $f(x,y)$ be a differentiable function.

Suppose that for every line, $l$ , that passes through $(0,0)$ the point $(0,0)$ is a local minimum of $f(x,y)\mid_ l$. Does that mean that $(0,0)$ is a local minimum of $f(x,y)$?

When I first saw the question I tried to prove it by letting $l=(a,b)\cdot t$ where a,b are constants. Then I got $x=at, y=bt$ and substituted that in the function $f(x,y)$ to get $f(at,bt)$. Then I used the chain rule to differentiate $f$ , but didn't know what to do from there.

Is this a good direction? or am I suppose to disprove the claim?

Any help will be appreciated!

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You can prove that the map $$\begin{array}{l|rcl} f : & \mathbb{R}^2 & \longrightarrow & \mathbb{R} \\ & (x,y) & \longmapsto & f(x,y)=3x^4-4x^2y+y^2 \end{array}$$

has a local minimum at zero along all lines passing through the origin.

However the origin is not a local minimum for $f$ as $f(x,2 x^2)=-x^4 $.

For more details, you can have a look at NO MINIMUM AT THE ORIGIN BUT A MINIMUM ALONG ALL LINES.

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  • $\begingroup$ Amazing! Thank you :) $\endgroup$
    – Eliads
    May 19 '18 at 17:08

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