Say we have a 2 variable function $f(x,y)$. At the point $P(x_0,y_0)$ it has a local minima along every line. What I mean by this is, if we choose any line $L$ that passes the point $P$ on the $xy$ plane, then the curve $f(L)$ will have a local minima at the point $P$.
My question is whether the function will have a local minima at the point?
I would say that $f$ will also have a local minima at the point $P$ because I am picturing the function to be like a bowl. The center of the bowl being the point $P$. The semicircles that we get as the curves of $f(L)$ all have a minima at the point $P$. I don't see another picture for this function that satisfies the condition. And according to this $f$ does infact have a minima at the point $P$.
But as tested this intuitive deduction on some functions, I came across this one:
$$ f(x,y) = 5 y ^ 4 − 6 x y ^ 2 + x ^ 2 $$
Notice that if we take the curve along the line $y=mx$. We get $$ g(x)=f(x,mx)=5m^4x^4-6m^2x^3+x^2 $$
So, $$g'(x)=20m^4x^3-18m^2x^2+2x =0$$ Obviously $x=0$ is a solution.
Now $$ g''(x)= 60m^4x^2-36m^2x + 2$$ $$ g''(0) > 0$$
This says that for the point $P(0,0)$ we get that $g(x)$ has a minima at $P$ for every value of $m$.
But for $f(x,y)$, let us consider $t>0$. Notice that
$$ f(0,t) > 0 $$ And $$ f(2t^2,t)<0 $$ We will always get points in the neighborhood of $f$ some of which are negative and some of which are positive. This means that even though $g(x)$ has a minima at $0$, the function $f$ doesnot.
No, $f$ need not have a local minimum at that point. Define $f(x,x^2)=x^3$ for $x\in \mathbb R,$ and set $f=0$ everywhere else. Then on each line through $(0,0),$ $f$ has a local minimum value of $0$ at $(0,0).$ However $(0,0)$ is not a local minimum point of $f$ as every neighborhood of $(0,0)$ contains points where $f<0.$
This example is actually differentiable at $(0,0).$