Local minimum of a function

Question

Let $f:\mathbb R^n \mapsto \mathbb R$ be a differentiable function. Let us assume that the point $\bar x$ is a local minimum of the function f on every line passing through $\bar x$ what means that the function $g(\alpha) = f(\bar x + \alpha d)$ has a local minimum at the point $\alpha = 0$ for every $d\in \mathbb R^n$.

I need to show that $\nabla f(\bar x) = 0$ and come up with an example showing $\bar x$ does not have to be a local minimum of $f$. I got a hint to this one - let $f(x,y) = (y-px^2)(y-qx^2)$, where $(0<p<q)$. Show that $(0,0)$ is a local minimum of $f$ on every line passing through $(0,0)$.

Answer 1

Note that $\lim_{t \to 0} {f(x+th)-f(x) \over t} = \langle \nabla f(x), h \rangle$.

If $\nabla f(x) \neq 0$, and set $h=-\nabla f(x) $ we have $\lim_{t \to 0} {f(x+th)-f(x) \over t} < 0$ and hence there are points on the ray $\{ x+t h \}_{t \ge 0 }$ arbitrarily close to $x$ such that $f(x+th) < f(x)$.

In particular, if $x$ is a local $\min$ we must have $\nabla f(x) = 0$.

The example you have shows that for any $h$, the function $\phi_h(t)= f(th)$ has a local $\min$ at $t=0$. It is straightforward to verify that $\phi_h'(0) = 0$. If $h_2=0$ then $\phi_h(t) = pq t^4$, then this is clear. Otherwise we have $\phi_h''(0) = 2h_2^2 >0$, hence $\phi_h$ is locally convex and $t=0$ is a local $\min$.

Finally, we note that if $y-q x^2 < 0 < y-p x^2$ we see that $f(x,y) <0$ and, for $x \neq 0$, this condition is equivalent to $p < {y \over x^2} < q$. Hence there are points $(x,y)$ arbitrarily close to $0$ such that $f(x,y)<0$. It follows that $(0,0)$ is not a local $\min.$ of $f$.