Any local minimum of convex function on convex set $U \subseteq \mathbb{R}^n$ is $\mathbf{global}$ minimum


Let $x \in U$ be a local minimum. Let $f: U \to \mathbb{R}$ be convex function. We want to show that $f(x) \leq f(z)$ for every $z \in U$. Suppose there is some $y \in U$ such that $f(y) < f(x)$. Now, since $f$ is convex, we know

$$ f( tx + ( 1 - t) y ) \leq t f(x) + (1 - t) f(y) $$

where $t \in [0,1]$. with $t = 1/2$, we see that

$$ f( (x+y)/2) \leq \frac{ f(x) + f(y) }{2} < \frac{ f(x) + f(x) }{2} = f(x) $$

but, here Im stuck. How can I obtain a contradiction from here? Is my approach correct?

  • $\begingroup$ Hint: Think about what a local minimum is. $\endgroup$ – grndl Feb 15 '17 at 12:39

Since $f$ is convex, then for all $t \in [0,1],$ we have $f(tx + (1-t)y) \leq tf(x) + (1-t)f(y)$.

If $f(y) < f(x)$, then $f(tx + (1-t)y) < tf(x) + (1-t)f(x) = f(x)$.

Hence, let $U$ be a neighborhood of $x$ for which it is the local minimum of $f$.

Then, by continuity of the function $g(t) = f(tx + (1-t)y)$, we have that there exists $t_0 > 0$ such that $ g(t_0) \in U$.

But then, we have already seen that for all $t > 0$, $g(t) = f(tx + (1-t)y) < f(x)$.

This means that $g(t_0) < f(x)$, and $g(t_0) \in U$, giving a contradiction.

Hence the result follows.

  • $\begingroup$ why there exists a $t_0 > 0$ such that $g(t_0) \in U$? $\endgroup$ – ILoveMath Feb 15 '17 at 12:48
  • $\begingroup$ This is for the following reason : because $g$ is continuous, it is true that $\lim_{t \to 0^+} g(t) = g(x)$. Now, what does limit mean? It means the following: given any neighbourhood $U$ of $x$, $g(t)$ is "eventually" in that neighbourhood. This means that there is $t_0$ small enough (but not zero) so that $g(t_0) \in U$. So we are using the definition of limit and the fact that $g$ is continuous to deduce this. $\endgroup$ – астон вілла олоф мэллбэрг Feb 15 '17 at 12:51

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