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Say does every differentiable $\mu$-strongly convex function $f:\mathbb{R}^n\mapsto\mathbb{R}$, with $\mu>0$ have a point where its gradient is $0$?

If not so which is the minimum you can impose on a convex function in order to guarantee the existence of a stationary point?

Thanks

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    $\begingroup$ Consider $f(x)=x^2+|x|$... strongly convex and not differentiable at its minimum. All strongly convex functions have a unique global minimum, however, even if they are not differentiable. $\endgroup$ – Michael Grant Nov 23 '14 at 16:54
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Yes, any differentiable strongly convex function has a single stationary point, and that point is its global minimum.

However, note that it is not necessary for a strongly convex function to be differentiable. Per my comment above, consider $f_1(x)=x^2+|x|$. It is strongly convex with $\mu=1$, but it is not differentiable at the origin, which is its global minimum.

Nor is strong convexity a necessary condition for the existence of a single stationary point. Consider $f_2(x)=x^4$ and $f_3(x)=x^4+|x|$. These functions are strictly convex, but not strongly convex. Even so, in both cases, the origin is the global minimum; and for $f_2$, it is a stationary point.

Strict convexity plus differentiability is not enough to guarantee the existence of a stationary point. For instance, $f_4(x)=e^x$ has no global minimum. But if a strictly convex function has a stationary point, it is guaranteed to be unique.

If you drop strict convexity you lose the guarantee of uniqueness as well. For instance, consider $f_5(x)=\max\{x,0\}^2$; $f_5'(x)=0$ for all $x\in(-\infty,0]$, and the global minimum is attained by any point on that interval.

There is another property known as uniform convexity that may suit your needs. A function $f$ is uniformly convex if its domain $\mathop{\textrm{dom}}(f)$ is a convex set, and $$f(tx+(1-t)y)\leq t f(x) + (1-t) f(y) + t(1-t)\phi(\|x-y\|) \quad \forall x,y\in\mathop{\textrm{dom}}(f),t\in[0,1]$$ where $\|\cdot\|$ is any norm, and $\phi$ is function that is strictly increasing on $\mathbb{R}_+$, and satisfies $\phi(0)=0$.

It's a mouthful, I know, but with the flexibility of being able to choose the norm and $\phi$ does make the class of uniformly convex functions larger than the class of strongly convex functions. In fact, strong convexity corresponds to the case where the norm is Euclidean and $\phi(z)=\mu z^2/2$.

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  • $\begingroup$ +1, excellent!! I am sure the OP wasn't expecting such a flurry of helpful information. $\endgroup$ – dineshdileep Nov 25 '14 at 3:45
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    $\begingroup$ What about convex functions over a compact set? Their restriction to the interior is continuous, hence they certainly have a minimum there, and my geometrical intuition is that they cannot degenerate at the boundary either. Is this correct? $\endgroup$ – DeM Dec 4 '15 at 9:48

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