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I am trying to come up with some examples of strictly convex function that is not strong convex.

Recall that a function is strictly convex if for $\theta \in [0,1]$, $w \neq w'$

$f(\theta w + (1-\theta) w') < \theta f(w) + (1-\theta) f(w')$

and strongly convex if for all $w, w' \in \mathbb{R}^n$

$f(\theta w + (1-\theta) w') \le \theta f(w) + (1-\theta) f(w') - \frac{\beta}{2}\theta(1-\theta)(w-w')^2$

The typical example of a strictly convex function is $f(x) = \|x\|^2_2$, however, it is also strongly convex. What would be an example of a strictly convex function but not strongly convex?

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    $\begingroup$ Stricly convex is $f(\theta w + (1-\theta) w') < \theta f(w) + (1-\theta) f(w')$ and strongly convex is some sort of uniform strict convexity. $\endgroup$ Commented Mar 19, 2017 at 23:51
  • $\begingroup$ $x \mapsto x^4$. If the $\inf$ of the eigenvalues of the Hessian is bounded away from zero then it is strongly convex, so a counterexample must have points where the $\min$ eigenvalue is arbitrarily close to zero. $\endgroup$ Commented Mar 20, 2017 at 0:16

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$f(x)=e^x$ is strictly convex, but the decay as $x \to -\infty$ is too slow to allow it to be strongly convex.

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    $\begingroup$ $e^x$ is strongly convex on $[a, \infty)$ but not on $(-\infty,\infty)$. $f''(x) > 0$ and $\lim_{x \to -\infty} f''(x) = 0$ should be sufficient for $f$ being stricly but not strongly convex $\endgroup$ Commented Mar 19, 2017 at 23:53
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If one has twice differentiable convex function. If the second derivative is zero at some point, but the first derivative strictly monotonically increases. That might be an indication.

Perhaps find a function, $F$, that is continuous and differentiable, that is monotonically strictly increasing, but has 'zero first derivative' at some point (such functions exist), and see what comes out as a result of $text{below}$

For example the question mark function of Minkowski, is an example of such a function.See https://en.wikipedia.org/wiki/Minkowski%27s_question_mark_function#Properties_of_.3F.28x.29

What I am about to describe $\text{Below}$, may not/presumably will not work, for this particular degenerate function. That is the Mink-owski question mark function.

As I am not sure that one can integrate it. It is presumably differ-entiable at very few points. So I would not pick that (the Minkowski Question Mark) function for what I am about describe.

But there are cases for of more standard, non-degenerate functions where this behavior occurs; that is where is F is strictly increasing but at some point its derivative vanishes to zero.

$\text{Below}$

For example, in any case if you can such a non-degenerate, function, $F$, and it can be considered to the derivative function of another function, $F1$,

Then in some cases, if $F$, is and continuously-differentiable, and integ-rable, $F$ ,may have to be at least, absolutely continuous for this, hence my concern.

One might be able to take the integral, (in some cases) of said function, $F$ to return, a 'strictly convex function' $F1$,in some cases. Where $F1$, whose first derivative corresponds to $F$, and whose second derivative (may)correspond to the first derivative of $F$ (in some cases); and may be strictly convex .

I presume that all strictly monotonic functions satisfy $F'(x) \geq 0$ as a bi-conditional (sufficient and necessary condition), as do monotonic increasing functions I think.

Although a strictly increasing function, does not have to satisfy just not $F'(x) > 0$ for all points in the domain.

Its a sufficient condition but not necessary.

The sufficient and necessary conditions for function to be strictly increasing , when diff-erentiable. That is, in terms of a first derivative test, are slightly more complex.

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An example is $f(x)=x^4$, which is strictly convex but not strongly convex. Since $\nabla^2 f(x) = 12x^2$ is not positive definite, the function is not strongly convex. For $x \neq y$, $\langle \nabla f(x) - \nabla f(y),x-y \rangle = 4(x-y)^2 (x^2+y^2 +xy) > 0$. Thus, the function is strictly convex.

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