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Let $f: U \rightarrow \mathbb{R}$ a function of $C^{2}$ in $U$. show that $f$ is $\alpha$-convexe on E if and only if $$ \forall x \in U, \forall h \in E, \quad D^{2} f(x)(h, h) \geq \alpha\|h\|^{2} $$ (f is $\alpha$-convexe if $f((1-t) x+t y) \leq(1-t) f(x)+t f(y)-\frac{\alpha}{2} t(1-t)\|x-y\|^{2}$) I'm not familiar with differentiability but I kow that $D f(x)= lim_ {t\to 0} \frac {f(x+h)-f(x)} {h}$. However is $D^{2} f(x)= \frac {Df(x+h)-Df(x)} {h}$ correct ? I

Arctic Char
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vemapo
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  • Given the appearance of norms $|\cdot|$ in your question, I suspect $U$ is an open subset of $\mathbb{R}^n$, and $Df$ denotes the gradient and $D^2 f$ denotes the Hessian. – angryavian Sep 11 '21 at 21:22
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    No, you need to learn the definition of derivative and Hessian when you’re in dimension $>1$. And your limit makes no sense, anyway. The derivative is a linear transformation and the second derivative is a bilinear form. – Ted Shifrin Sep 11 '21 at 21:38

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If $U \subseteq \mathbb{R}$ then the statement is asking for the equivalence between $$f((1-t)x+ty) \le (1-t)f(x) + tf(y) - \frac{\alpha}{2} t(1-t) |x-y|^2, \qquad \forall x,y,\ \forall t\in[0,1]$$ and $$f''(x) \ge \alpha, \qquad \forall x$$ for $f\in C^2$.

Suppose $f''(x) \ge \alpha$. Then by the mean value form for the remainder of the first-order Taylor polynomial, we have $$f(y) \ge f((1-t)x+ty) + f'((1-t)x + ty)(1-t)(y-x) + \frac{\alpha}{2}(1-t)^2(y-x)^2$$ and $$f(x) \ge f((1-t)x+ty) - f'((1-t)x + ty) t(y-x) + \frac{\alpha}{2}t^2(y-x)^2.$$ These two inequalities together imply \begin{align} &(1-t)f(x) + tf(y) - \frac{\alpha}{2}t(1-t)|x-y|^2 \\ &\ge f((1-t)x+ty) + \frac{\alpha}{2}|x-y|^2 [t^2 + (1-t)^2 - t(1-t)]. \end{align} The expression in square brackets equals $1-3t(1-t) \ge 0$, so the second term is nonnegative.

Suppose $f$ is $\alpha$-convex. Then for $h > 0$, $$f(x) \le \frac{1}{2} f(x-h) + \frac{1}{2} f(x+h) - \frac{\alpha}{2} h^2.$$ By the second symmetric derivative, $$f''(x) = \lim_{h \to 0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2} \ge \alpha.$$

The more general case where $U \subseteq \mathbb{R}^n$ can be reduced to the one-dimensional case, since $D^2f(x)(h,h)$ is the same as $\|h\|^2 g''(0)$ with $g:\mathbb{R} \to \mathbb{R}$ defined as $g(t)=f(x+th/\|h\|)$.

angryavian
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  • Thank you so much brother !! This helps me understand But how u get $f(x) \le \frac{1}{2} f(x-h) + \frac{1}{2} f(x+h) - \frac{\alpha}{2} h^2.$ i know $h= x-y$ – vemapo Sep 11 '21 at 22:33
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    @vemapo In the original definition of $\alpha$-convexity, replace $x$ with $x-h$, and replace $y$ with $x+h$. – angryavian Sep 12 '21 at 00:04