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math.stackexchange.com/a/65587
Lemma: Let $f$ be a convex function on an open interval $I$. For all $x \in I$, $$ g(x) = \lim_{h \to 0} \frac{f(x+h) - f(x-h)}{2h} $$ exists and $f(y) \geq f(x) + g(x) (y-x)$ for all $y \in I$.
In particular, $g$ is a subderivative of $f$.
math.stackexchange.com/questions/1820848/prov … n-a-n-under-some-con
Similar problem:
math.stackexchange.com/questions/817944/limit … unction?noredirect=1
$g: \mathbb{R} \rightarrow \mathbb{R}$, differentiable at some $x$. Let $(u_n), (v_n)$ be sequences s.t. $u_n \leq x \leq v_n$ for all $n$, $u_n \neq v_n$ and $u_n,v_n \rightarrow x$ as $n \rightarrow \infty$. Prove that
$$\lim_{n \rightarrow \infty} \frac{g(v_n)-g(u_n)}{v_n-u_n}$$exists.
$$\frac{g(v_n)-g(u_n)}{v_n-u_n}=λ\frac{g(v_n)-g(x)}{v_n-x}+(1-λ)\frac{g(u_n)-g(x)}{u_n-x}$$hence$$ \min\left\{\frac{g(v_n)-g(x)}{v_n-x},\frac{g(u_n)-g(x)}{u_n-x}\right\}\le \frac{g(v_n)-g(u_n)}{v_n-u_n}\le \max\left\{\frac{g(v_n)-g(x)}{v_n-x},\frac{g(u_n)-g(x)}{u_n-x}\right\}$$Let $n→∞$,$$g'(x)\le\lim_{n→∞} \frac{g(v_n)-g(u_n)}{v_n-u_n}\le g'(x)$$
Similar problem:
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Straddle Lemma
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Presenting the Straddle Lemma in an introductory Real Analysis course
Presenting the Straddle Lemma in an introductory Real Analysis course.pdf
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青青子衿
Posted 2022-6-6 12:43
