3
$\begingroup$

Let $f$ be a real function with left and right derivatives $f'_-$ and $f'_+$ on the open interval $(a,b)$, and continuous on $[a,b]$ (e.g., let $f$ be convex on $[a,b]$). Then,

Is there something like the mean value theorem for $f$?

$\endgroup$
  • 1
    $\begingroup$ If $f:[a,b]\to\mathbb{R}$ is a convex function let \begin{equation*} \partial f(x)=\left\{p\in\mathbb{R}\bigm\vert f(y)\ge f(x)+p(y-x)\right\}. \end{equation*} We refer to this as the subdifferential of $f$. Now suppose, in addition to the above, that we know $f(a)=0=f(b)$ and that $f$ is continuous. Then I claim there is $c\in(a,b)$ so that \begin{equation*} 0\in\partial f(c). \end{equation*} $\endgroup$ – user71352 Mar 9 '17 at 5:41
  • 1
    $\begingroup$ To see this, note that we may assume that $f$ is non-constant. Then by extreme value theorem a minimum is obtained somewhere on $[a,b]$. By the previous assumption it occurs at $c\in(a,b)$. It follows by definition that $0\in\partial f(c)$. To conclude a mean value theorem note that \begin{equation*} g(x)=f(x)-\left[\frac{f(b)-f(a)}{b-a}(x-a)+f(a)\right] \end{equation*} is convex, satisfies $g(a)=0=g(b)$, and $\partial v(x)=\partial f(x)-\frac{f(b)-f(a)}{b-a}$. $\endgroup$ – user71352 Mar 9 '17 at 5:41
  • $\begingroup$ I think if $f$ is convex then there exits a $c$ in $(a,b)$ such that $f'_-(c)\leq \frac{f(b)-f(a)}{b-a}\leq f'_+(c)$, is it true? Now, what can we say about the original case? $\endgroup$ – M.H.Hooshmand Mar 9 '17 at 7:17
  • $\begingroup$ I believe this statement about convex functions follows from my previous comments and the notion of the subdifferential. Observe that $p\in\partial f(c)$ means that for $y<x$ we have \begin{equation*} \frac{f(y)-f(x)}{y-x}\le p \end{equation*} and so taking limits gives $f_{-}'(c)\le p$. A similar conclusion follows for $f_{+}'(c)$. Were you trying to prove a similar statement for the original case? $\endgroup$ – user71352 Mar 9 '17 at 15:29
  • 1
    $\begingroup$ I think the result you mentioned is valid after a slight modification. The proof follows from a reduction to Rolle's Theorem. The statement I'm thinking of is: Let $f:[a,b]\to\mathbb{R}$ be a continuous function. Suppose that the right and left derivatives, $f_{+}'(x),f_{-}'(x)$ exist everywhere on (a,b). Then there is $c\in(a,b)$ so that the following inequality holds: \begin{equation*} \min\left\{f_{-}'(c),f_{+}'(c)\right\}\le\frac{f(b)-f(a)}{b-a}\le\max\left\{f_{-}'(c),f_{+}'(c)\right\}. \end{equation*} $\endgroup$ – user71352 Mar 9 '17 at 17:07
2
$\begingroup$

Not quite an answer:

For locally Lipschitz functions $f$ there is a notion of generalised gradient $\partial f$ that you can view as a generalisation of the subdifferential in convex analysis. The (generalised) mean value theorem then states that there is some $ t \in (a,b)$ such that ${f(b)-f(a) \over b-a } \in \partial f (a + t(b-a))$.

This is not as strong as the corresponding differentiable result.

As an example, take $f(x) = |x|$, $a=-1, b=1$, then $\partial f(x) = \begin{cases} \{-1\}, & x <0 \\ [-1,1], & x = 0 \\ \{1\}, & x > 0 \end{cases}$, so we see that ${|1|-|-1| \over 1 - (-1) } = 0 \in \partial f(0) = [-1,1]$.

$\endgroup$
0
$\begingroup$

By the mean value version of Taylor's theorem we have: \begin{align} f(y) &=f(x)+f'(x)(y-x)+\dfrac{1}{2}f''(z)(y-x)^2, \text{for some }z\in[x,y].\\ \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.