In relation to the problem:

Let $A$ be a subset of $\mathbb{R}^k$ such that $A$ is both an open subset, is convex, and that a function $h \in C^1(A)$.

Now show that for any choice of $(x,y) \in A^2$, there exists a value $\lambda \in (0,1)$ satisfying the equality: $$h(y)-h(x) = \big{(} \nabla h((1-\lambda)x + \lambda y ) \big{)} \cdot \big{(}y-x \big{)}$$

I know that the convexity of $A$ implies that for any choice of $(x,y)$ the line joining $x$ to $y$ is contained in $A$. Here we define this line to be:

$$L := \{ \lambda x + (1-\lambda )y \space \space | \space \lambda \in (0,1) \} $$

I also know the Mean Value Inequality which states that for a convex set $B$ satisfying $\lvert \lvert \partial h(x) \rvert \rvert \le W $ for all $x$, then we conclude the following inequality:

$$\lvert h(x) - h(y) \rvert \le W \lvert x - y \rvert $$

As both of these results are relevant to convex sets, it feel as though they are likely to be involved in some way (to make use of the convexity assumption), however, it is unclear to me, how this helps us with this particular problem and would be grateful for any guidance on how to proceed.

  • $\begingroup$ For your theorem I think the only thing that is relevant is that the line segment between $x$ and $y$ is contained in $A$. $\endgroup$
    – Mason
    Commented Nov 15, 2022 at 6:07

1 Answer 1


Let $g(t):=tx+(1-t)y$ for $t \in [0,1]$. Since $A$ is covex, we have $g([0,1] \subseteq A.$

Then consider the function $f(t):=h(g(t))$ for $t \in [0,1]$.Then we have $f(0)=h(y)$ and $f(1)=h(x).$

Now apply the Mean value Theorem to $f$.

  • $\begingroup$ For vector-valued function, the mean value theorem is in general not valid. Consider the function $f(t)= ( \cos t, \sin t)$ for $t \in [0, 2 \pi].$ We have $f(0)=f(2 \pi).$ But there is no $t_0 \in [0, 2 \pi]$ such that $f'(t_0)=0,$ since this means that $ \sin t_0= \cos t_0 =0,$ which is impossible. $\endgroup$
    – Fred
    Commented Nov 16, 2022 at 9:48
  • $\begingroup$ Suppose that for the function $f$ in my first comment, the mean value theorem holds. Then there would be $t_0 \in [0,2 \pi]$ with $(0,0)=f(2 \pi)-f(0)= 2 \pi f'(t_0) .$ Hence $(0,0)= f'(t_0)= (-\sin t_0, \cos t_0),$ thus $ \sin t_0= \cos t_0=0$, which contadicts $ \cos^2 t_0+ \sin^2 t_0=1,$ $\endgroup$
    – Fred
    Commented Nov 16, 2022 at 9:57
  • $\begingroup$ No, $f$ is real valued. $\endgroup$
    – Fred
    Commented Nov 16, 2022 at 15:53

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