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This is just a question that popped up in my head while going through basic real analysis. The ordinary Mean Value Theorem (MVT) is given as follows.

Let $f:[a,b] \to \mathbb{R}$ be a function satisfying the following conditions: $(i)$ $f$ is continuous on $[a,b]$ and $(ii)$ $f$ is differentiable on $(a,b)$. Then $\exists c \in (a,b),$ such that $$f(b)-f(a)=f'(c) \cdot (b-a)$$

Of course, we can restrict $f$ to any interval $[x,y] \subset [a,b]$ and apply the theorem on $f|_{[c,d]}$ to state that $\exists z \in (x,y)$ such that $f(y)-f(x)=f'(z) \cdot (y-x)$

Can I formulate a converse proposition as follows?

Let $f:[a,b] \to \mathbb{R}$ be a function satisfying the following conditions: $(i)$ $f$ is continuous on $[a,b]$ and $(ii)$ $f$ is differentiable on $(a,b)$. Then $\forall c \in (a,b), \exists x,y \in [a,b]$ such that $$f(y)-f(x)=f'(c) \cdot (y-x)$$

Does it hold?

If yes, how do I prove it? (How does one find out the points $x$ and $y$, which need not be unique, that work for a given point $c?)$ Does it hold under weaker assumptions on the function?

If no, can you provide me a counter example? Can I impose stronger conditions to make it work?

Any help would be much appreciated. Thank you.

EDIT : As pointed out by @Henrik, the proposition does not hold with it's current assumptions. On the other hand, @MANMAID claims that it does hold with the additional assumption that $f$ has no point of inflection. It seems very interesting. I require a formal proof, though.

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    Okk, this is not that bad proposition. You just have to add $f$ has no point of inflection. The answer given below actually shows that if $f$ has point of inflection, you are wracked, otherwise ok. – MAN-MADE Jun 28 '17 at 14:54
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    In fact more is true. If $c$ is not a point of inflection then we can find the points $x, y$ as desired. The idea is that we can draw a line parallel to the tangent at $(c, f(c)) $ which is at a small distance from the tangent and this will intersect the curve on two points $x, y$. This fails if the point $c$ is a point of inflection. You need to use the fact that the portion of the curve near $c$ lies either wholly below the tangent or wholly above the tangent. – Paramanand Singh Jun 28 '17 at 20:19
  • You should see an answer to a related question: https://math.stackexchange.com/a/2274171/72031 Here I discuss ways to find the interersection of a line with the curve which will be helpful here. The situation in that question is not exactly the same but you can utilize a similar method here. – Paramanand Singh Jun 28 '17 at 20:42
  • Btw +1 for a good question. – Paramanand Singh Jun 28 '17 at 20:43
  • @ParamanandSingh Thanks for the idea. And for the link too. It's really helpful. –  Jun 28 '17 at 20:46

2 Answers2

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I don't think so.

Consider $f(x)=x^3$ on $[-1,1]$. It's continuous and differentiable, but for $c=0\in ]-1,1[$, $f'(c)=0$ so the $x$ and $y$ you want would have to satisfy $$ y^3-x^3=0 $$ which implies $x=y$, but since you haven't required that $x\neq y$, this isn't technically a counterexample.

Henrik supports the community
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    In fact, $x=y$ always trivially satisfies the requirements, so any meaningful version of this theorem would exclude that possibility. Hence your answer indeed shows it isn't true. – Jason Jun 28 '17 at 14:46
  • You may think of the problem now after the edit. – MAN-MADE Jun 28 '17 at 15:10
  • @Jason is it possible for the proposition to hold with the added assumption that $f$ has no point of inflation? –  Jun 28 '17 at 15:15
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In fact, your result is almost true. To see it, Put first $C=\{(x,y); a\leq x<y\leq b\}$. Then $C$ is the interior of a triangle (with some side accepted). Let for $(x,y)\in C$, $g(x,y)=\frac{f(y)-f(x)}{y-x}$. Then clearly $g$ is continuous from $C$ to $\mathbb{R}$. As $C$ is connected, $I=g(C)$ is a connected subset of $\mathbb{R}$, hence an interval. Now the Mean value theorem say that the image $J$ of $]a,b[$ by $f^{\prime}$ contains $I$; it is well known that a derivative has the intermediate value property, so $J$ is an interval too. Now each $f^{\prime}(x)$ is the limit as $n\to +\infty$ of $g(x,x+1/n)$; hence $J\subset \overline{I}$. As you can see, as $I\subset J \subset \overline{I}$, $I$ and $J$ differ by at most two points.

Kelenner
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  • Very Nice(+1) : one question , Does it show that $c$ is between $x$ and$y$. – Red shoes Jun 28 '17 at 17:09
  • @Kelenner Thanks for the answer. Is it possible to remove the word almost with the additional condition that $f$ does not have any point of inflection (as proposed by @MANMAID)? –  Jun 28 '17 at 18:52
  • can you explain why $g$ is continuous? Take $f(x)=x^3$, $a=-1,b=2$. Is in this case $g(C)$ connected? – MAN-MADE Jun 29 '17 at 03:16
  • @manmaid Le two functions $(x,y)\to f(y)-f(x)$ and $x,y)\to y-x$ are continuous, and on $C$, $y-x\not =0$, so their quotient is continuous. In the case $f(x)=x^3$, $g(x,y)=x²+xy+y^2$. – Kelenner Jun 29 '17 at 12:52
  • @Kelenner this looks fine to me(I totally got confused and mixed up nature of $f$ and $g$.). Now $C$ do not contain $(x,x)$. $\phi :g(C)\rightarrow f'(a,b)$ is not surjective. – MAN-MADE Jun 29 '17 at 13:36