something like MVT

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something like MVT

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71 posts

#1 h Mar 29, 2025, 11:37 AM

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If $F$ is a continuous function on $[0,1]$ such that $F(0) = F(1)$ , then there exists a $c \in (0,1)$ such that:

$F(c) = \frac{1}{c} \int_0^c F(x) \,dx$

This post has been edited 2 times. Last edited by mqoi_KOLA, Mar 29, 2025, 11:39 AM

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#2 h Mar 29, 2025, 3:48 PM

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withdrawn

This post has been edited 3 times. Last edited by Mathzeus1024, Mar 30, 2025, 10:21 AM

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#3 h Mar 29, 2025, 4:50 PM

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Hopefully not fakesolve
Apply integral MVT (justified as $F$ is continuous)

$F(c) = \frac{1}{c} \int_0^c F(x) \,dx = F(d)$
So we need to show that there exists $0 < d < c < 1$ such that $F(c) = F(d)$

Let $g(x) = F(x+0.5) - F(x)$ for $x \in [0, 0.5]$
$g(0) = F(0.5) - F(0)$
$g(0.5) = F(1) - F(0.5) = -g(0)$

Applying IVT on $g$ ,
$g(0)g(0.5) < 0$ , so there exists a $\eta \in (0, 0.5)$ such that $g(\eta) = 0$ , ie $F(\eta + 0.5) = F(\eta)$

Call $\eta + 0.5 = c$ , $\eta = d$ and we are done.

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#4 h Mar 29, 2025, 6:22 PM

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Mathzeus1024 wrote:

If $F:[0,1] \rightarrow \mathbb{R}$ is a continuous function such that $F(0)=F(1)$ , then by Rolle's Theorem $\exists c \in (0,1)$ such that $F'(c)=0$ . Suppose that $F'(x) = A(x-c) = 0 \Rightarrow F(x) = \frac{A}{2}(x-c)^2 + B$ for some $A,B \in \mathbb{R}$ . If $F(0)=F(1)$ , then:

$\frac{A}{2}(0-c)^2 + B = \frac{A}{2}(1-c)^2+B \Rightarrow c^2=(1-c)^2 \Rightarrow c = \frac{1}{2}$ .

If $F(c) = \frac{1}{c}\int_{0}^{c} F(x) dx$ holds, then we obtain:

$F(1/2) = 2\int_{0}^{1/2} F(x) dx \Rightarrow B = 2\left[\frac{A}{6}(x-1/2)^3+Bx\right]_{0}^{1/2} \Rightarrow B = B +\frac{A}{24} \Rightarrow A=0$ ;

or $F(x) = B \Rightarrow \exists c \in (0,1)$ such that $F(c)=\frac{1}{c}\int_{0}^{c}F(x) dx$ for constant function $F$ .

sorry but i dont think your proof is on right lines.

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#5 h Mar 29, 2025, 6:33 PM

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Mathzeus1024 wrote:

If $F:[0,1] \rightarrow \mathbb{R}$ is a continuous function such that $F(0)=F(1)$ , then by Rolle's Theorem $\exists c \in (0,1)$ such that $F'(c)=0$ . Suppose that $F'(x) = A(x-c) = 0 \Rightarrow F(x) = \frac{A}{2}(x-c)^2 + B$ for some $A,B \in \mathbb{R}$ . If $F(0)=F(1)$ , then:

$\frac{A}{2}(0-c)^2 + B = \frac{A}{2}(1-c)^2+B \Rightarrow c^2=(1-c)^2 \Rightarrow c = \frac{1}{2}$ .

If $F(c) = \frac{1}{c}\int_{0}^{c} F(x) dx$ holds, then we obtain:

$F(1/2) = 2\int_{0}^{1/2} F(x) dx \Rightarrow B = 2\left[\frac{A}{6}(x-1/2)^3+Bx\right]_{0}^{1/2} \Rightarrow B = B +\frac{A}{24} \Rightarrow A=0$ ;

or $F(x) = B \Rightarrow \exists c \in (0,1)$ such that $F(c)=\frac{1}{c}\int_{0}^{c}F(x) dx$ for constant function $F$ .

i think you understood the question wrong we dont have to find such function which obey the condition(there are other functions too),

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#6 h Mar 29, 2025, 6:50 PM

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quasar_lord wrote:

Hopefully not fakesolve
Apply integral MVT (justified as $F$ is continuous)

$F(c) = \frac{1}{c} \int_0^c F(x) \,dx = F(d)$
So we need to show that there exists $0 < d < c < 1$ such that $F(c) = F(d)$
Let $g(x) = F(x+0.5) - F(x)$ for $x \in [0, 0.5]$
$g(0) = F(0.5) - F(0)$
$g(0.5) = F(1) - F(0.5) = -g(0)$

Applying IVT on $g$ ,
$g(0)g(0.5) < 0$ , so there exists a $\eta \in (0, 0.5)$ such that $g(\eta) = 0$ , ie $F(\eta + 0.5) = F(\eta)$

Call $\eta + 0.5 = c$ , $\eta = d$ and we are done.

its a fakesolve.

This post has been edited 1 time. Last edited by mqoi_KOLA, Mar 29, 2025, 11:41 PM

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#7 h Mar 29, 2025, 7:27 PM

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@Mathzeus1024: You keep making random assumptions in most of your posts. For example here you assume that $F$ is differentiable, then you assume that $F'(x)=A(x-c),$ and worst of all, you assume exactly what you are supposed to prove. Please reconsider the way you are contributing, because this type of posts doesn't help anyone, but actually creates confusions.

@quasar_lord: Unfortunately, once you apply MVT, that $c$ that you get becomes fixed, so you cannot choose it to be $\eta + 0.5.$

This post has been edited 2 times. Last edited by Filipjack, Mar 29, 2025, 8:33 PM

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#8 h Mar 29, 2025, 8:33 PM • 1 Y

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According to Weierstrass Theorem, $F$ has minimum and maximum, so let $a= \sup \{ s : F(s)= \min_{x \in [0,1]} F(x) \}$ and $b= \sup \{ s : F(s)= \max_{x \in [0,1]} F(x) \}.$ By continuity, we get $F(a)=\min_{x \in [0,1]} F(x)$ and $F(b)=\max_{x \in [0,1]} F(x).$

Notice that if $a=0,$ then $F(0)= \min_{x \in [0,1]} F(x),$ so $F(1)= \min_{x \in [0,1]} F(x),$ contradicting the definition of $a.$ Therefore $a>0,$ and similarly we get $b>0.$

Consider the function $G:(0,1] \to \mathbb{R},$ $G(x)=F(x)- \frac{1}{x} \int\limits_0^x F(t) \mathrm{d}t.$ We have $G(a) \le 0.$ If $G(a)=0,$ then $\int\limits_0^a (F(t)-F(a)) \mathrm{d}t=0,$ which combined with $F(t)-F(a) \ge 0$ and the continuity of $F(t)-F(a)$ implies $F(t)=F(a), \forall t \in [0,a].$ In this case we can choose any $c \in (0,a),$ for example $c=a/2.$

Likewise, $G(b) \ge 0,$ and if $G(b)=0,$ then $F(t)=F(b), \forall t \in [0,b],$ and we can choose $c=b/2.$

Finally, if $G(a)<0$ and $G(b)>0,$ then by IVT there is $c \in (a,b)$ such that $G(c)=0.$

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#9 h Mar 30, 2025, 4:20 AM

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@Filipjack Nice idea!

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