I tried to solve this problem but I could not continue

Let $f: [a, b​​]\to \mathbb{R}$ be a continuous function. Prove that if $x_1, x_2,\ldots,x_n \in (a,b​)$, then exists $x_0 \in (a, b​​)$ such that $$f (x_0) =\frac{f (x_1) + f (x_2) +\cdots + f (x_n)}{n}$$

the function is continuous, so $\forall\ x_1, x_2,\ldots,x_n \in(a, b​​)$ $$\begin{align*} \lim_{x \to \ x_1}f(x)&=f (x_1)=L_1\\ \lim_{x \to \ x_2}f(x)&=f (x_2)=L_2\\ \lim_{x \to \ x_3}f(x)&=f (x_3)=L_3\\ &\vdots\\ \lim_{x \to \ x_n}f(x)&=f (x_n)=L_n\\ \end{align*}$$

We have that:

$$\lim_{x \to \ x_0}f(x)=f (x_0)=\frac{f (x_1) + f (x_2) +\cdots + f (x_n)}{n}= \frac{1}{n}\sum_{j=1}^n f(x_j)$$ Now $$\min\left \{ L_1,L_2,L_3,\cdots,L_n \right \}\le\frac{1}{n}\sum_{j=1}^n f(x_j)\le\max\left \{ L_1,L_2,L_3,\cdots,L_n \right \}$$

and at this point I'm lost ...

  • 4
    $\begingroup$ Just prove that $\frac1n\sum f(x_j)\in \left[\min\limits_{x\in[a,b]}f(x),\max\limits_{x\in[a,b]}f(x)\right]$ and use the Intermediate Value Theorem. $\endgroup$ – Ilya Dec 6 '11 at 12:18
  • $\begingroup$ ok ... but how?! $\endgroup$ – FrConnection Dec 6 '11 at 19:23

Let $f: [a, b​​]\to \mathbb{R}$ be a continuous function. Prove that if $x_1, x_2,\ldots,x_n \in (a,b​)$, then exists $x_0 \in (a, b​​)$ such that $$f (x_0) =\frac{f (x_1) + f (x_2) +\cdots + f (x_n)}{n}$$

Well, First we'll relable $x_1, ..., x_n$ so that $x_1 \leq x_2 \leq ... \leq x_n$ then the first important thing to notice is that $[x_1,x_n] \subset \mathbb{R}$ is closed and bounded, so we know that it's compact. The second thing is the important fact about continuous functions, they map compact sets to compact sets.

So, we know that $f([x_1,x_n])$ is also compact, so it is closed and bounded. Bounded: since we're in $\mathbb{R}$ we know that $\sup(f([x_1,x_n])$ and $\inf(f([x_1,x_n]))$ exist. Closed: Since the set is closed we now know that $\sup(f([x_1,x_n])), \inf(f([x_1,x_n]))\in f([x_1,x_n])$.

This tells us that we have $x, y\in [x_1,x_n]$ such that $f(x) = \sup(f([x_1,x_n]))$ and $f(y) = \inf(f([x_1,x_n]))$.

From here, we can see how to solve the problem, since $$A =\frac{f (x_1) + f (x_2) +\cdots + f (x_n)}{n} \leq \frac{f(x) + f(x) + \cdots f(x)}{n} = f(x)$$ and $$A =\frac{f (x_1) + f (x_2) +\cdots + f (x_n)}{n} \geq \frac{f(y) + f(y) +\cdots f(y)}{n} = f(y)$$ Since $x_1, ..., x_n \in [x_1,x_n]$

So now we have $y,x\in [x_1,x_n]$ such that $f(y) < A < f(x)$. (Note: if $A = f(x)$ or $A = f(y)$ then we're done since $x,y \in (a,b)$ by assumption, so we're supposing they aren't equal)

Moreover WLOG we'll assume $x < y$, and so $f$ is continuous on $[x,y]$ (since $[x,y] \subset [x_1,x_n]$)

Thus we use IVT, which tells us that that we have $x_0\in (x,y)\subset (a,b)$ for which $f(x_0) = A$.


Even though this question has been answered informally, I'll put this up so it doesn't appear unanswered. Let $S:=\{x_1,x_2,.....,x_n\}$. Then consider

$$\max_{x \in S} \, \,f(x) = f(x_i)$$

for some $1 \leq i \leq n$. Now it is clear that

$$\frac{f(x_1)+f(x_2)+.......+f(x_n)}{n} \leq \frac{n\cdot f(x_i)}{n} = f(x_i). $$

Similarly we can take

$$\min_{x\in S} f(x) = f(x_j)$$

for some $1\leq j \leq n$ and have

$$f(x_j) \leq \frac{f(x_1)+f(x_2)+.......+f(x_n)}{n}.$$

Hence by taking

$$f(x_0):= \frac{f(x_1)+f(x_2)+.......+f(x_n)}{n} $$

we have $f(x_0)\in [f(x_j),f(x_i)]\subset (a,b)$.

Therefore, as $f$ is continuous on $(a,b)$ it is continuous on $[f(x_j),f(x_i)]$ and the intermediate value theorem guarantees the existence of $x_0$ within this interval.


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