# Show that there exists $x_0\in(a,b)$ such that $f(x_0)=\frac{1}{n}(f(x_1)+f(x_2)+\cdots+f(x_n)).$

Question: Suppose that $$f:[a,b]\to\mathbb{R}$$ is continuous. Let $$x_1,x_2,\cdots, x_n$$ be any $$n$$ points in $$(a,b).$$ Show that there exists $$x_0\in(a,b)$$ such that $$f(x_0)=\frac{1}{n}(f(x_1)+f(x_2)+\cdots+f(x_n)).$$

Solution: Let $$g:[a,b]\to\mathbb{R}$$ be such that $$g(x)=nf(x)-\sum_{k=1}^nf(x_k), \forall x\in[a,b].$$ Observe that to prove the statement of the problem it is enough to show that $$g(x_0)=0$$ for some $$x_0\in(a,b)$$.

Now note that by the 3rd form of the Pigeon Hole principle we can conclude that there exists $$1\le i,j\le n$$ such that $$f(x_i)\le \frac{1}{n}\sum_{k=1}^nf(x_k)\le f(x_j)\\\implies nf(x_i)\le \sum_{k=1}^nf(x_k)\le nf(x_j).$$ Thus, $$g(x_i)=nf(x_i)-\sum_{k=1}^nf(x_k)\le 0$$ and $$g(x_j)=nf(x_j)-\sum_{k=1}^nf(x_k)\ge 0.$$ Now if $$g(x_i)=0$$ or $$g(x_j)=0$$, then we are done. Thus, let us assume that $$g(x_i)<0$$ and $$g(x_j)>0$$. Now since $$f$$ is continuous on $$[a,b]$$, implies that $$g$$ is continuous on $$[a,b]$$. Therefore, by IVT we can conclude that there exists $$x_0\in(x_i,x_j)$$ or $$x_0\in(x_j,x_i)$$ such that $$g(x_0)=0$$. This completes the proof.

Is this solution correct and rigorous enough and is there any other way to solve the problem?

• What is the PHP? Aug 26, 2020 at 19:22
• @MartinR, it's the "Pigeon Hole Principle". Sorry, I will expand that term in a bit. Aug 26, 2020 at 19:24

Your proof looks fine to me. There is no need however to introduce the function $$g$$. You know that $$f(x_i)\le \frac{1}{n}\sum_{k=1}^nf(x_k)\le f(x_j)$$ for some indices $$i, j$$, so you can just apply the intermediate value theorem to $$f$$ on the interval $$I = [\min(x_i, x_j), \max(x_i, x_j)]$$ and conclude that $$\frac{1}{n}\sum_{k=1}^nf(x_k) = f(x)$$ for some $$x \in I$$.

Instead of using the pigeon hole principle you can also apply the mean value theorem to $$f$$ on the interval $$J= [\min_k x_k, \max_k x_k] \subset (a, b)$$ because $$m\le \frac{1}{n}\sum_{k=1}^nf(x_k)\le M$$ with $$m = \min_J f(x)$$ and $$M = \max_J f(x)$$.

Given a continuous $$f(x)$$, an iterated application of the Intermediate value Theorem gives \eqalign{ & \exists x_{1,2} \in \left[ {x_1 ,x_2 } \right]:f(x_{1,2} ) = t\;f(x_1 ) + \left( {1 - t} \right)f(x_2 )\quad \left| {\,0 \le t \le 1} \right. \cr & \exists x_{2,3} \in \left[ {x_2 ,x_3 } \right]:f(x_{2,3} ) = u\;f(x_2 ) + \left( {1 - u} \right)f(x_3 )\quad \left| {\,0 \le u \le 1} \right. \cr} which express the possibility of finding a point corresponding to the weighted mean within each interval.

Putting $$t=2/3, \, u=1/3$$, we can write \eqalign{ & \exists x_{1,2} \in \left[ {x_1 ,x_2 } \right]:f(x_{1,2} ) = {2 \over 3}\;f(x_1 ) + {1 \over 3}f(x_2 )\quad \left| {\,0 \le t \le 1} \right. \cr & \exists x_{2,3} \in \left[ {x_2 ,x_3 } \right]:f(x_{2,3} ) = {1 \over 3}\;f(x_2 ) + {2 \over 3}f(x_3 )\quad \left| {\,0 \le u \le 1} \right. \cr & \exists x_{1,3} \in \left[ {x_1 ,x_2 } \right] \cup \left[ {x_2 ,x_3 } \right]:f(x_{1,3} ) = {1 \over 2}\,f(x_{1,2} ) + {1 \over 2}f(x_{2,3} ) = \cr & = {{f(x_1 ) + f(x_2 ) + f(x_3 )} \over 3} \cr} and the extension to n points is clear.

Pick $$i$$ with \begin{align} f(x_i) &\le f(x_k) & \text{for all k = 1, \ldots, n.} \tag{1} \end{align}
Pick $$j$$ with \begin{align} f(x_j) &\ge f(x_k) & \text{for all k = 1, \ldots, n.} \tag{2} \end{align}
If $$i = j$$, then all $$x_k$$ are equal, and $$x_0 = x_i$$ solves the problem.
Consider the case $$i < j$$; the $$i > j$$ case is almost identical. But equation $$1$$, we have $$n f(x_i) \le \sum_k f(x_k)$$ By equation 2, similarly $$n f(x_j) \ge \sum_k f(x_k)$$.
Then by the Intermediate value theorem, there's an $$x_0 \in [x_i, x_j]$$ such that $$f(x_0) = \frac{1}{n} \sum_k f(x_k).$$