Prove there exists $\xi \in (a,b)$ such that $f(\xi)=f^{(n+1)}(\xi)$.

Solution 1:

Failed attempt:

Consider the function $$g(x)=(f(x)+f'(x)+\ldots+f^{(n)}(x))e^{-x} $$ that is continuous(!) over $[a,b]$ and differentiable over $(a,b)$ with derivative $$\begin{align}g'(x)&=(f'(x)+f''(x)+\ldots+f^{(n+1)}(x))e^{-x}-(f(x)+f'(x)+\ldots+f^{(n)}(x))e^{-x}\\&=(f^{n+1}(x)-f(x))e^{-x}.\end{align}$$ As $g(a)=g(b)=0$, the Rolle theorem tells us that here exists $\xi\in(a,b)$ with $g'(\xi)=0$.

Correction:

Unfortunately, $g$ need not be continuous at $a$ or $b$, so the Rolle theorem cannot be applied. $g$ will only be continuous (and differentiable) over $(a,b)$. In other words, the Rolle theorem holds under somewhat weaker assumptions, but it is usually not presented that way - for the simple reason that Rolle is one of the first theorems about derivatives one encounters. Here's how to mend the mistake above:

Assume $g$ is injective over $(a,b)$. Then $g$ must be strictly monotonic over $(a,b)$ (and in particular not constant). Pick $c\in(a,b)$ with $g(c)\ne 0$, wlog. $g(c)>0$ (or else replace $f$ with $-f$). Then either $g(x)>g(c)$ for all $x\in(a,c)$ or for all $x\in (c,b)$; wlog the former (or else replace $f$ with $x\mapsto f(a+b-x)$). As $e^x$ is bounded over $[a,c]$, it follows that there exist $\delta>0,\epsilon>0$ such that $$\tag1h(x):=e^xg(x)>\epsilon$$ for all $a<x<a+\delta$. But $ h(x)=f(x)+f'(x)+\ldots+f^{(n)}(x)$ is the derivative of $x\mapsto \int_a^xf(t)\,\mathrm dt+f(x)+\ldots +f^{(n-1)}(x)$ and hence has the Darboux (aka. Inermediate Value) property. In particular, as $h(a)=0$, there must exist $x\in(a,a+\delta)$ with $h(x)=\frac12\epsilon$, contradicting $(1)$.

We conclude that $g$ cannot be injective over $(a,b)$. Hence, there are $u,v$ with $a<u<v<b$ and $g(u)=g(v)$. By Rolle, there exixts $\xi\in(u,v)$ with $g(\xi)=0$. This implies that $f^{(n+1)}(\xi)=f(\xi)$.

Solution 2:

Proof

At first, let's introduce and prove a lemma, which could be stated as follows.

Lemma Let $f(x)$ be differentiable over $[a,b]$, and $f'(x)$ be continuous over $(a,b)$. $f'(a)=f'(b)=0$. Then there exist $x_1,x_2$ satisfying $a<x_2<x_1<b$ such that $f'(x_1)=f'(x_2),$ namely, $f'(x)$ can not be injective over $(a,b)$.

Consider proving by contradiction. Assume the conclusion does not hold. Then $f'(x)$ is injective over $(a,b)$. Taking the fact $f(x)$ is continuous over $(a,b)$ into account, we may obtain $f'(x)$ is strictly monotonic over $(a,b)$. Without loss of generality, we assume $f'(x)$ is strictly increasing over $(a,b)$. Obviously, there exsits $c \in (a,b)$ such that $f'(c) \neq 0$. Without loss of generality, we assume $f'(c)>0$ and consider the interval $(c,b)$. (If $f'(c)<0$, we may consider the interval $(a,c)$. The reasoning is similar.) Notice that $f'(c)>0$ and $f'(b)=0$. According to Darboux's theorem, there exists $\xi \in (c,b)$ such that $f'(\xi)=\dfrac{1}{2}f'(c)$. But, since $f'(x)$ is strictly increasing over $(c,b)$, we may have $\dfrac{1}{2}f'(c)=f'(\xi)>f'(c)>0$, which contradicts.

Now, we can go back to deal with the target problem. We must restrain that $n \geq 1$. Otherwise, the statement need not hold, and I have given a conterexample before when $n=0$. Denote$$F(x):=(n+1)\int_a^x e^{-t}f(t){\rm d}t+\sum_{k=0}^{n-1}(n-k)e^{-x}f^{(k)}(x), x\in [a,b].$$Then $F(x)$ is differentiable over $[a,b]$, and$$F'(x)=e^{-x}\sum_{k=0}^{n}f^{(k)}(x).$$It's easy to verify that $F'(x)$ is continuous over $(a,b)$ and $F'(a)=F'(b)=0$. Thus，by Lemma，$$\exists x_1,x_2(a<x_1<x_2<b):F'(x_1)=F'(x_2).$$Further， by Rolle's theorem，$$\exists \xi \in (x_1,x_2)\subset (a,b):F''(\xi)=e^{-\xi}[f^{(n+1)}(\xi)-f(\xi)]=0,$$which implies$$f(\xi)=f^{(n+1)}(\xi).$$