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Question: Let $f:[0,1]\in\mathbb{R}$ be a continuous function satisfying $\int_0^1f(x)dx=0$ and $f(0)f(1)>0$. Assume further that $f'$ is continuous on $(0,1)$. Show that $\exists c\in (0,1)$ such that $$e^{f'(c)}=f(c)+1.$$

My solution: It is given that $f(0)f(1)>0\implies f(0)>0,f(1)>0$ or $f(0)<0, f(1)<0$. Let us assume WLOG, that $f(0)>0,f(1)>0$.

Now since $\int_0^1f(x)dx=0$ and $f(0)>0,f(1)>0$, thus we must have a point $a\in (0,1)$ such that $f(a)<0$. Now since $f$ is continuous on $[0,1]$, therefore by IVT we can conclude that $\exists c_1\in (0,a)$, such that $f(c_1)=0$ and $\exists c_2\in(a,1),$ such that $f(c_2)=0$. From this we can conclude, there exists at least two distinct roots of $f$ in $(0,1)$. Therefore we are sure to have a point $b,0<b<1$ such that $f(x)>0, \forall x\in (0,b)$ and $f(b)=0$. Also we are sure to have a point $a,0<a<e<1$, such that $f(x)<0, \forall x\in(a,e)$ and $f(e)=0$.

Now consider the interval $[0,b]$. We have $f(0)>0$ and $f(b)=0$. Thus by MVT, we can conclude that $\exists c_3\in (0,b)$, such that $$f'(c_3)=\frac{f(b)-f(0)}{b-0}=-\frac{f(0)}{b}<0.$$ Now since $c_3\in (0,b)\implies f(c_3)>0.$

Again consider the interval $[a,e]$. We have $f(a)<0$ and $f(e)=0$. Thus by MVT, we can conclude that $\exists c_4\in (a,e)$, such that $$f'(c_4)=\frac{f(e)-f(a)}{e-a}=-\frac{f(a)}{e-a}>0.$$ Now since $c_4\in (a,e)\implies f(c_4)<0.$

Now let $g(x):=e^{f'(x)}-f(x)-1, \forall x\in(0,1).$ Observe that $g$ is continuous on $(0,1)$.

We have $g(c_3)=e^{f'(c_3)}-f(c_3)-1.$

Now $f'(c_3)<0\implies e^{f'(c_3)}<1$ and $f(c_3)>0\implies f(c_3)+1>1\implies -f(c_3)-1<-1.$

Thus $e^{f'(c_3)}-f(c_3)-1<0\implies g(c_3)<0.$

A similar reasoning shows that $g(c_4)>0$.

Thus by IVT, we can conclude that $\exists c\in(c_3,c_4)\subseteq(0,1)$, such that $g(c)=0\implies e^{f'(c)}=f(c)+1.$

Does this solution works fine? And is there a more better solution?

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  • $\begingroup$ Looks fine. You seem to have an impressive supply of questions based on standard theorems of differential calculus. +1 $\endgroup$
    – Paramanand Singh
    Apr 16 '20 at 8:09
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WLOG, we suppose $f(0)>0,f(1)>0$. Define $$F(x):=e^{f'(x)}-f(x)-1, \forall x\in(0,1).$$ On the one hand, $\int_0^1f(x)dx=0$ implies $f$ can attain its Minimum value at some point $x_0\in(0,1)$, such that $f(x_0)<0,$ and $f'(x_0)=0$(Fermat's lemma). So $$F(x_0)=-f(x_0)>0.$$

On the other hand, $f(0)>0,f(x_0)<0$ implies that: the zeros set $$\{x\mid x\in[0, x_0], f(x) =0\}\ne \emptyset.$$ (by intermediate value theorem)

Thanks @ Paramanand Singh for pointing out that: the zeros set $$\{x\mid x\in[0, x_0], f(x) =0\}$$ has minimum and maximum element no matter it is finite set and infinite set.(Proof can be found Intermediate value theorem and supremum)

Let $$x_1=\max\{x\mid x\in[0, x_0], f(x) =0\},$$ and obviously $x_1\in(0, x_0)$. So $f(x_1)=0$ and $f(x)<0$ for $x\in(x_1,x_0)$ . Consider the derivative $f'(x_1)$, we know that $$f'(x_1)=\lim_{x\to x_1^+}\frac{f(x)-f(x_1)}{x-x_1} =\lim_{x\to x_1^+}\frac{f(x)}{x-x_1}\leq 0.$$

If $f'(x_1)=0$, take $c=x_1$, we can get $e^{f'(c)}=f(c)+1$.

If $f'(x_1)<0$, then $F(x_1)=e^{f'(x_1)}-1<0$, by intermediate value theorem, we can conclude that $\exists\ c\in(x_1,x_0)\subset(0,1)$, such that $$F(c)=0\iff e^{f'(c)}=f(c)+1.$$

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  • $\begingroup$ Why did you treat the case of infinitely many zeroes separately? If $f$ is continuous on $[a, b] $ with $f(a) f(b) <0$ then the set $A=\{x\mid x\in[a, b], f(x) =0\}$ is non-empty and possesses a least as well as a greatest member. Some proofs of IVT show this without mentioning it explicitly. $\endgroup$
    – Paramanand Singh
    Apr 19 '20 at 4:34
  • $\begingroup$ It is possibile there are infinite zeros in $[a,b]$, and the following proof is different when finite zero between infinite zeros! IVT only give the existence of zero, for this question, only existence is not enlugh for the proof! and $\max\{x\mid x\in[a, b], f(x) =0\}$ and $\min\{x\mid x\in[a, b], f(x) =0\}$ may not exist! @Paramanand Singh $\endgroup$
    – Riemann
    Apr 19 '20 at 6:20
  • $\begingroup$ well the truth is that those max and min always exist. some of the proofs of IVT show the existence of these min and max $\endgroup$
    – Paramanand Singh
    Apr 19 '20 at 6:24
  • $\begingroup$ Maybe you are right! If likes you say, the proof is easy and no need to discuss the infinite zers case!! But in general, the textbook only gives the existence of zero! And can you give me some references?? $\endgroup$
    – Riemann
    Apr 19 '20 at 6:29
  • $\begingroup$ there is no need to prove the existence of max and min. rather you can just state that there is a biggest zero $x_1$ in interval $(0,x_0)$. thus your case 1 actually is sufficient. I will give a +1 anyway. $\endgroup$
    – Paramanand Singh
    Apr 19 '20 at 6:30
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If such $c$ doesn't exist, either always $e^{f'}>f+1$ or always $e^{f'}<f+1$. Whenever $f=0$, the former means $f'>0$ always and the latter means $f'<0$ always. However $f$ must change sign at least twice, and this is impossible.

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  • $\begingroup$ Why $f$ changes sign at least twice is impossible? Give you an example $f(x)=\cos(2\pi x),x\in[0,1],$ this function changes sign twice!! $\endgroup$
    – Riemann
    Apr 19 '20 at 6:15
  • $\begingroup$ @Riemann: $f$ has same signs at $0,1$ and integral of $f$ is $0$ so it has a sign different from $f(0)$ somewhere in $(0,1)$. Thus $f$ changes sign at least twice. If $a, b$ are first and last zero of $f$ then $f'$ has same sign at $a, b$ and this is not possible. $\endgroup$
    – Paramanand Singh
    Apr 19 '20 at 9:47
  • $\begingroup$ Nice and short proof. +1 I hope my understanding (mentioned in previous comment of mine) of your proof is correct. $\endgroup$
    – Paramanand Singh
    Apr 19 '20 at 9:48
  • $\begingroup$ Thank you for your nice comment @Paramanand Singh $\endgroup$
    – Riemann
    Apr 19 '20 at 11:06
  • $\begingroup$ @ParamanandSingh That's right. If $f'$ maintains the same sign everytime $f=0$, then $f$ cannot change sign twice. $\endgroup$
    – user632577
    Apr 19 '20 at 16:56

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