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Consider the following Putnam question:

Consider a smooth function $f:\Bbb{R}\to\Bbb{R}$ such that $f\geq 0$, and $f(0)=0$ and $f(1)=1$. Prove that there exists a point $x$ and a positive integer $n$ such that $f^{(n)}(x)<0$.

This is a problem from the 2018 Putnam, and only 10 students were able to solve it completely. I spent a day thinking about it, and my "proof" differs a lot from the official solutions, and is really a heuristic. Could you tell me if it is correct?

My proof: Assume that there does not exist any $x$ and $n$ such that $f^{(n)}(x)<0$. We will compare $f$ with functions of the form $x^n$ in $[0,1]$. We will prove that $f\leq x^n$ on $[0,1]$. Because $x^n\to 0$ on $[0,1)$ as $n\to\infty$, we will have proven that $f=0$ on $[0,1)$ and $f(1)=1$. Hence, $f$ cannot be smooth.

Why is $f\leq x^n$? Let us first analyze what $f$ looks like. It is easy to see that $f(x)=0$ for $x\leq 0$. This is because as $f\geq 0$, if $f(x)>0$ for $x<0$, when $f$ will have to decrease to $0$ at $x=0$. Hence, there will be a negative derivative involved, which is a contradiction. Hence, $f(x)=0$ for $x\leq 0$, and by continuity of derivatives for smooth functions, all derivatives at $x=0$ are also $0$.

Now consider the functions $x^n$, which are $0$ at $x=0$ and $1$ at $x=1$. These are the same endpoints for $f(x)$ in $[0,1]$. If $f(x)$ ever crosses $x^n$ in $[0,1)$, then it will have a higher $n$th derivative than $x^n$ at the point of intersection. As its $(n+1)$th derivative is also non-negative, $f$ will just keep shooting above $x^n$, and hence never "return" to $x^n$ at $x=1$. This contradicts the fact that $f(1)=1$. Hence, $f$ will always be bounded above by $x^n$ in $[0,1]$. As this is true for all $n$, $f=0$ on $[0,1)$ and $f(1)=1$. This contradicts the fact that $f$ is continuous.

Is my proof correct?

1 Answers1

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I disagree with

If $f(x)$ ever crosses $x^n$ in $[0,1)$, then it will have a higher $n$th derivative than $x^n$ at the point of intersection.

What you are claiming is that let $a > 0$ be the first time (assuming a "first" exists) that $ f(a) = a^n$, then $ f^{(n)}a > n!$.

It is clear that the 1st derivative is higher (in order to cross), but you have very little control over the subsequent derivatives (without much further work).

We might be able to show that this statement is true if we also use the condition that all of the derivatives leading up to $x$ are non-negative, but that's a different argument than what you have here.

Suppose there is a function with non-negative derivatives and $f(0)=0, f(1) = 1)$.

Claim: $f(x) \leq x^n$ at every point $x$.

Proof: Suppose not, then let $a$ be any point where $f(a) > a^n$.

Smaller claim: $f^{(n) } (a) \geq n!$.
Suppose not, then for any $ x \in [0, a ]$, since $f^{(n+1) } (x) \geq 0$, hence $f^{(n) } (x) \leq f^{(n) } (a) < n!$. Integrating this $n$ times, we conclude that $ f(a) < a^n$, which is a contradiction.

Back to proof: Then, for $ y \in [a, 1]$, we have $ f^{(n)}(y) \geq f^{(n)} (a) = n!$.
Integrating this $n$ times, we conclude that $f(1) \geq 1-a^n + f(a) > 1$, which is a contradiction.

Corollary: Hence $f(x) = 0 $ at every point $ x \in [0,1)$.

Calvin Lin
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  • The fact that all of the derivatives leading up to $x$ are non-negative is true. That is how the function is defined: we are assuming that $f^{(n)}(x)\geq 0$ for all $n\in\Bbb{N}$ and $x\in\Bbb{R}$. Couple this with the fact that all derivatives at $x=0$ are $0$, and I think my assertion become believable? –  Apr 07 '20 at 15:13
  • One way to convince yourself that this is true is the following: take a function with first and second derivatives $0$ at $x=0$. Now compare it with the graph of $y=x^2$. Can the graph ever cross $y=x^2$ in $[0,1)$ if its second derivative is never greater than $2$? The slope of $f$ will not increase as fast as the slope of $x^2$ increases, and hence $f$ will never be able to intersect $x^2$ –  Apr 07 '20 at 15:18
  • To be clear, I'm not saying that "the second derivative is never greater than 2 $(f''(x) \geq 2)$". I'm saying that "the second derivative at the crossing point $a$ need not be greater than 2" IE "at the point of intersection". Your solution requires that the "second derivative is greater than 2 over a significant region", but I don't think it need be true even "just around the crossing point". Side note: I agree that the second derivative at some point will have to be greater than 2 to cross (just not "at every point"). – Calvin Lin Apr 07 '20 at 18:22
  • If the second derivative at some point has to be greater than 2, then because the third derivative at every point is non negative, the second derivative at the point of intersection also has to be greater than 2. –  Apr 07 '20 at 18:28
  • My solution only requires that the second derivative at the point of intersection be greater than 2 –  Apr 07 '20 at 18:29
  • What you commented is in addition to what you've written.

    I've written up how your solution could work, where you're missing a proper explanation of "then it will have a higher $n$th derivative than $x^n$ at the point of intersection". My objection was exactly that this remains to be proven (which was done in the "smaller claim" using your comment above). Based on what I could read (and guess what you meant), your solution is incomplete.

     – Calvin Lin Apr 07 '20 at 18:52
  • Thanks for writing it up formally! The comments that I made, at least in my mind, are not in addition to what I've written in my answer, but I suppose I didn't flesh out the details completely. –  Apr 07 '20 at 19:23
  • I suppose now you could modify the "I disagree with" part, so that I could accept your answer? The claim that you disagreed with has turned out to be correct? –  Apr 07 '20 at 19:25
  • My statements are correct, and do not need to be modified. I disagree with your solution as written and it is incomplete / not completely correct. For the Putnam, your solution as written would likely be worth 2 points at the most (It's not worth 8, and they didn't give out 3 to 7 points for this question). Part of why the Putnam is so brutal is that you need to writeup a complete solution in order to get 3+ points out of 10. – Calvin Lin Apr 07 '20 at 19:45
  • It's confusing to me, as you said you disagree with the statement that "if $f$ ever crosses the $x^n$ in $[0,1)$ then its nth derivative has to be higher than that of $x^n$." The statement that you claim to disagree with is obviously true, as you yourself have proven after our extended discussion. So maybe you could rephrase that as "you haven't offered a proof of this statement"? –  Apr 07 '20 at 20:39
  • The statement is true, but you have not demonstrated that it is true in your solution. That is my whole point. You asked if your proof is correct. It is not correct. It could be tweaked to become correct. In the context of the Putnam, this is a serious enough gap that the solution is unlikely to score above $2$. I'm not sure why you want me to rephrase it. – Calvin Lin Apr 07 '20 at 21:16