# Spivak Calculus: $f$ satisfies $f''(x)+f'(x)g(x)-f(x)=0$, for some g. Prove that if 𝑓 is 0 at two points, then 𝑓 is 0 on the interval between them.

The following is a problem from chapter 11, "Significance of the Derivative" from Spivak's Calculus

1. Suppose that $$f$$ satisfies $$f''(x)+f'(x)g(x)-f(x)=0\tag{1}$$ for some function $$g$$. Prove that if $$f$$ is $$0$$ at two points, then $$f$$ is $$0$$ on the interval between them. Hint: Use Theorem 6.

Here is the theorem referred to above.

Theorem 6: Suppose $$f''(a)$$ exists. If $$f$$ has a local minimum at $$a$$, then $$f''(a) \geq 0$$; if $$f$$ has a local maximum at $$a$$, then $$f''(a) \leq 0$$.

The solution manual solution is as follows

Suppose $$f(a)=f(b)=0$$. If $$x$$ is a local maximum point of $$f$$ on $$[a,b]$$, then $$f'(x)=0$$ and $$f''(x) \leq 0$$; from the equation $$f''(x)+f'(x)g(x)-f(x)=0$$ we can conclude that $$f(x) \leq 0$$. Similarly, $$f$$ cannot have a negative local minimum on $$(a,b)$$.

My question regards the above proof, which seems terse to the point that I don't understand how the desired result ensues from it.

Let me go through it in more words and steps.

If we assume that $$(1)$$ is true for all $$x$$, then $$f$$ is (twice) differentiable at all $$x$$.

The interval $$[a,b]$$ is closed, and $$f$$ is continuous on this interval. Therefore any max or min on $$[a,b]$$ must either be a critical point, one of the endpoints, or points where $$f$$ isn't differentiable.

Suppose there is a local max in $$x_1 \in [a,b]$$. Theorem 6 tells us that $$f''(x_1)<0$$ and $$(1)$$ tell us that $$f''(x_1)=f(x_1)<0\tag{2}$$

The proof above stops here and seems to imply that whatever $$(2)$$ means or implies is clear. It is not, however, clear to me.

Could we not have a situation such as

Though I can't quite finish the reasoning, it seems the situation above also leads to a contradiction.

Reasoning 1

Since $$x_1$$ is a local max, there is an interval around it where $$f$$ is smaller than $$f(x_1)$$.

In particular, $$\exists x_2, x_2 < x_1 \land f(x_2). The Intermediate Value Theorem tells us that there is some $$x_3 \in [a,x_2]$$ such that $$f(x_3)=f(x_1)$$.

Hence, by Rolle's Theorem, there is some $$x_4 \in [x_3, x_1]$$ such that $$f'(x_4)=0$$. But then $$(1)$$ tell us that $$f''(x_4)=f(x_4) \leq 0$$. So $$x_4$$ is a local max.

I can't quite finish this reasoning, but it seems to imply that there are infinite local maxima in $$(a,b)$$, at every local max $$f$$ is negative, and every critical point at which $$f$$ is negative is a local max.

The reasoning when we assume there is a local min in $$(a,b)$$ should be analogous.

Is the reasoning above on the right track, and if so how do I finish it (ie explicitly conclude that $$f=0$$ in $$[a,b]$$)?

Reasoning 2

Since $$f''(x_1)<0$$, and $$f'(x)<0$$ in an interval around $$x_1$$, $$f$$ is decreasing as we move to the left of $$x_1$$ and decreasing as we move to the left of $$x_1$$. But since $$f(a)=f(b)=0$$, and $$f$$ is continuous, it will need to increase at some point between $$x_1$$ and $$a$$, and between $$x_1$$ and $$b$$.

But then $$f'$$ needs to be positive, and hence $$f''$$ needs to be positive so that $$f'$$ increases from a negative value to a positive value. But then $$f$$ is positive at such points by $$(1)$$, which contradicts the fact that actually $$f$$ is negative whenever the sign of $$f''$$ changes because $$f$$ is negative and decreasing when that happens.

Again, this seems to mean contradiction, which means $$f$$ can't have a local max at a point where $$f<0$$.

I can't seem to satisfy myself with the details though. The reasoning doesn't seem rigorous enough.

$$f$$ attains $$0$$ on the endpoints of $$[a,b]$$.

One possibility is that $$f$$ is the constant zero function on the whole interval $$[a,b]$$.

The task is to show that the other possibility, that $$f$$ is not the constant zero function, leads to a contradiction.

$$\\$$

Suppose $$f$$ attains a positive value somewhere on $$(a,b)$$.

Then there exists $$c\in(a,b)$$ be such that $$f(c)\geq f(x)$$ for all $$x\in[a,b]$$. (You can work this out, starting from the Extreme Value Theorem.) Choose such a $$c$$.

Clearly $$f(c)>0$$.

Theorem $$6$$ implies that $$f'(c)=0$$ and that $$f''(c)\leq 0$$.

But hold on! We would then have $$f''(c)-f(c)<0$$ which is a contradiction.

$$\\$$

Do something similar after supposing that $$f$$ attains a negative value somewhere on $$(a,b)$$.

Suppose $$f$$ attains a local maximum $$f(x_1)>0$$ at $$x_1\in(a,b)$$ (it must if $$f(x)\neq0$$ for all $$x\in(a,b)$$, by the Extreme Value Theorem). Then $$f'(x_1)=0$$ and $$f''(x_1)\leq0$$, meaning $$f''(x_1)=f(x_1)\leq0,$$ which is a contradiction. Conversely, suppose $$f$$ has a local minimum $$f(x_1)<0$$ at $$x_1\in(a,b)$$ (again by the EVT). Then $$f'(x_1)=0$$ and $$f''(x_1)\geq0$$, meaning $$f''(x_1)=f(x_1)\geq0,$$ which is again a contradiction. Thus, since $$f$$ can't take on positive or negative values, we must have $$f(x)=0$$.

• Is the equals sign a typo in the first displayed equation? Shouldn't it be $f''(x_1)-f(x_1) < 0$, which contradicts the assumption of the problem that says that since $f'(x_1)=0$ we have $f''(x_1)-f(x_1)=0$?
– xoux
Commented May 15, 2022 at 17:07
• No, I set up the contradiction using the equality. It means the same thing as what you're saying: We get $f''(x_1)=f(x_1)\leq0$ from the equation, which contradicts that $f(x_1)>0$. This is equivalent to your statement. Commented May 15, 2022 at 21:50