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Let $f(x)$ be differentiable over $[0,+\infty)$, and $f'(x)$ be increasing and convex over $[0,+\infty).$ If $f(0)=f'(0)=0$, then there exist at most two roots of $f(x)=f'(x)$ over $[0,+\infty).$

Apparently, $x=0$ is already a root, hence we only need prove there exists at most one nonzero root else. Maybe we may construct an auxiliary function $F(x):=e^{-x}f(x)$, then $F'(x)=e^{-x}(f'(x)-f(x))$. This will help? Besides, notice that we are not told that $f''(x)$ exists.

mengdie1982
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    Sounds like an intermediate value theorem problem to me. The auxiliary function you have constructed is also an interesting idea. – Certainly not a dog Oct 18 '19 at 17:22
  • I think that the idea of the auxiliary function requires implementing the key fact that $f'$ is increasing and concave. Since there is no access to $f''$ I think that a key part of the problem is reformulating concaveness just in terms of $f$ and $f'$. I think that $f'$ increasing and concave and $f'(c)=f(c)$ implies $f'(x)>f(x)$ for $x>c$ but no idea how to prove it. – Fernando Chu Oct 18 '19 at 18:20

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Suppose that there exists $0 < a < b$ such that $f(a) = f'(a)$ and $f(b) = f'(b)$.

Since $f'$ is concave we have

$$\frac{f'(b) - f'(a) }{b-a} < \frac{f'(b)-f'(0)}{b-0}< \frac{f'(a)-f'(0)}{a-0}=\frac{f'(a)}{a}$$

Since $f'$ is increasing, $f$ is convex and

$$\frac{f(b)-f(a)}{b-a} > \frac{f(a)-f(0)}{a-0}= \frac{f(a)}{a}$$

Thus, we arrive at a contradiction

$$\frac{f(a)}{a} < \frac{f(b)-f(a)}{b-a} = \frac{f'(b) - f'(a) }{b-a} < \frac{f'(a)}{a} = \frac{f(a)}{a}$$

Edit: The original question included the assumption that $f’$ is concave.

RRL
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    sir, it should be $$\frac{f'(b) - f'(a) }{b-a} > \frac{f'(b)-f'(0)}{b-0}> \frac{f'(a)-f'(0)}{a-0}=\frac{f'(a)}{a}$$? – mengdie1982 Oct 19 '19 at 07:35
  • Since $f′$ is increasing, $f$ is concave. – mengdie1982 Oct 19 '19 at 07:39
  • @mengdie1982: See here. A function is convex if and only if the derivative is increasing. The essential point here is that the derivative is concave and the function is convex. Just by drawing a graph you can see that the two curves can meet at only one other point other than zero and then diverge from each other. – RRL Oct 19 '19 at 10:53
  • The first chain of inequalities in your comment are for $f’$, the derivative. The hypothesis in the problem statement is that $f’$ is itself “concave”. The inequalities should be the reverse of what you wrote. Even if they were the reverse, the inequalities go the other way for the function which must be convex and we still would get a contradiction. – RRL Oct 19 '19 at 11:04
  • Take a look at this example – RRL Oct 19 '19 at 12:02
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    maybe, there exists a different concept on the concave and convex between us. In your opinion, a rainbow in the sky is concave or convex? – mengdie1982 Oct 19 '19 at 13:14
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    I changed my problem. Can you prove the new version? – mengdie1982 Oct 19 '19 at 13:20
  • Rainbow would be considered concave. – Minus One-Twelfth Oct 19 '19 at 13:29