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Can we have a smooth function, say $f: [0,1] \mapsto [0,1],$ such that $f^{(k)}(0)=0$ and $f^{(k)}(x)>0$ for all $k \ge 0,$ $x>0?$ I suspect this is impossible, but I'm not sure how to prove this.

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  • $\begingroup$ Take the standard mollifier. $\endgroup$
    – Svetoslav
    Mar 5, 2016 at 22:08
  • $\begingroup$ I fail to see how a mollifier satisfies the properties I asked for. $\endgroup$ Mar 5, 2016 at 22:18
  • $\begingroup$ Yes, you are right. The second derivative of the mollifier $f(x)=e^{\frac{-1}{1-(x-1)^2}}$ already gets negative: derivative-calculator.net $\endgroup$
    – Svetoslav
    Mar 5, 2016 at 22:42
  • $\begingroup$ While I am too sick to spend much time on this, I stongly suspect you are mistaken. I can see no reason it would be impossible. $\endgroup$ Mar 6, 2016 at 0:37
  • $\begingroup$ I certainly hope I'm mistaken. I've been trying to construct such a function. $\endgroup$ Mar 6, 2016 at 0:43

1 Answer 1

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$f^{(n)}(x)$ should be convex for all $n\ge 0$ on $[0,1]$. This means in particular that $$f^{(n)} \left( \frac{t}{2}\right)\leq \frac{1}{2}f^{(n)}(t),\,\forall n\in\mathbb N$$ Integrating this we get $$\int\limits_{0}^{x}{f^{(n)} \left( \frac{t}{2}\right)dt}\leq \int\limits_{0}^{x}{\frac{1}{2}f^{(n)}(t)dt}$$ $$\Leftrightarrow f^{(n-1)}\left(\frac{x}{2}\right)\leq \frac{1}{2^2}f^{(n-1)}(x)$$ We integrate this $n-1$ more times to get $$f\left(\frac{x}{2}\right)\leq \frac{1}{2^{n+1}}f(x)$$ Since $n$ was arbitrary it follows that $f\left(\frac{x}{2}\right)$ should be $0$ for all $x\in[0,1]$, i.e $f(x)\equiv 0$ for $x\in [0,1/2]$. But we can also apply convexity with arbitrary $\lambda \in (0,1)$ to get $f(\lambda x)\leq \lambda^n f(x),\forall n\in\mathbb N,\,\forall \lambda\in(0,1)$ and therefore we get that $f(x)\equiv 0,\,\forall x\in [0,1)$ and by continuity also $f(1)=0$.

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