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Let $f: \mathbb{R} \longrightarrow(0, \infty)$ be an infinitely differentiable function with $\int_{-\infty}^{\infty} f(t) d t=1$. Then prove/disprove that

(i) There exists $t_0 \in \mathbb{R}$ such that $f\left(t_0\right) \geq f(t)$ for all $t \in \mathbb{R}$,

(ii) $f^{\prime \prime}(a)=0$ for some $a \in \Bbb R$.

It is clear that $f$ should be bounded, otherwise $\lim_{|t| \to \infty}f(t) \not \lt\infty$ which makes trouble for the integrability on $\Bbb R$. But how to say $f$ attains its maximum? where does the differentiability is used?

Messi Lio
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    Are you really sure this is correct? There are examples of unbounded continuous functions $f$ that satisfy $f(x)\ge 0$ and $\int_{\mathbb R} f=1$. (classical, but see for example here) You require $f>0$ and $f$ to be $C^\infty$ smooth, but those examples can be modified to prove that your (i) fails. – Giuseppe Negro Dec 14 '22 at 13:48
  • I got such crazy integrable graphs, thanks a lot..., and what happens when $f''$ does not vanish on $\Bbb R$? – Messi Lio Dec 14 '22 at 14:08
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    Then your function is either concave or convex, and this is surely incompatible with that integrability (unless $f=0$ of course). – Giuseppe Negro Dec 14 '22 at 14:29

1 Answers1

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You have an answer in the comments for the first part, so I will show that there must exist some $a \in \mathbb{R}$ such that $f''(a) = 0$. Suppose to the contrary that there is no such $a$. Then by continuity of $f''$, either $f''(x) > 0$ for all $x$ or $f''(x)<0$ for all $x$. Assume WLOG that $f''(x) > 0$ for all $x$. This implies $f$ is convex, so $f(x) \ge f(y) + f'(y)(x-y)$ for all $x,y \in \mathbb{R}$. Now $f'$ cannot be identically $0$ because $f''$ is not identically $0$, so assume WLOG that there exists a $y$ with $f'(y) > 0$. This implies $\lim_{x \rightarrow \infty} f(x) \ge \lim_{x \rightarrow \infty} (f(y)+f'(y)(x-y)) = \infty$, which contradicts the integrability of $f$. Hence we conclude that we must in fact have $f''(a) = 0$ for some $a \in \mathbb{R}$.

user6247850
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