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I am struggling to solve this exercise.

Let $I\subset\mathbb{R}$ be an interval (whether open or closed, bounded or unbounded, etc.), and let $\phi:\mathbb{R}\rightarrow\mathbb{R}$ be a twice differentiable function such that $\forall x\in \mathbb{R}\setminus I\;\;\phi(x)=0$, $\forall x\in I\;\;\phi(x)\ge 0$. Prove that $$\forall x\in I\;\;\phi(x)>0\Rightarrow\frac{(\phi'(x))^{2}}{2\phi(x)}\le \max_{y\in\mathbb{R}}(|\phi''(y)|)$$

Because $\phi$ and $\phi'$ are continuous in $\mathbb{R}$, I know that both $\phi$ and $\phi'$ valued at the lower/upper limit(s) of $I$ return $0$.

I guess I have to use either Cauchy's, Taylor's, or L'Hôpital's theorem somehow but I am stuck in a rut. No other information is given regarding $\phi$.

Domenico Fossaceca
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1 Answers1

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I will consider only the case where $I$ is bounded or $\lim_{|x|\rightarrow{\infty}}\phi(x)=0$.

Define $g(x)=\frac{(\phi'(x))^2}{2\phi(x)}$ on $I$. Being continuous it achieves a maximum. As $\phi(x)=0$ outside $I$, the maximum must be attained inside $I$, say at $x_0$. There, $g'(x)=0$.

$$g'(x_0)=\frac12\frac{2\phi(x_0)\phi'(x_0)\phi''(x_0)-(\phi'(x_0))^3}{\phi^2(x_0)}$$ and so $\phi'(x_0)\big(2\phi(x_0)\phi''(x_0)-(\phi'(x_0))^2\big)=0$. If $\phi'(x_0)=0$ there is nothing to prove. If not, then $(\phi'(x_0))^2=2\phi(x_0)\phi''(x_0)$. Putting things together $$g(x)=\frac{(\phi'(x))^2}{2\phi(x)}\leq g(x_0)=\phi''(x_0)\leq\max_{x\in\mathbb{R}}|\phi''(x)|$$

Another case that can be handle is one in which $I$ is bounded on one side and infinite on the other ($[a,\infty)$ or $(-\infty,a]$) and $\phi'\neq0$ inside $I$ for in this case, the general mean value theorem yields a point $\xi\in I$ between $a$ and $x$ such that $$2\phi'(\xi)\left((\phi'(x))^2-(\phi'(a))^2\right)=2\phi'(\xi)\phi''(\xi)\left(\phi(x)-\phi(a)\right)$$ Hence $$\frac{(\phi'(x))^2-(\phi'(a))^2}{2(\phi(x)-\phi(a))}=\frac{2\phi'(\xi)\phi''(\xi)}{2\phi'(\xi)}=\phi''(\xi)\leq\max_{x\in\mathbb{R}}|\phi''(x)|$$

Mittens
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  • I thought that $g$ could achieve a maximum only if $I$ was bounded. How can $\lim_{|x|\rightarrow +\infty} \phi(x)=0$ imply the existence of a maximum for a continuous function (which in our case is $g$)? – Domenico Fossaceca Jan 17 '25 at 00:01
  • @DomenicoFossaceca: It is another case that I can deal with, that is $\lim_{|x|\rightarrow\infty}\phi(x)=0$ is an assumption not a consequence of the assumptions. The long and the short of it, if either $I$ is a bounded interval OR if $\lim_{|x|\rightarrow\infty}\phi(x)=0$ holds, then the argument I presented works. – Mittens Jan 17 '25 at 00:03
  • Say $I$ is bounded from below but unbounded above. Let us define a closed interval $I'\subseteq I$. Does your argument work because (by definition of infinite limit) the following proposition holds: $\exists K\in\mathbb{R}:\forall x\in I\cap(K,+\infty);;\phi(x)<\max_{y\in I'}(\phi(y))$ ? – Domenico Fossaceca Jan 17 '25 at 00:16
  • Can the general mean value theorem be applied if $I=\mathbb{R}$? I mean, in that case could we simply pick an interval $[x_{0}-\delta,x_{0}+\delta]$ where $\phi'(x)\neq 0$and apply the mean value theorem there? – Domenico Fossaceca Jan 17 '25 at 00:29
  • @DomenicoFossaceca: under the assumptions I laid out, yes, the argument holds. In the case you are describing, suppose the function $\phi$ has maximum value $M$ on $[a,a+1]$ ($I=[a,\infty)$) then the $\varepsilon=M/2$ and choose $A>a$ such that $\phi(x)<\varepsilon$ for all $x\geq A$. $\phi$ has a maximum in $[a,\infty)$ and is achieve at some point in $(a,A)$ and so, it must be a critical point. In believe that a combination of the three scenarios I laid out provide an answer in the general case. One goes by pieces considering $\phi$ restricted to $(-\infty, n]$. – Mittens Jan 17 '25 at 00:45
  • Suppose $\exists x \in I : \phi(x)=0$. If $\exists y \in I : \phi(y)\ne 0$, then it is still possible to find an interval where the function is strictly positive, because \phi is continuous. The same can be done for $\phi’$, it being continuous and there existing a point where the derivative is not equal to $0$ because $\phi$ is not a constant function. Therefore the hypothesis that $\phi’\ne 0$ inside I is not needed. – Domenico Fossaceca Jan 17 '25 at 09:03