# Elementary inequality generalizing convexity of a function on a segment

I am looking for a proof of the following statement which is known to be true as far as I heard.

Let $$g\colon [a,b]\to \mathbb{R}$$ be a smooth function. Assume that $$b-a< \pi.$$ Assume also $$g(a)\geq 0,g(b)\geq 0,$$ $$g''+g\leq 0 \mbox{ on } [a,b].$$ Then $$g\geq 0$$ on $$[a,b]$$.

• False. Take $g(x)=\sin x$, $a=\pi$, $b=2\pi$. Jul 13, 2020 at 14:41
• @Mizar: Thanks, corrected. The inequality $b-a<\pi$ has to be strict.
– asv
Jul 13, 2020 at 15:22

Write $$g=g^+-g^-$$ in $$[0,\ell]$$, multiply $$g''+g \le 0$$ by $$g^-$$ (which vanishes at the the endpoints) and integrate. Then we get with $$v=g^-$$ $$\int_0^l v'^2- \int_0^l v^2 \le 0.$$ Since the first eigenvalue of the Dirichlet laplacian in $$[0,\ell]$$ is $$\pi^2/\ell^2$$ we have also $$\int_0^l v^2 \le \frac{\ell^2}{\pi^2} \int_0^l v'^2$$ and then $$v=g^-=0$$, since $$\ell <\pi$$.

Here's another argument using Sturm-Liouville theory.

Given $$b - \pi < a \le x \le b$$, let $$f(x) = \sin (x - b + \pi).$$ Observe that $$f > 0$$ and $$f'' + f = 0$$ on $$[a,b)$$, $$f(a) > 0$$, $$f(b) = 0$$, and $$f'(b) = -1$$.

Since \begin{align*} (fg'-f'g)' &= (g'' + g)f - (f''+f)g \le 0\\ (fg'-f'g)(b) &= f(b)g'(b) - f'(b)g(b) = g(b) \ge 0, \end{align*} it follows that, on $$[a,b]$$, $$fg' - f'g \ge 0.$$ and therefore, on $$[a,b)$$, \begin{align*} \left(\frac{g}{f}\right)' &= \frac{fg' - gf'}{f^2} \ge 0. \end{align*} Since $$\frac{g(a)}{f(a)} \ge 0,$$ it follows that, on $$[a,b)$$, $$\frac{g}{f} \ge 0.$$

• What is "Sturm comparison theorem"? Jul 13, 2020 at 17:46
• @FedorPetrov, it compares solutions to self-adjoint 2nd order ODEs. I never actually remember the precise statement of the theorem, so I adapt the relatively elementary argument to each particular situation. Jul 13, 2020 at 17:51
• And where is it used? The argument looks self-contained. Jul 13, 2020 at 17:52
• @FedorPetrov, variants of this argument are commonly used to compare the behavior of geodesics on a Riemannian manifold to those on a manifold with constant curvature. But the argument itself was originally developed in what's known as Sturm-Liouville theory. It appears in many ODE textbooks. Jul 13, 2020 at 17:56
• my question was: where is it used in your answer? Probably the answer is that you include the proof but not the formulation. Jul 13, 2020 at 19:31

Suppose the contrary, so that $$g(s)<0$$ for some $$s\in(a,b)$$. Replacing now $$a$$ and $$b$$ by $$\max\{t\in[a,s)\colon g(t)\ge0\}$$ and $$\min\{t\in(s,b]\colon g(t)\ge0\}$$, respectively, we see that without loss of generality (wlog) $$\begin{equation} g(a)=0=g(b). \end{equation}$$ Also, by a horizontal shift, wlog $$\begin{equation} a=0,\quad b\in(0,\pi). \end{equation}$$

Let $$\begin{equation} h:=-(g''+g)\ge0. \end{equation}$$ Then $$\begin{equation} G(t):=G_b(t):=g(t)\sin b=\sin t\,\int_0^b du\,h(u)\sin(b-u) -\sin b\,\int_0^t du\,h(u)\sin(t-u). \end{equation}$$ We have to show that $$G\ge0$$ on $$[0,b]$$. Because the nonnegative function $$h$$ can be however closely approximated in $$L^1$$ by conical combinations of the indicators of intervals, wlog $$h=1_{[c,d]}$$ for some $$c,d$$ such that $$0, in which case $$\begin{equation} G(t)=\sin t\, (\cos (b-d)-\cos (b-c))-\sin b\, (\cos (t-\max (c,\min (d,t)))-\cos (c-t)). \end{equation}$$ So, if $$0\le t\le c$$, then $$G(t)=\sin t\, (\cos (b-d)-\cos (b-c))\ge0$$. The case $$d\le t\le b$$ is similar, by the left-to-right symmetry.

It remains to consider the case when $$c\le t\le d$$. Then $$\begin{equation} G''(t)=\sin t\, (\cos b\, (\cos c-\cos d)-\sin b\, \sin d)-\sin b \cos c\,\cos t =A\sin(t+C)=:f(t), \end{equation}$$ where $$A,C$$ depend only on $$b,c,d$$. The function $$f$$ will be $$\le0$$ on the interval $$[c,d]$$ of length $$<\pi$$ iff $$f(c)\le0$$ and $$f(d)\le0$$. In our case, we have $$\begin{equation} 2f(c)=\sin (b-c-d)-\sin (b+c-d)-\sin (b-2 c)-\sin (b) \end{equation}$$ and hence $$(f(c))'_d=-\sin c\, \sin (b-d)\le0$$, so that $$f(c)$$ decreases in $$d$$. So, wlog $$d=c$$, in which case $$f(c)=-\sin b\le0$$. Thus, $$f(c)\le0$$ always, that is, for all $$d\in[c,b]$$. Similarly, by the left-to-right symmetry, $$f(d)\le0$$, which completes the proof.

This is sometimes called a maximum principle "on thin domains". My answer below is technically equivalent to @Deane Yang's answer in this specific context, but the scope is more general so I thought I'd still give it a shot (in particular the argument applies to higher dimensions as well, see the comments below).

Note first that if the eliptic operator under consideration had nonnegative zeroth-order coefficient we could immediately conclude from the standard maximum principle that $$-g''+g\geq 0$$ implies $$g\geq 0$$, given that $$g\geq 0$$ on the boundaries. The problem here is of course that your operator $$\mathcal L[g]=-g''-g$$ has negative zeorth-order coefficient, $$c(x)\equiv -1$$.

Here comes the trick now: assume that you can find some particular function $$f(x)\geq C>0$$ (up to the boundary) such that $$\mathcal L[f]\geq 0$$. (The fact that you can find such a function relies on the thinness of the domain, I will come back to this later on. In your specific example you can take $$f(x)=\sin(x-b+\pi)$$ as in @Deane Yang's answer.) Think then of $$g$$ as $$uf$$ for some $$u$$. (Here the strict positivity of $$f$$ is important so that $$u$$ is smooth enough, no funny business can arise from this change of variables) Define then $$\tilde{\mathcal{L}}[u]:=\mathcal{L}[uf].$$ Here you can compute expliticly $$\mathcal L$$, but the key is that in whole generality this new elliptic operator $$\tilde{\mathcal{L}}[u] =\Big(\mbox{1st & 2nd order}\Big) + \Big(\mathcal L[f]\Big) u$$ always has a zeroth-order coefficient with the right sign, i-e $$\tilde c(x)=\mathcal L[f](x)\geq 0$$ . Now if $$g$$ is a supersolution of $$\mathcal L[g]\geq 0$$ you have by definition that $$u:=\frac{g}{f}$$ is a supersolution of $$\tilde{\mathcal L}[u]\geq 0$$. Applying the usual maximum principle (with $$\tilde c(x) \geq 0$$) you conclude that $$u\geq 0$$, hence $$g\geq 0$$.

Comment 1: here you see that the key point is the existence of a well-chosen $$f(x)$$, which is not to be taken for granted. The reason why you can actually find such a function is that your domain is small enough in terms of the lowest eigenvalue of the Dirichlet problem for the homogeneous problem, without zeroth-order terms: notice obviously that this mysterious function $$f(x)=\sin(x-b+\pi)$$ is indeed the principal eigenvalue of $$-\frac{d^2}{dx^2}$$ on the domain $$(b-\pi,b)$$ with zero boundary conditions. The trick is here that your domain $$(a,b)\subset (b-\pi,b)$$ is $$b-a>0$$ is small enough. (This reminds me of a classical exercise in elliptic PDEs where one is asked to apply the Lax-Milgram theorem for operator whose zeroth-order term has the wrong sign, but not too large in modulus compared to the first eigenvalue, see also @Giorgio Metafune's answer)

Comment 2: The trick works exactly the same in higher dimensions. For example if you were working in two dimensions on an infinite but thin enough strip, then plugging in a well-chosen $$\sin$$ function depending only on the thin coordinate automatically gives a suitable $$f$$. I can't remember where I learnt this trick, and also I'm pretty sure that I also read somewhere a completely general statement that, regardless of the initial coefficients, sufficiently narrowing down the domain always gives a suitable $$f$$ (this is natural if one thinks of resonant frequencies: in the absence of zeroth-order terms, thinness in any direction of a given object/domain gives a high-pitch natural frequency, hence a large principal eigenvalue that can effectively dominate any zeroth-order coefficient that you might want to add to the free resonance). The trick also works sometimes for nonlinear operators, but the "change of variables" to go from $$g$$ to $$u$$ can be very delicate to find, it's not just products and quotients anymore.

Comment 3: this is called "thin domains" because the argument is usually applied as above (in a systematic way, since thinness is a classical sufficient condition), but in fact everything relies on the existence of a "nice" $$f$$. So even in the broader context of not necessarily thin domains (balls, or whatever) one might get lucky by guessing the right $$f$$ (if any, of course!)

• I like your discussion. I think basic elliptic PDE theory be presented first in the 1-dimensional case. Jul 14, 2020 at 21:02
• What do you mean by "think of $g$ as $uf$"? Jul 14, 2020 at 21:58
• I mean "change variables" and write $g=uf$, so the new primary variable is $u$ Jul 15, 2020 at 5:06
• @Deane Yang: Thanks, me too, and this is how I always try to introduce and motivate the theory when I teach elliptic equations. But I guess this is really a matter of taste. Jul 15, 2020 at 5:09

Another approach in the spirit of viscosity solutions/maximum principle:

Assume wlog $$a>0$$ and $$b<\pi$$. Let $$I:=[a,b]$$ and define $$f(x):=\sin x,\quad g_\lambda:=g+\lambda f.$$ If $$g=g_0<0$$ somewhere on $$I$$, then we can then find $$\lambda>0$$ such that $$\min_I g_\lambda=0$$ (observe that $$f\ge c$$ on $$I$$ for some $$c>0$$ and look at $$\lambda\mapsto\min_I g_\lambda$$).

But then $$g_\lambda''\le g_\lambda''+g_\lambda\le 0$$ on $$I$$, so $$g_\lambda$$ is concave. Since $$g_\lambda>g$$ is positive at the boundary of $$I$$, it must be positive on all of $$I$$, contradiction.