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Inspired by this question, I would like pose this following problem.

Let $E$ be the set of all nonnegative continuous functions $f:[0,1]\to \mathbb{R}$ such that $$f(x)\,f(y)\ge |x-y|\qquad\forall\{x,y\}\subset [0,1]$$

Find $$\min_{f\in E}\left(\int_0^1f(x) \,dx\right)$$

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  • $\begingroup$ Do you have an example of a function $f \in E$? $\endgroup$ – user159517 Nov 17 '17 at 13:04
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    $\begingroup$ @user159517 $f(x) > \sup_{x,y} |x-y| = 1$ will do it. $\endgroup$ – Calvin Khor Nov 17 '17 at 13:23
  • $\begingroup$ @CalvinKhor ah, of course. That also shows that $E$ is unbounded, hence not compact which makes the existence of a minmizing element tricky. $\endgroup$ – user159517 Nov 17 '17 at 13:47
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Integrating both sides with respect to x and y we get

$$\left(\int_0^1f(x) dx\right)^2= \int_0^1\int_0^1f(x)f(y)dxdy \ge\int_0^1\int_0^1 |x-y|dxdy= \frac13.$$ that is $$\int_0^1f(x)dx \ge \frac{1}{\sqrt3.}$$

I am still searching whether the min is attain or not.

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  • $\begingroup$ +1 for thinking of the double integral. But @kimchi lover 's bound is the tight one, since any function with the required quality must dominate $\sqrt{|1-2x|}$. $\endgroup$ – Mathemagical Nov 11 '17 at 3:03
  • $\begingroup$ My bound(s) might be tighter but I don't have a clue if the most recent version is the tight one. $\endgroup$ – kimchi lover Nov 11 '17 at 21:19
  • $\begingroup$ i am not sure whether the minimizer exists for this problem. $\endgroup$ – Guy Fsone Nov 11 '17 at 21:46
  • $\begingroup$ @GuyFsone Good question. I've been grasping for straws of compactness for this problem, without luck. $\endgroup$ – kimchi lover Nov 12 '17 at 0:02
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Let $A=A(f)=\int_0^1f(x)\,dx$. By the AGM inequality, $$A=\frac12\int_0^1 \big(f(x)+f(1-x)\big) dx \ge \int_0^1 \sqrt{f(x)f(1-x)}\,dx \ge \int_0^1 \sqrt{|1-2x|}dx = \frac23.$$

Similarly, $$A=2\int_0^{1/2} \frac{f(x)+f(x+1/2)}2\, dx $$ $$\ge 2\int_0^{1/2} \sqrt{f(x)f(x+1/2)} \, dx\ge 2\int_0^{1/2}\sqrt{1/2}\,dx=\frac1{\sqrt2}.$$

Added 14 Nov.
This might or might not have been obvious all along, but not explicitly so to me until just now. Note that if $g_1$ and $g_2$ satisfy the constraints, then so does $g(x) = \sqrt{g_1(x)g_2(x)}$, and moreover, by the AGM inequality, $$\int_0^1 \sqrt{g_1(x)g_2(x)}\,dx\le \int_0^1 \frac {g_1(x)+g_2(x)} 2\, dx.$$ Given any admissible $h$, the functions $h_1(x) = h(x)$, $h_2(x)=h(1-x)$, and $f(x)=\sqrt{h_1(x)h_2(x)} = \sqrt{h(x)h(1-x)}$ are admissible, and $\int_0^1 f\le \int_0^1 h$. Hence one cannot beat functions of the form $\sqrt{h(x)h(1-x)}$, that is, functions that depend only on $|2x-1|$.

Another consequence of the AGM inequality is that if the minimum is attained, it is attained uniquely: Suppose $A(f)=A(g)$ where $f\ne g$. Then $A(h)<A(f)$, where $h(x)=\sqrt{f(x)g(x)}$.

Added 17 Nov.
Not only can one not beat functions of form $f(x)=\phi(|2x-1|)$ , but one cannot beat such functions where continuous $\phi:[0,1]\to\mathbb R^+$ is also required to be nondecreasing. The $f(x)f(y)\ge|x-y|$ constraint translates to a $\phi(u)\phi(v)\ge(u+v)/2$ constraint for $u,v\in[0,1]$. If $\phi$ obeys these constraints, so does monotonic $\phi^*(u)=\min\{\phi(t): t\in[u,1]\}$, for which the corresponding $A(f^*)\le A(f)$.

(I'm hoping to use a compactness argument somehow to then prove the minimum $A(f)$ is attained. I think that such an argument succeeds in the restricted problem where one sticks in an additional constraint $f(0)=\gamma$, but don't yet see yet why the restricted optimal value of $A$ depends continuously on $\gamma$.)

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    $\begingroup$ The function $f(x)=\sqrt{|1-2x|}$ doesn't satisfy the assumption that $\sqrt{|1-2x|\cdot|1-2y|}\ge |x-y|$ for all $x,y\in[0,1]$ since you could choose $x=1/2$ and $y = 1$, and you would find $f(1/2)f(1) = 0 < |1/2-1| = 1/2$. $\endgroup$ – Alex Ortiz Nov 11 '17 at 3:01
  • $\begingroup$ Any response to @AOrtiz's objection? $\endgroup$ – Hans Nov 11 '17 at 18:09
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    $\begingroup$ @Hans: kimchi lover's argument is a correct proof that the minimum in question is $\geq{2\over3}$. He/She didn't state that $f(x)=\sqrt{|1-2x|}$ is optimal, or even admissible. $\endgroup$ – Christian Blatter Nov 11 '17 at 18:34
  • $\begingroup$ @ChristianBlatter: I am well aware of that it is only a lower bound, just like Guy Fsone's but tighter. I want to ascertain if he is aware of it and whether he has any further idea on finding the minimum. $\endgroup$ – Hans Nov 11 '17 at 18:45
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    $\begingroup$ It's an interesting problem, and I've not given up on it. $\endgroup$ – kimchi lover Nov 11 '17 at 18:57
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NOTE: Still not a complete answer, just improving on the lower bound.
EDIT: Added an improvement on the upper bound (a better function which "works").

I decided to combine the two ideas in kimchi lover's bounds, by putting a bound on $$ \int_0^1 f(x)dx = \int_0^{1/4} \left[f(x) + f(1-x) + f(1/2+x) + f(1/2-x)\right]dx $$ Considering the four values $(u,v,s,t) = (f(x), f(1-x), f(1/2-x), f(1/2+x))$ as independent, but satisfying the constraints $$ \begin{align} &uv \geq 1-2x \\ &ut \geq 1/2 \\ &sv \geq 1/2 \\ &st \geq 2x, \end{align} $$ I found a lower bound for $u+s+v+t$. (There are two more inequalities, of course, but the minimizing set of values turns out to be symmetric, $s = t$ and $u = v$, so the $us$ and $tv$ inequalities are weaker than the $ut$ and $sv$ ones.) The approach is rather brute-force: pick $u$ and $s$ as independent, consider the different cases for which of the inequalities involving $t$ and $v$ are the stronger ones, then minimize an expression in terms of $u$ and $s$. The calculations are tedious, and since this is not a complete answer anyway, I'll omit them. I got the following: $$ u + s + t + v \geq \frac{3-4x}{\sqrt{1-2x}}, $$ minimum attained at $u = v = \sqrt{1-2x}$ and $s = t = {1\over2u}$ (these are not too surprising, of course). Thus $$ \int_0^1 f(x) dx \geq \int_0^{1\over4} \frac{(3-4x)dx}{\sqrt{1-2x}} = {5-2\sqrt2\over 3} \approx 0.72386 $$ If we try to piece together a function from the $(u,s,t,v)$'s for $x \in [0,1/4]$, it still doesn't satisfy some of the constraints, such as for example $f(x)f(1) \geq 1-x$.

A function which does meet the constraints everywhere is $$ f(x) = \frac{e^{|2x-1|}}{\sqrt{2e}}. $$ That it does work can be verified using the triangle inequality $$ f(x)f(y) = \frac{e^{|2x-1|+|2y-1|}}{2e} \geq \frac{e^{2|x-y|}}{2e} \geq |x-y| $$ where the last one holds because $e^{2t} - 2et$ has a global minimum $0$ at $t = 1/2$. Its integral is $$ \int_0^1 \frac{e^{|2x-1|}}{\sqrt{2e}}dx = 2\int_{1/2}^1 \frac{e^{2x-1}}{\sqrt{2e}}dx = \frac{e-1}{\sqrt{2e}} \approx 0.73694 $$

EIDT: Following @kimchilover's idea to generalize the exponential function to $\beta e^{\alpha|2x-1|}$, the constraints will be satisfied if and only if the minimum of $$ \beta^2 e^{\alpha|2x-1|}e^{\alpha|2y-1|} - |x-y| $$ over $[0,1]\times[0,1]$ is non-negative. The strongest constraints are obtained using $x \in [0,1/2], y \in [1/2,1]$ so that we need $$ \beta^2 e^{2\alpha(y-x)} - (y-x) = \beta^2 e^{2\alpha t} - t \geq 0 $$ for $t \in [0,1]$. We can find the minimum, then if the minimum is in $[0,1]$, require that it is non-negative, or if it is outside, require that the closest "cut-off" is non-negative. Long story short, the optimal ones I found to be the root of $e^\alpha(\alpha - 3/2) + 3/2 = 0$, $\alpha \approx 0.874217$ (by Wolfram), $\beta = 1/\sqrt{2e\alpha}$, and for the integral $\approx 0.733001$.

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  • $\begingroup$ Good! Sorry I didn't notice this post earlier. $\endgroup$ – kimchi lover Nov 14 '17 at 14:20
  • $\begingroup$ @kimchilover That's also good to know. I've been trying to get some functional/differential equation to describe a family of functions that could be optimal, but I don't get anything useful. Basically I can prove rigorously that for every $x$, there must exist a $y$ such that $f(x)f(y) = |x-y|$, then differentiate with respect to either $x$ or $y$, but getting nowhere after that. It's how I came up with the exponential function in the first place, actually. By the way, this last exponential is definitely not optimal either, because for $x \in [1/2, 1/2\alpha)$ no such $y$ exists. $\endgroup$ – Nick Pavlov Nov 17 '17 at 12:39
  • $\begingroup$ I had to retract my most recent claim: what I wrote was obviously false, and what I meant to write is more slippery than I thought. $\endgroup$ – kimchi lover Nov 17 '17 at 12:58
  • $\begingroup$ @kimchilover I must admit I didn't actually read your argument; monotonicity is another property that intuitively makes sense, so it was good enough for me that you claimed to have proven it. I was much more motivated to find a mistake when you claimed to beat my exponential (which at the time seemed so elegant that I truly hoped it was the answer). I only crunched the numbers on the generalized exponential because I was so convinced I would find that $\alpha = 1$ is best. Now that the illusion is dispelled, I feel way out of my depth here - I haven't even heard of this theorem you mention! $\endgroup$ – Nick Pavlov Nov 17 '17 at 13:19
  • $\begingroup$ The mistake was minor, and I have now repaired it. The Helly-Bray theorem was big news in the 1920 or whenever, but is nowadays subsumed into the "Portmanteau Theorem" or "Prokhorov's Theorem" in probability theory. My idea was, bounded monotone functions on $[0,1]$ are in effect finite measures on $[0,1]$, the set of which is compact, by H-B. $\endgroup$ – kimchi lover Nov 17 '17 at 13:51
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Extending kimchi lover's idea:

$$I = n \int_0^{1/n} \frac{1}{n} \sum_{k=0}^{n-1} f(x+k/n) dx \geq n \int_0^{1/n} \left(\prod_{k=0}^{n-1} f(x+k/n) \right)^{1/n} dx$$

Pairing up opposite edges, then we can pick pairs $(x+k/n, x+(n-k+1)/n)$, with $n=2m$, such that

$$\lim_{n \to \infty} I \geq \lim_{n \to \infty} 2m \int_0^{1/2m} \left( \prod_{k=1}^{m} \left(\frac{2k-1}{2m} \right) \right)^{1/2m} dx = \lim_{m \to \infty} \frac{[(2m-1)!!]^{1/2m}}{(2m)^{1/2}} = \sqrt{\frac{1}{e}}$$

where we compute the limit using wolfram alpha.

This bound isn't the best one found so far, but perhaps my approach will be inspiring. Loved thinking about the problem!

EDIT: More information. Using a continuous approximate (from above) to the step function $f(x)=a^{1/2} \chi_{[0,a]} + a^{-1/2} \chi_{(a,1]}$, you should be able to obtain a value below 1 for the integral if I did by calculation correctly. It is clear that $f=1$ means the minimum is less than or equal to 1.

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  • $\begingroup$ Interesting. I check your $a\approx.767$ yields $I\approx.937774$. I tried an $f(x)=1$ for $|x-1/2|>c/2$, otherwise $f(x)=b$, found $b=2/3$ and $c=b^2$ seemed optimal, and got $I=23/27\approx.852$. $\endgroup$ – kimchi lover Nov 13 '17 at 23:53
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    $\begingroup$ @kimchilover Guys, not to be immodest, but in my answer there's a function which has a significantly lower $I$ then these. Not to mention even $|x-1/2|+1/2$ produces $I = 3/4$. $\endgroup$ – Nick Pavlov Nov 14 '17 at 13:04
  • $\begingroup$ @NickPavlov Sorry about that: I suffer tunnel vision on this web site, and the most recent thing I see flushes out knowledge of better things I've seen before. $\endgroup$ – kimchi lover Nov 14 '17 at 13:18
  • $\begingroup$ No, thanks for making us aware @NickPavlov! $\endgroup$ – abnry Nov 14 '17 at 18:30
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Numerical experiment

The following numerical experiment yields, as we will see later, the correct assumptions for getting a good candidate for the optimum. Convexity will help us show that this candidate is indeed the optimum.

We plot the active set of the optimum of a discretization: active set

This is the output of the following code: activeset[k_] := Module[{h, constraints, vars, objective, optresult, result, pts}, h = 1/(2 k - 2); constraints = Flatten[Table[ Table[x[i] x[j] >= (2 k - i - j) h, {j, i, k}], {i, 1, k}]]; constraints = Join[constraints, Table[x[i] >= 0, {i, 1, k}]]; vars = Table[x[i], {i, 1, k}]; objective = Sum[x[i], {i, 1, k}] + Sum[x[i], {i, 2, k - 1}]; optresult = FindMinimum[{objective, constraints}, vars, Method -> "InteriorPoint"]; result = Table[{(i - 1) h, x[i]}, {i, 1, k}] /. optresult[[2]]; pts = Select[Flatten[Table[{i, j}, {i, 1, k}, {j, 1, k}], 1], Abs[x[#[[1]]] x[#[[2]]] - (2 k - #[[1]] - #[[2]]) h] < 10^(-5) /. optresult[[2]] &]; ListPlot[(pts - 1)/(k - 1), AspectRatio -> 1, PlotStyle -> Red, PlotTheme -> "Detailed"] ]; activeset[71]

Full solution

Set $S:=\{(x,y) \in [0,1]^2 : x+y \leq 1\}$. We define the maps \begin{alignat*}{2} &J : C[0,1] \rightarrow \mathbb{R},\quad &&J(u) := \int_0^1 \! \exp(u(x)) \, \mathrm{d}x, \\[1em] &I : C[0,1] \rightarrow C(S),\quad &&I(u)(x,y) := u(x)+u(y)-\log(2-x-y). \end{alignat*} We will study the optimization problems \begin{align} & \text{minimize} && J(u) \nonumber\\ & \text{subject to} && I(u) \geq 0, \tag{1} \\[1em] \text{and}\nonumber \\[1em] & \text{minimize} && \int_0^1 \! f(x) \, \mathrm{d}x \nonumber\\ & \text{subject to} && f \in C[0,1], \tag{2}\\ &&& f \geq 0, \nonumber\\ &&& f(x)f(y) \geq |x-y| \text{ for all } x,y \in [0,1]. \nonumber \end{align}

We will show that:

The unique solution $\bar{u}$ of $(1)$ is given by \begin{align*} \exp(\bar{u}(x)) &= \sqrt{2 \left(1+\sqrt{2}\right)}-\sqrt{2 x+\sqrt{2}-1}, \\[1em] J(\bar{u}) &= \frac{2}{3}\sqrt{1+\sqrt{2}}. \end{align*} The unique solution $\bar{f}$ of $(2)$ is given by \begin{align*} \bar{f}(x) &= \begin{cases} \dfrac{\exp(\bar{u}(2x))}{\sqrt{2}}&\text{if}\quad 0 \leq x \leq \frac{1}{2},\\ \dfrac{\exp(\bar{u}(2(1-x)))}{\sqrt{2}}&\text{if}\quad \frac{1}{2} < x \leq 1, \end{cases} \\[1em] \int_0^1 \! \bar{f}(x) \, \mathrm{d}x &= \frac{1}{3} \sqrt{2 \left(1+\sqrt{2}\right)}\quad (\approx 0.732456) \end{align*} where $\bar{u}$ is the unique solution of $(1)$.

Theorem 1 (Generalized Kuhn-Tucker Theorem).

Let $X$ be a normed space and $Y$ a normed space having positive cone $P$. Assume that $P$ contains an interior point.

Let $J$ be a Fréchet differentiable real-valued functional on $X$ and $I$ a Fréchet differentiable mapping from $X$ into $Y$. Suppose $\bar{u}$ minimizes $J$ subject to $I(\bar{u}) \geq 0$ and that $\bar{u}$ is a regular point of the inequality $I(\bar{u}) \geq 0$ i.e. there exists an $h \in X$ such that $I(\bar{u}) + I'(\bar{u})(h) > 0$. Then there is a positive functional $\psi \in Y^*$ such that \begin{align} J'(\bar{u}) &= \psi \circ I'(\bar{u}), \tag{3} \\\psi(I(\bar{u})) &= 0. \tag{4} \end{align}

Proof. See [Luenberger, David G, Optimization by vector space methods, Wiley, New York, 1969, p. 249]. $$\tag*{$\Box$}$$

Note: The following assumption (iii) is inspired by our numerical experiment.

Claim 1. If

(i) $\bar{u}$ is the solution of $(1)$

(ii) $\bar{u} \in C[0,1]$

(iii) there exists $g \in C^1[0,1]$ such that $$A:=\{(x,g(x)) : x \in [0,1]\} = \left\{(x,y) \in S : I(\bar{u})(x,y) = 0\right\},$$

then $g$ is an involution and $$\exp(\bar{u}(x)) = -g'(x)\exp(\bar{u}(g(x)))$$ for all $x \in [0,1]$.

Proof. By (iii) we have $I(\bar{u})(x,g(x)) = 0$ for all $x \in [0,1]$.

Since $I(\bar{u})(x,y) = I(\bar{u})(y,x)$ for all $x,y \in [0,1]$, we have $I(\bar{u})(g(x),x) = 0$ and thus $g(g(x)) = x$ by (iii) for all $x \in [0,1]$. Therefore $g$ is an involution.

We will now apply Theorem 1. The point $\bar{u}$ is regular, since $I'(\bar{u})(\mathbf{1}_{[0,1]}) = 2\cdot \mathbf{1}_S.$

Thus there exists a positive functional $\psi \in C(S)^*$ such that $(3)$ and $(4)$ are satisfied. Since $S$ is compact, there is a unique regular Borel measure $\nu$ on $\mathcal{B}(S)$ by the Riesz-Markov-Kakutani representation theorem such that $$\psi(v) = \int_S v(x,y) \, \mathrm{d}\nu(x,y)$$ for all $v \in C(S)$. We define a measure on $\mathcal{B}(S)$ by setting $$\mu(E) := \lambda\left(\{x \in [0,1] : (x,g(x)) \in E\}\right)$$ for all $E \in \mathcal{B}(S)$, where $\lambda$ is the Lebesgue measure on $[0,1]$. We will show that $\nu \ll \mu$, i.e. that $\nu$ is absolutely continuous with respect to $\mu$. Let $N \in \mathcal{B}(S)$ such that $\mu(N)=0$. By $(4)$ we have $\psi(I(\bar{u})) = 0$. Therefore $$\int_{S} I(\bar{u})(x,y) \, \mathrm{d}\nu(x,y) = 0,$$ and since $I(\bar{u}) \geq 0$, we have $\nu(S\setminus A)=0$ and $\nu(N \cap (S\setminus A))=0$. By $(3)$ we have $$\int_0^1 \exp(\bar{u}(x)) h(x) \, \mathrm{d}x = \int_S h(x) + h(y) \, \mathrm{d}\nu(x,y)$$ for all $h \in C[0,1]$. Set $N' = \{x \in [0,1] : (x,g(x)) \in N\}$. By definition of $\mu$ we have $\lambda(N') = 0$. Let $\epsilon > 0$. Then there exists $\bar{h} \in C[0,1]$ such that $\bar{h}(x) = 1$ for all $x \in N'$, $\bar{h}\geq 0$, and $\int_0^1 \exp(\bar{u}(x)) \bar{h}(x) \, \mathrm{d}x<\epsilon$. Now we have \begin{align*} \nu(N \cap A) &= \int_S \mathbf{1}_{N \cap A}(x,y) \, \mathrm{d}\nu(x,y) \\[1em] &= \int_S \mathbf{1}_{N'}(x) \, \mathrm{d}\nu(x,y) \\[1em] &\leq \int_S \bar{h}(x) + \bar{h}(y) \, \mathrm{d}\nu(x,y) \\[1em] &= \int_0^1 \exp(\bar{u}(x)) \bar{h}(x) \, \mathrm{d}x<\epsilon. \end{align*} Therefore $\nu(N \cap A) = 0$, thus $\nu(N) = 0$ and $\nu \ll \mu$. By the Radon–Nikodym theorem there exists a measurable function $w : S \rightarrow [0,\infty)$ such that $$\psi(v) = \int_S v(x,y)\, \mathrm{d}\nu(x,y) = \int_S v(x,y)\,w(x,y) \, \mathrm{d}\mu(x,y)$$ for all $v \in C(S)$. Since $\mu$ is the pushforward measure under $x \mapsto (x,g(x))$ we have $$\psi(v) = \int_0^1 v(x,g(x))\,w(x,g(x)) \, \mathrm{d}x$$ for all $v \in C(S)$. By $(4)$ we now have $$\int_0^1 \exp(\bar{u}(x)) h(x) \, \mathrm{d}x = \int_0^1 (h(x) + h(g(x)))\,w(x,g(x)) \, \mathrm{d}x$$ for all $h \in C[0,1]$. Since $g$ is a $C^1$-involution on $[0,1]$, we can substitute $g(x)$ for $x$ in $\int_0^1 h(g(x))\,w(x,g(x)) \, \mathrm{d}x$. We get $$\int_0^1 \exp(\bar{u}(x)) h(x) \, \mathrm{d}x = \int_0^1 h(x) (w(x,g(x))-w(g(x),x)g'(x)) \, \mathrm{d}x$$ for all $h \in C[0,1]$. Therefore $$\exp(\bar{u}(x)) = w(x,g(x))-w(g(x),x)g'(x)$$ for almost all $x \in [0,1]$. Since $g$ is an involution, we have \begin{align*} \exp(\bar{u}(g(x))) &= w(g(x),g(g(x)))-w(g(g(x)),g(x))g'(g(x)) \\[1em] &= w(g(x),x)-w(x,g(x)) \frac{1}{g'(x)} \\[1em] &= -\frac{1}{g'(x)}(w(x,g(x))-w(g(x),x)g'(x)) \\[1em] &= -\frac{1}{g'(x)}\exp(\bar{u}(x)) \end{align*} for almost all $x \in [0,1]$. Since $g$, $g'$ and $\bar{u}$ are continuous we have \begin{align} \exp(\bar{u}(x)) = -g'(x)\exp(\bar{u}(g(x))) \tag{5} \end{align} for all $x \in [0,1]$. $$\tag*{$\Box$}$$

Claim 2. If

(i) $\bar{u}$ is the solution of $(1)$

(ii) $\bar{u} \in C^1[0,1]$

(iii) there exists $g \in C^2[0,1]$ such that $$\{(x,g(x)) : x \in [0,1]\} = \left\{(x,y) \in S : I(\bar{u})(x,y) = 0\right\}$$ and $$\{(x,g(x)) : x \in (0,1)\} \subseteq \operatorname{int}(S),$$ then $$g(x) = 1+\sqrt{2}+x-\sqrt{4 \left(1+\sqrt{2}\right) x+2},$$ and $$\exp(\bar{u}(x)) = \sqrt{2 \left(1+\sqrt{2}\right)}-\sqrt{2 x+\sqrt{2}-1}.$$

Proof. Let $x \in (0,1)$. We have $I(\bar{u})(x,g(x)) = 0$. Since $I(\bar{u}) \geq 0$, we have that $(x,g(x))$ is a minimum point of $I(\bar{u})$. Since $I(\bar{u})$ is a $C^1$-function by $(ii)$, we have that $I(\bar{u})'(x,g(x)) = 0$. Therefore \begin{align} \bar{u}'(x) = -\frac{1}{2-x-g(x)}. \tag{6} \end{align} By Claim 1 we have $(5)$. We will see that $(5)$ and $(6)$ determine $g$ and $\bar{u}$ uniquely. The following calculations hold for all $x \in [0,1]$.

Since $I(\bar{u})(x,g(x)) = 0$, we have \begin{align*} \bar{u}(g(x)) = \log(2-x-g(x)) - \bar{u}(x), \end{align*} and thus with $(5)$ we have \begin{align} \bar{u}(x) = \frac{1}{2}\left(\log(-g'(x)) + \log(2-x-g(x))\right), \tag{7} \end{align} and thus \begin{align} \bar{u}'(x) = \frac{1}{2}\left(\frac{g''(x)}{g'(x)} + \frac{-1-g'(x)}{2-x-g(x)}\right). \tag{8} \end{align} Putting $(6)$ and $(8)$ together yields $$g''(x)(2-x-g(x)) -g'(x)^2 +g'(x) = 0.$$ Therefore \begin{align} g'(x)(2-x-g(x)) +2g(x) = a. \tag{9} \end{align} for some $a \in \mathbb{R}$. In particular, we have \begin{align} g'(0)(2-g(0)) +2g(0)= g'(1)(1-g(1)) +2g(1). \tag{10} \end{align} Since $g$ is an involution we have $g(0) = 1$, $g(1) = 0$, $1 = g'(g(0))g'(0)= g'(1)g'(0)$ and $g' < 0$. We put this into $(10)$ and get $a = 1-\sqrt{2}$. Thus \begin{align} g'(x) = \frac{1-\sqrt{2}-2g(x)}{2-x-g(x)}. \tag{11} \end{align} Now we replace $x$ with $g(x)$ in $(9)$ and then use $g'(g(x)) = \frac{1}{g'(x)}$. We get \begin{align} g'(x) = \frac{2-x-g(x)}{1-\sqrt{2}-2x}. \tag{12} \end{align} Putting $(11)$ and $(12)$ together yields a quadratic equation for $g(x)$. Only the solution \begin{align} g(x) = 1+\sqrt{2}+x-\sqrt{4 \left(1+\sqrt{2}\right) x+2} \tag{13} \end{align} satisfies $g(0)=1$. We also have \begin{align*} \exp(\bar{u}(g(x)))^2 &\underset{\hphantom{(12)}}{=} \frac{2-x-g(x)}{\exp(\bar{u}(x))}\exp(\bar{u}(g(x))) \\[1em] &\underset{(5)\hphantom{0}}{=} -\frac{2-x-g(x)}{g'(x)} \\[1em] &\underset{(12)}{=} 2 x+\sqrt{2}-1, \end{align*} and thus \begin{align*} \exp(\bar{u}(x)) &\underset{\hphantom{(12)}}{=} \frac{2-x-g(x)}{\sqrt{2 x+\sqrt{2}-1}} \\[1em] &\underset{(13)}{=} \frac{1-2x-\sqrt{2}+\sqrt{4 \left(1+\sqrt{2}\right) x+2}}{\sqrt{2 x+\sqrt{2}-1}} \\[1em] &\underset{\hphantom{(12)}}{=} -\sqrt{2 x+\sqrt{2}-1}+\frac{\sqrt{4 \left(1+\sqrt{2}\right) x+2}}{\sqrt{2 x+\sqrt{2}-1}} \\[1em] &\underset{\hphantom{(12)}}{=} -\sqrt{2 x+\sqrt{2}-1}+\sqrt{2 \left(1+\sqrt{2}\right)}. \end{align*} $$\tag*{$\Box$}$$

We define \begin{align*} \bar{u} : [0,1] \rightarrow \mathbb{R}, \quad \bar{u}(x) &:= \log\left(\sqrt{2 \left(1+\sqrt{2}\right)}-\sqrt{2 x+\sqrt{2}-1}\right), \\[1em]g : [0,1] \rightarrow [0,1], \quad g(x) &:= 1+\sqrt{2}+x-\sqrt{4 \left(1+\sqrt{2}\right) x+2}. \end{align*}

Claim 3. $g$ is an involution. We have \begin{align*} I(\bar{u})(x,g(x)) &= 0, \\[0.5em] \exp(\bar{u}(x) + \bar{u}(y)) &\geq 2-x-y \end{align*} for all $x,y \in [0,1]^2$, and thus $I(\bar{u}) \geq 0$.

Proof. We solve $g(x)=y$ for $x$ and see that $x=g(y)$. Thus $g(g(x))=x$ and $g$ is an involution on $[0,1]$. We define $$F : [0,1]^2 \rightarrow \mathbb{R}, \quad F(x,y) := \exp(\bar{u}(x)+\bar{u}(y)) - (2-x-y).$$ For all $x \in [0,1]$ we have $x + g(x) \leq 1$ and thus $(x,g(x)) \in S$. We see that $\exp(\bar{u}(x))^2 = 2 g(x)+\sqrt{2}-1$, thus $\exp(\bar{u}(g(x)))^2 = 2 x+\sqrt{2}-1$ and by putting this into $F(x,g(x))$ we get $F(x,g(x)) = 0$ and $I(\bar{u})(x,g(x)) = 0$. For all $y \in [0,1]$ we have
\begin{align*} (\partial_{y y} F)(x,y) &= \exp(\bar{u}(x))\, \partial_{y y} \exp(\bar{u}(y)) \\ &= \exp(\bar{u}(x))\, \frac{1}{\left(2 y+\sqrt{2}-1\right)^{3/2}} \\ &> 0. \end{align*} Therefore $y \mapsto F(x,y)$ is convex and since $(\partial_y F)(x,g(x))= 0$, we have that $(x,g(x))$ is the global minimizer of $y \mapsto F(x,y)$. Since $F(x,g(x)) = 0$ we have $F \geq 0$ and thus $I(\bar{u}) \geq 0$. $$\tag*{$\Box$}$$

Claim 4. $\bar{u}$ is the unique solution of $(1)$.

Proof. We define \begin{alignat*}{2} &w : [0,1] \rightarrow [0,\infty), \quad &&w(x) := \frac{\exp(\bar{u}(x))}{1-g'(x)}, \\[1em] &\psi : C(S) \rightarrow \mathbb{R}, \quad &&\psi(v) := \int_0^1 v(x,g(x)) \, w(x) \, \mathrm{d}x, \\[1em] &L : C[0,1] \rightarrow \mathbb{R}, \quad &&L(u) := J(u) - \psi(I(u)). \end{alignat*} $w$ is well-defined since $g$ being an involution and not the identity implies $g'\leq0$.

$\psi$ is well-defined since $(x,g(x)) \in S$ for all $x \in [0,1]$. We have $w(g(x))=w(x)$ and thus \begin{align*} L'(\bar{u})(h) &= J'(\bar{u})(h) - \psi'(I(\bar{u}))(I'(h)) \\[1em] &= J'(\bar{u})(h) - \psi(I'(h)) \\[1em] &= \int_0^1 \exp(\bar{u}(x))\, h(x) \, \mathrm{d}x - \int_0^1 (h(x)+h(g(x))) \, w(x) \, \mathrm{d}x \\[1em] \text{by substitution:} \\[0.5em] &= \int_0^1 \exp(\bar{u}(x))\, h(x) \, \mathrm{d}x - \int_0^1 h(x) \, (1-g'(x)) \,w(x)\, \mathrm{d}x \\[1em] &= \int_0^1 \exp(\bar{u}(x))\, h(x) \, \mathrm{d}x - \int_0^1 \exp(\bar{u}(x)) \, h(x) \, \mathrm{d}x \\[1em] &= 0. \end{align*} for all $h \in C[0,1]$. Since $J$ and $-(\psi \circ I)$ are convex, $L$ is convex. We have $L'(\bar{u})=0$ and thus $\bar{u}$ is a global minimizer of $L$. Since $\psi$ is a positive functional, we have $L(u) \leq J(u)$ for all $u \in C[0,1]$ with $I(u) \geq 0$. But $\psi(I(\bar{u}))= 0$ since $I(\bar{u})(x,g(x))=0$ by Claim 3 for all $x \in [0,1]$. Therefore $L(\bar{u})=J(\bar{u})$ and thus $J(\bar{u}) \leq J(u)$ for all $u \in C[0,1]$ with $I(u) \geq 0$. By Claim 3 we have $I(\bar{u}) \geq 0$. Therefore $\bar{u}$ is a solution of $(1)$.

Assume there is a $\tilde{u} \in C[0,1]$ such that $I(\tilde{u}) \geq 0$ and $J(\tilde{u}) \leq J(\bar{u})$. We set $$\widehat{u} := \frac{\tilde{u} + \bar{u}}{2}.$$ We have $I(\widehat{u})\geq 0$. Since $J$ is strictly convex, $\bar{u} = \widehat{u} = \tilde{u}$. Therefore $\bar{u}$ is the unique solution of $(1)$. $$\tag*{$\Box$}$$

Claim 5. The unique solution of $(2)$ is given by $$\bar{f}(x)= \begin{cases} \dfrac{\exp(\bar{u}(2x))}{\sqrt{2}}&\text{if}\quad 0 \leq x \leq \frac{1}{2},\\ \dfrac{\exp(\bar{u}(2(1-x)))}{\sqrt{2}}&\text{if}\quad \frac{1}{2} < x \leq 1. \end{cases}$$

Proof. Since $\exp > 0$, we have $\bar{f} > 0$. Let $x,y \in[0,1]$ such that $x \leq y$.

If $x \leq \frac{1}{2} \leq y$, then \begin{align} \bar{f}(x)\bar{f}(y)&=\frac{1}{2}\exp\left(\bar{u}(2x)+\bar{u}(2(1-y))\right) \nonumber \\ &\geq \frac{1}{2}(2- 2x - 2(1-y)) \qquad \text{(by Claim 3)} \nonumber \\ &=y-x \tag{14} \\ &= |x-y|. \nonumber \end{align} If $x \leq y \leq \frac{1}{2}$, then $$\bar{f}(x)\bar{f}(y)=\bar{f}(x)\bar{f}(1-y) \underset{(14)}{\geq} 1-y-x \geq y-x = |x-y|.$$ If $\frac{1}{2} \leq x \leq y$, then $$\bar{f}(x)\bar{f}(y)=\bar{f}(1-x)\bar{f}(y) \underset{(14)}{\geq} y-(1-x) \geq y-x = |x-y|.$$ Thus $\bar{f}(x)\bar{f}(y) \geq |x-y|$ for all $x,y \in [0,1]$.

Assume there exists $\tilde{f}$ such that $\tilde{f}$ satisfies the constraints of (2) and $$\int_0^1 \tilde{f} \, \mathrm{d}x \leq \int_0^1 \bar{f} \, \mathrm{d}x.$$ We set $$\widehat{f}(x):= \left(\tilde{f}(x)\tilde{f}(1-x)\bar{f}(x)^2\right)^\frac{1}{4}.$$ $\widehat{f}$ satisfies the constraints of (2) and \begin{align} \int_0^1 \widehat{f}(x) \, \mathrm{d}x &= \int_0^1 \left(\tilde{f}(x)\tilde{f}(1-x)\bar{f}(x)^2\right)^\frac{1}{4} \, \mathrm{d}x \nonumber \\[1em] &\leq \int_0^1 \frac{\tilde{f}(x)+\tilde{f}(1-x)+2\bar{f}(1-x)}{4} \, \mathrm{d}x \tag{15} \\[1em] &= \int_0^1 \frac{\tilde{f}(x)+\bar{f}(x)}{2} \, \mathrm{d}x \nonumber \\[1em] &\leq \int_0^1 \bar{f}(x) \, \mathrm{d}x. \nonumber \end{align} Since $\widehat{f}(x) = \widehat{f}(1-x)$ and $\bar{f}(x) = \bar{f}(1-x)$ for all $x \in [0,1]$, we thus have \begin{align} \int_0^{\frac{1}{2}} \widehat{f}(x) \, \mathrm{d}x \leq \int_0^{\frac{1}{2}} \bar{f}(x) \, \mathrm{d}x. \tag{16} \end{align} We set $$\widehat{u}(x) := \log\left(\sqrt{2}\, \widehat{f}\left(\frac{x}{2}\right)\right),$$ and by $(16)$ we have $J(\widehat{u}) \leq J(\bar{u})$. We also have $I(\widehat{u}) \geq 0$ since \begin{align*} \exp(\widehat{u}(x)+\widehat{u}(y)) &= 2\widehat{f}\left(\frac{x}{2}\right)\widehat{f}\left(\frac{y}{2}\right) \\[1em] &= 2\widehat{f}\left(\frac{x}{2}\right)\widehat{f}\left(1-\frac{y}{2}\right) \\[1em] &\geq 2\left|1-\frac{x}{2}-\frac{y}{2}\right| \\[1em] &= 2-x-y \end{align*} for all $x,y \in [0,1]$. By Claim 4, $\widehat{u} = \bar{u}$. Since $$\widehat{f}(x)=\frac{\exp(\widehat{u}(2x))}{\sqrt{2}}$$ for all $x \in [0,\frac{1}{2}]$ by definition of $\widehat{u}$, we have $\widehat{f} = \bar{f}$. The inequality at $(15)$ must be an equality and thus $\tilde{f} = \widehat{f} = \bar{f}$. Therefore $\bar{f}$ is the unique solution of $(2)$. $$\tag*{$\Box$}$$

$\endgroup$
  • $\begingroup$ By Golly, this is impressive, cafaxo! But I won't have time to check it out until two days later. $\endgroup$ – Hans Jan 14 '18 at 9:33
  • $\begingroup$ it is not possible to reduce this ?\ $\endgroup$ – Guy Fsone Jan 15 '18 at 12:36
  • $\begingroup$ There might be an elementary argument for $(5)$. I'm sorry, I could not find one. $\endgroup$ – cafaxo Jan 15 '18 at 13:02
  • $\begingroup$ I will have many questions. Here are the first two. 1) To arrive at the equation right beneath (15), are you assuming $\tilde f(x)=\tilde f(1-x)$? If so, it seems we have not proved the evenness of the minimizer of Problem (2) with respect to $x=\frac12$ or $f(x)=f(1-x)$. 2) What is your motivation for devising the equivalent Problem (1) of (2)? $\endgroup$ – Hans Jan 16 '18 at 5:47
  • $\begingroup$ 1) I'm using $\int_0^1 \tilde{f}(x) \, \mathrm{d}x = \int_0^1 \tilde{f}(1-x) \, \mathrm{d}x$, which is true by substitution ($x \mapsto 1-x$). 2) For me, Problem (1) was far easier to work with. The main reasons are that $J$ and $-I$ are convex and $I(u)$ is $C^1$ for all $u \in C^1[0,1]$. $\endgroup$ – cafaxo Jan 16 '18 at 10:31

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