Minimum Problem
Show that the function $bar y(x)=1-|x|$ solves the minimum problem :
$min( int_(-1)^1 e^(1-|x|-y(x))dx : y in C^0([-1,1],RR) , int_(-1)^1 y(x)dx =1 )$.
$min( int_(-1)^1 e^(1-|x|-y(x))dx : y in C^0([-1,1],RR) , int_(-1)^1 y(x)dx =1 )$.
Risposte
I was reviewing old threads while gathering some of my problems into a single PDF file, when I came to this one.
Like some months ago, I was fascinated by Rigel's post because his argument works so well when one wants to prove the symmetry properties of the minimizers [tex]$\overline{y}$[/tex] of:
[tex]$\min \left\{ \mathcal{I} [y]:=\int_{-1}^1 f(x,y(x))\ \text{d} x,\ \text{con $y\in C([-1,1])$ e $\lVert y\rVert_1=1$}\right\}$[/tex]
without assuming strong hypothesis on [tex]$f(x,y)$[/tex]. On the contrary, the symmetrization technique I used does work only in the special case when the integrand comes in the form [tex]$f(x,y)=\phi (x)\ \psi (y)$[/tex], but it allows to prove also the monotonicity properties of [tex]$\overline{y}$[/tex] (in fact, they come for free from the identity [tex]$\overline{y} =\overline{y}^\star$[/tex]).
Nevethless it seems that, with little more work, one can succeed in proving the monotonicity of the minimizers also in the general setting of Rigel (hence without the aid of "mysterious tools" like symmetrization)...
***
Sooo, here's a new challenge to you, young mathematicians:
Assuming the Rigel's hypothesis on the integrand [tex]$f(x,y)$[/tex] (see here), show that a function [tex]$\overline{y} \in C([-1,1])$[/tex] minimizing [tex]$\mathcal{I}[y]$[/tex] under the constraint [tex]$\lVert y\rVert_1=1$[/tex] does not increase in [tex]$[0,1]$[/tex] (hence it does not decrease in [tex]$[-1,0]$[/tex]).
Like some months ago, I was fascinated by Rigel's post because his argument works so well when one wants to prove the symmetry properties of the minimizers [tex]$\overline{y}$[/tex] of:
[tex]$\min \left\{ \mathcal{I} [y]:=\int_{-1}^1 f(x,y(x))\ \text{d} x,\ \text{con $y\in C([-1,1])$ e $\lVert y\rVert_1=1$}\right\}$[/tex]
without assuming strong hypothesis on [tex]$f(x,y)$[/tex]. On the contrary, the symmetrization technique I used does work only in the special case when the integrand comes in the form [tex]$f(x,y)=\phi (x)\ \psi (y)$[/tex], but it allows to prove also the monotonicity properties of [tex]$\overline{y}$[/tex] (in fact, they come for free from the identity [tex]$\overline{y} =\overline{y}^\star$[/tex]).
Nevethless it seems that, with little more work, one can succeed in proving the monotonicity of the minimizers also in the general setting of Rigel (hence without the aid of "mysterious tools" like symmetrization)...
***
Sooo, here's a new challenge to you, young mathematicians:
Assuming the Rigel's hypothesis on the integrand [tex]$f(x,y)$[/tex] (see here), show that a function [tex]$\overline{y} \in C([-1,1])$[/tex] minimizing [tex]$\mathcal{I}[y]$[/tex] under the constraint [tex]$\lVert y\rVert_1=1$[/tex] does not increase in [tex]$[0,1]$[/tex] (hence it does not decrease in [tex]$[-1,0]$[/tex]).
How nice!
Thank you, Rigel; it's a nice proof.
***
Here's mine.
Assume [tex]$\overline{y} \in C([-1,1])$[/tex] minimizes [tex]$\mathcal{I}[y]$[/tex] under the constraint [tex]$||y||_{L^1([-1,1])}=1$[/tex]; then [tex]$\overline{y}$[/tex] also minimize [tex]$\mathcal{J}[y]:=\mathcal{I}[y]-2(e-1)=\int_{-1}^1 e^{1-|x|} (e^{-\overline{y}(x)}-1) \text{ d} x$[/tex].
We have:
[tex]$\mathcal{J}[\overline{y}] =\int_{-1}^1 e^{1-|x|} \ (e^{-\overline{y} (x)}-1) \text{ d} x$[/tex]
[tex]$=\int_{-1}^1 e^{1-|x|} \Big\{ \int_0^{\overline{y}(x)} -e^{-t} \text{ d} t \Big\} \text{ d} x$[/tex]
[tex]$=-\int_{-1}^1 e^{1-|x|} \Big\{ \int_0^{+\infty} e^{-t}\ \chi_{\{\overline{y} > t\} }(x) \text{ d} t\Big\} \text{ d} x$[/tex]
where [tex]$\chi_{\{ \overline{y} >t\}} (x)$[/tex] is the characteristic function of the set [tex]$\{ \overline{y} >t\} :=\{ x\in [-1,1]:\ \overline{y} (x)>t\}$[/tex], i.e.:
[tex]$\chi_{\{ \overline{y} >t\}} (x) :=\begin{cases} 1 &\text{, if $\overline{y}(x) >t$}\\ 0 &\text{, otherwise}\end{cases}$[/tex];
by Fubini we can interchange the order of integration to get:
(a) [tex]$\mathcal{J}[\overline{y}] =-\int_0^{+\infty} e^{-t} \Big\{ \int_{-1}^1 e^{1-|x|} \ \chi_{\{ \overline{y} >t\}} (x) \text{ d} x\Big\} \text{ d} t$[/tex].
Now rearrangement and Hardy-Littlewood inequality come into play: in fact, since H-L inequality holds, Schwarz symmetrization [tex]$\star$[/tex] increases the [tex]$L^2$[/tex] scalar product and we have:
[tex]$\int_{-1}^1 e^{1-|x|} \ \chi_{\{ \overline{y} >t\}} (x) \text{ d} x \leq \int_{-1}^1 (e^{1-|x|})^\star \ \left( \chi_{\{ \overline{y} >t\}} (x)\right)^\star \text{ d} x$[/tex];
but [tex]$(e^{1-|x|})^\star=e^{1-|x|}$[/tex] (because [tex]$e^{1-|x|}$[/tex] is spherically decreasing) and [tex]$\left( \chi_{\{ \overline{y} >t\}} (x)\right)^\star =\chi_{\{ \overline{y} >t\}^\star} (x) =\chi_{\{ \overline{y}^\star >t\}}$[/tex] (for the Schwarz rearrangement of the characteristic function [tex]$\chi_{\{ \overline{y} >t\}}$[/tex] is the characteristic function of the open symmetric interval [tex]$\{ \overline{y} >t\}^\star$[/tex] with the same length of [tex]$\{ \overline{y} >t\}$[/tex]; the latter interval is by definition the [tex]$t$[/tex]-level set of [tex]$\overline{y}^\star$[/tex]), hence the previous inequality reads:
(b) [tex]$\int_{-1}^1 e^{1-|x|} \ \chi_{\{ \overline{y} >t\}} (x) \text{ d} x \leq \int_{-1}^1 e^{1-|x|}\ \chi_{\{ \overline{y}^\star >t\}} (x) \text{ d} x$[/tex].
Taking into account the minus sign in (a), from (a) and inequality (b) follows:
[tex]$\mathcal{J}[\overline{y}]=-\int_0^{+\infty} e^{-t} \Big\{ \int_{-1}^1 e^{1-|x|} \ \chi_{\{ \overline{y} >t\}} (x) \text{ d} x\Big\} \text{ d} t \geq -\int_0^{+\infty} e^{-t} \Big\{ \int_{-1}^1 e^{1-|x|} \ \chi_{\{ \overline{y}^\star >t\}} (x) \text{ d} x\Big\} \text{ d} t =\mathcal{J}[\overline{y}^\star]$[/tex];
moreover [tex]$||\overline{y}^\star||_{L^1([-1,1])} =||\overline{y}||_{L^1([-1,1])}=1$[/tex] and [tex]$\overline{y}^\star \in C([-1,1])$[/tex].
Hence [tex]$\overline{y}^\star$[/tex] minimizes [tex]$\mathcal{J}[\cdot ]$[/tex] and [tex]$\mathcal{I}[\cdot ]$[/tex] on the constraint [tex]$||y||_{L^1([-1,1])} =1$[/tex]; but [tex]$\mathcal{I}[\cdot ]$[/tex] has an unique minimizer on the constraint, therefore [tex]$\overline{y}=\overline{y}^\star$[/tex].
Because [tex]$\overline{y}$[/tex] coincides with its Schwarz rearrangement, it is spherically decreasing, i.e. it is an even function which does not increase in [tex]$[0,1]$[/tex] (and obviously does not decrease in [tex]$[-1,0]$[/tex]).
Pretty cool, isn't it?
Thank you, Rigel; it's a nice proof.
***
Here's mine.
Assume [tex]$\overline{y} \in C([-1,1])$[/tex] minimizes [tex]$\mathcal{I}[y]$[/tex] under the constraint [tex]$||y||_{L^1([-1,1])}=1$[/tex]; then [tex]$\overline{y}$[/tex] also minimize [tex]$\mathcal{J}[y]:=\mathcal{I}[y]-2(e-1)=\int_{-1}^1 e^{1-|x|} (e^{-\overline{y}(x)}-1) \text{ d} x$[/tex].
We have:
[tex]$\mathcal{J}[\overline{y}] =\int_{-1}^1 e^{1-|x|} \ (e^{-\overline{y} (x)}-1) \text{ d} x$[/tex]
[tex]$=\int_{-1}^1 e^{1-|x|} \Big\{ \int_0^{\overline{y}(x)} -e^{-t} \text{ d} t \Big\} \text{ d} x$[/tex]
[tex]$=-\int_{-1}^1 e^{1-|x|} \Big\{ \int_0^{+\infty} e^{-t}\ \chi_{\{\overline{y} > t\} }(x) \text{ d} t\Big\} \text{ d} x$[/tex]
where [tex]$\chi_{\{ \overline{y} >t\}} (x)$[/tex] is the characteristic function of the set [tex]$\{ \overline{y} >t\} :=\{ x\in [-1,1]:\ \overline{y} (x)>t\}$[/tex], i.e.:
[tex]$\chi_{\{ \overline{y} >t\}} (x) :=\begin{cases} 1 &\text{, if $\overline{y}(x) >t$}\\ 0 &\text{, otherwise}\end{cases}$[/tex];
by Fubini we can interchange the order of integration to get:
(a) [tex]$\mathcal{J}[\overline{y}] =-\int_0^{+\infty} e^{-t} \Big\{ \int_{-1}^1 e^{1-|x|} \ \chi_{\{ \overline{y} >t\}} (x) \text{ d} x\Big\} \text{ d} t$[/tex].
Now rearrangement and Hardy-Littlewood inequality come into play: in fact, since H-L inequality holds, Schwarz symmetrization [tex]$\star$[/tex] increases the [tex]$L^2$[/tex] scalar product and we have:
[tex]$\int_{-1}^1 e^{1-|x|} \ \chi_{\{ \overline{y} >t\}} (x) \text{ d} x \leq \int_{-1}^1 (e^{1-|x|})^\star \ \left( \chi_{\{ \overline{y} >t\}} (x)\right)^\star \text{ d} x$[/tex];
but [tex]$(e^{1-|x|})^\star=e^{1-|x|}$[/tex] (because [tex]$e^{1-|x|}$[/tex] is spherically decreasing) and [tex]$\left( \chi_{\{ \overline{y} >t\}} (x)\right)^\star =\chi_{\{ \overline{y} >t\}^\star} (x) =\chi_{\{ \overline{y}^\star >t\}}$[/tex] (for the Schwarz rearrangement of the characteristic function [tex]$\chi_{\{ \overline{y} >t\}}$[/tex] is the characteristic function of the open symmetric interval [tex]$\{ \overline{y} >t\}^\star$[/tex] with the same length of [tex]$\{ \overline{y} >t\}$[/tex]; the latter interval is by definition the [tex]$t$[/tex]-level set of [tex]$\overline{y}^\star$[/tex]), hence the previous inequality reads:
(b) [tex]$\int_{-1}^1 e^{1-|x|} \ \chi_{\{ \overline{y} >t\}} (x) \text{ d} x \leq \int_{-1}^1 e^{1-|x|}\ \chi_{\{ \overline{y}^\star >t\}} (x) \text{ d} x$[/tex].
Taking into account the minus sign in (a), from (a) and inequality (b) follows:
[tex]$\mathcal{J}[\overline{y}]=-\int_0^{+\infty} e^{-t} \Big\{ \int_{-1}^1 e^{1-|x|} \ \chi_{\{ \overline{y} >t\}} (x) \text{ d} x\Big\} \text{ d} t \geq -\int_0^{+\infty} e^{-t} \Big\{ \int_{-1}^1 e^{1-|x|} \ \chi_{\{ \overline{y}^\star >t\}} (x) \text{ d} x\Big\} \text{ d} t =\mathcal{J}[\overline{y}^\star]$[/tex];
moreover [tex]$||\overline{y}^\star||_{L^1([-1,1])} =||\overline{y}||_{L^1([-1,1])}=1$[/tex] and [tex]$\overline{y}^\star \in C([-1,1])$[/tex].
Hence [tex]$\overline{y}^\star$[/tex] minimizes [tex]$\mathcal{J}[\cdot ]$[/tex] and [tex]$\mathcal{I}[\cdot ]$[/tex] on the constraint [tex]$||y||_{L^1([-1,1])} =1$[/tex]; but [tex]$\mathcal{I}[\cdot ]$[/tex] has an unique minimizer on the constraint, therefore [tex]$\overline{y}=\overline{y}^\star$[/tex].
Because [tex]$\overline{y}$[/tex] coincides with its Schwarz rearrangement, it is spherically decreasing, i.e. it is an even function which does not increase in [tex]$[0,1]$[/tex] (and obviously does not decrease in [tex]$[-1,0]$[/tex]).
Pretty cool, isn't it?
I propose a simple argument in order to show that the solution $\bar{y}$ is even.
It seems that this argument works in the following setting: $\mathcal{I}[y] = \int_{-1}^{1} f(x,y(x)) dx$, where
$f: [-1,1]\times RR\to RR$ satisfies $f(x, y) = f(-x,y)$ for every $(x,y)\in [-1,1]\times RR$, and $f(x, \cdot)$ is strictly convex and monotone decreasing for (a.e.) $x\in [-1,1]$.
Assume that $\bar{y}$ is a solution.
Then also the function $\hat{y}(x) := \bar{y}(-x)$ is a solution, since $\mathcal{I}[\bar{y}] = \mathcal{I}[\hat{y}]$.
As discussed in previous posts, we know that $\bar{y}$ and $\hat{y}$ are non-negative functions.
Let us consider the function $z := \frac{\bar{y} + \hat{y}}{2}$. Since $\bar{y}$ and $\hat{y}$ are non-negative admissible functions, then also $z\geq 0$ and $||z||_1 = 1$.
From the convexity of $f(x, \cdot)$ we have that
$\mathcal{I}[z] = \int_{-1}^1 f(x, z(x)) dx \le \frac{\mathcal{I}[\bar{y}] + \mathcal{I}[\hat{y}]}{2}$.
Moreover, the strict convexity of $f(x,\cdot)$ implies that the strict inequality holds unless $\bar{y} = \hat{y}$. But the minimality of $\bar{y}$ and $\hat{y}$ implies that equality must hold and that $\bar{y} = \hat{y}$, i.e. $\bar{y}$ is an even function.
It seems that this argument works in the following setting: $\mathcal{I}[y] = \int_{-1}^{1} f(x,y(x)) dx$, where
$f: [-1,1]\times RR\to RR$ satisfies $f(x, y) = f(-x,y)$ for every $(x,y)\in [-1,1]\times RR$, and $f(x, \cdot)$ is strictly convex and monotone decreasing for (a.e.) $x\in [-1,1]$.
Assume that $\bar{y}$ is a solution.
Then also the function $\hat{y}(x) := \bar{y}(-x)$ is a solution, since $\mathcal{I}[\bar{y}] = \mathcal{I}[\hat{y}]$.
As discussed in previous posts, we know that $\bar{y}$ and $\hat{y}$ are non-negative functions.
Let us consider the function $z := \frac{\bar{y} + \hat{y}}{2}$. Since $\bar{y}$ and $\hat{y}$ are non-negative admissible functions, then also $z\geq 0$ and $||z||_1 = 1$.
From the convexity of $f(x, \cdot)$ we have that
$\mathcal{I}[z] = \int_{-1}^1 f(x, z(x)) dx \le \frac{\mathcal{I}[\bar{y}] + \mathcal{I}[\hat{y}]}{2}$.
Moreover, the strict convexity of $f(x,\cdot)$ implies that the strict inequality holds unless $\bar{y} = \hat{y}$. But the minimality of $\bar{y}$ and $\hat{y}$ implies that equality must hold and that $\bar{y} = \hat{y}$, i.e. $\bar{y}$ is an even function.
I hope dissonance will find this post interesting, because it shows a nice application of what we were discussing here.
***
Let us consider the modified Camillo's problem:
[tex]$\min \left\{ \mathcal{I} [y],\ \text{with } y\in C^0([-1,1]) \text{ and } ||y||_{L^1([1,1])} =1 \right\}$[/tex]
where [tex]$\mathcal{I}[y]:=\int_{-1}^1 e^{1-|x|-y(x)} \text{ d} x$[/tex].
We've just seen that if a minimizing function [tex]$\overline{y}$[/tex] exists then it is nonnegative in [tex]$[-1,1]$[/tex]; moreover, if a minimizing function exists, then it is unique (because the integrand in [tex]$\mathcal{I}$[/tex] is strictly convex in [tex]$y$[/tex] and the minimizers are [tex]$\geq 0$[/tex]).
Now we want to investigate a priori the monotony of the minimizing functions; in particular, we want to prove the following lemma:
[tex]$\text{If $\overline{y}$ is a minimizing function for the \emph{modified Camillo's problem}, then $\overline{y}$ is an even function in $[-1,1]$}$[/tex]
[tex]$\text{which does not increase in $[0,1]$.}$[/tex]
The proof I got involves a nice measure-theoretic tool called Schwarz symmetrization (or spherically decreasing rearrangement).
Using symmetrization, the classical Hardy-Littlewood inequality, Fubini's theorem, a result quoted in the post linked at the begining and the uniqueness of minimizing function, it becomes easy to show that [tex]$\overline{y}$[/tex] coincide with its Schwarz rearrangement, so it is an even function.
I don't know if there is a more elementary way to prove the lemma, but IMHO my proof is very cool.
I will post my proof (in spoiler) tonight, because today I'm just too busy...
I'm going to appreciate any elementary proof of the lemma that you will post.
P.S.: It seems to me that my symmetrization method works in the more general case of integrands in the form [tex]$f(x,y)=\varphi (x)\ \psi(y)$[/tex] with [tex]$\varphi ,\ \psi$[/tex] non negative measurable and [tex]$\psi$[/tex] absolutely continuous in [tex]$[-1,1]$[/tex].
***
Let us consider the modified Camillo's problem:
[tex]$\min \left\{ \mathcal{I} [y],\ \text{with } y\in C^0([-1,1]) \text{ and } ||y||_{L^1([1,1])} =1 \right\}$[/tex]
where [tex]$\mathcal{I}[y]:=\int_{-1}^1 e^{1-|x|-y(x)} \text{ d} x$[/tex].
We've just seen that if a minimizing function [tex]$\overline{y}$[/tex] exists then it is nonnegative in [tex]$[-1,1]$[/tex]; moreover, if a minimizing function exists, then it is unique (because the integrand in [tex]$\mathcal{I}$[/tex] is strictly convex in [tex]$y$[/tex] and the minimizers are [tex]$\geq 0$[/tex]).
Now we want to investigate a priori the monotony of the minimizing functions; in particular, we want to prove the following lemma:
[tex]$\text{If $\overline{y}$ is a minimizing function for the \emph{modified Camillo's problem}, then $\overline{y}$ is an even function in $[-1,1]$}$[/tex]
[tex]$\text{which does not increase in $[0,1]$.}$[/tex]
The proof I got involves a nice measure-theoretic tool called Schwarz symmetrization (or spherically decreasing rearrangement).
Using symmetrization, the classical Hardy-Littlewood inequality, Fubini's theorem, a result quoted in the post linked at the begining and the uniqueness of minimizing function, it becomes easy to show that [tex]$\overline{y}$[/tex] coincide with its Schwarz rearrangement, so it is an even function.
I don't know if there is a more elementary way to prove the lemma, but IMHO my proof is very cool.
I will post my proof (in spoiler) tonight, because today I'm just too busy...
I'm going to appreciate any elementary proof of the lemma that you will post.
P.S.: It seems to me that my symmetrization method works in the more general case of integrands in the form [tex]$f(x,y)=\varphi (x)\ \psi(y)$[/tex] with [tex]$\varphi ,\ \psi$[/tex] non negative measurable and [tex]$\psi$[/tex] absolutely continuous in [tex]$[-1,1]$[/tex].
I am glad that the exercise I have selected has produced such an interesting discussion with different approaches for solving the problem.
The exercise comes out from Admission Test to Politecnico di Milano for Doctorate in Mathematical Engineering .
May be I will use the same source for proposing other problems.
The exercise comes out from Admission Test to Politecnico di Milano for Doctorate in Mathematical Engineering .
May be I will use the same source for proposing other problems.
@mathematico: In general I agree with you, it is better an elementary proof (i.e. a proof that does not use powerful tools).
I have only stressed one important point (at least for me). In many cases one can see an elementary proof only looking from a higher perspective.
In this particular case the higher perspective comes from convexity (embedded in Jensen's inequality).
I have only stressed one important point (at least for me). In many cases one can see an elementary proof only looking from a higher perspective.
In this particular case the higher perspective comes from convexity (embedded in Jensen's inequality).
For elgiovo: Yes, with the strong EL equations you find minimum points (extremals) in $C^2$; unfortunately in this exercise the minimum point is $C^2$ everywhere unless in $x=0$, and for this reason it seems that EL equations works.
"gac":
Jensen's inequality offers a "higher" point of view (and so it also requires a "higher" mathematical maturity).
Maybe you're right, Jensen's inequality requires a "higher" mathematical maturity, but i think that is not necessary to use "cannons" when it'exists an easy way.
Everytime I have to solve a problem, I search for a elementary solution (elementary$\ne$ easy), i.e a solution that does not require "powerful" tools.
_____
Mi esprimo in Italiano, non credo di aver reso l'idea in inglese
Ogni volta che ho di fronte un problema cerco di trovare una soluzione che utilizzi strumenti matematici che maneggiano un po' tutti, perchè trovo sia più divertente così
. Sinceramente, non ho lontanamente pensato alla disuguaglianza di Jensen, e forse da qui si vede la mia immaturità matematica che non ho mai nascosto di possedere 
So what you want to say is that using the EL equation in strong form I'm excluding possible solutions belonging to [tex]C^0[(-1,1)]/C^2[(-1,1)][/tex], right?
The strong EL equation only "scans" the space of twice differentiable functions, so the EL method is not correct in this case (even though the solution is actually found by EL). Ok, now I'm convinced.
The strong EL equation only "scans" the space of twice differentiable functions, so the EL method is not correct in this case (even though the solution is actually found by EL). Ok, now I'm convinced.
No, obviously you can't use the fact that you have to show that $\bar y$ is a solution; if you want to use the Euler Lagrange approach then you have to find $\bar y$ as a solution of the Euler Lagrange equations, but since the admissible functions are not-differentiable you cannot write Euler Lagrange equations in the strong form, as you wrote.
"Luca.Lussardi":
You don't know that $\bar y$ solves the problem...
The problem claims that [tex]\overline{y}[/tex] is the solution, and one can observe that [tex]\overline{y}[/tex] is absolutely continuous, so in case it actually solves the minimum problem, one should be able to prove it using the E-L equation.
You don't know that $\bar y$ solves the problem... The point is that you want to write Euler Lagrange equations for a functional defined only on continuous functions, and then you only can write Euler Lagrange equations in the weak form.
"Luca.Lussardi":
But since we don't know a priori that $\bar y$ is a minimum point for $F$, we cannot assume this kind of regularity, and then we are able only to derive the Euler Lagrange equations in the weak form (integral form).
Why do you say that we don't know the solution a priori?
Yes, you are right.
Neverthless, in this way it seems that the estimate $e^x \ge 1+x$ is a trick that works without any apparent reason.
Jensen's inequality offers a "higher" point of view (and so it also requires a "higher" mathematical maturity).
Neverthless, in this way it seems that the estimate $e^x \ge 1+x$ is a trick that works without any apparent reason.
Jensen's inequality offers a "higher" point of view (and so it also requires a "higher" mathematical maturity).
I agree but for me it is not necessary to use a tool like Jensen's inequality; the most simple way is the Mathematico's proof, $e^x\ geq x+1$ for any $x \in \RR$ (which is the convexity of $e^x$, I know...), this does not involve integral functionals, it is an easy exercise of calculus in one variable...
In my opinion cirasa's solution is in fact equivalent to the other one.
Namely, setting $\phi(t) = e^t$, for any admissible $y$ one has
$\frac{1}{2} F(\bar y) = \phi(0) = \phi(\frac{1}{2}\int_{-1}^{1} (1-|x|-y(x))dx) \leq \frac{1}{2} F(y)$.
Morover, thanks to the strict convexity of $\phi$, equality holds if and only if
$1-|x|-y(x) = 0$ for every $x\in [-1,1]$.
The equivalence with mathematico's solution follows from the fact that Jensen's inequality is in turn a consequence of the fact that the graph of a convex function lies above every tangent plane (and, in this case, we are considering the tangent plane of $\phi$ at $x_0=0$).
Namely, setting $\phi(t) = e^t$, for any admissible $y$ one has
$\frac{1}{2} F(\bar y) = \phi(0) = \phi(\frac{1}{2}\int_{-1}^{1} (1-|x|-y(x))dx) \leq \frac{1}{2} F(y)$.
Morover, thanks to the strict convexity of $\phi$, equality holds if and only if
$1-|x|-y(x) = 0$ for every $x\in [-1,1]$.
The equivalence with mathematico's solution follows from the fact that Jensen's inequality is in turn a consequence of the fact that the graph of a convex function lies above every tangent plane (and, in this case, we are considering the tangent plane of $\phi$ at $x_0=0$).
For the original problem proposed by camillo:
In my opinion the most simple correct proof is the Mathematico's proof: to di this it is sufficient to observe that $e^x \geq x+1$ for any $x \in RR$, and then $F(y) \geq F(1-|\cdot|)$ for each $y$ admissible. Notice that the solution of this problem is unique since $F$ is strictly convex.
The cirasa's solution, by means of the Jensen's inequality, is correct but it is a little bit complicated than the previous one: indeed first you have to show that $F(y) \geq 2$ for any $y$ admissibile and then you have to observe that $F(\bar y)=2$.
Finally the elgiovo's solution is not completely correct since you are working with non-differentiable (in general) functions. It seems that it works only because actually the minimum point for $F$ is absolutely continuous, and for this kind of extremals the classical methods of the Calculus of Variations work. But since we don't know a priori that $\bar y$ is a minimum point for $F$, we cannot assume this kind of regularity, and then we are able only to derive the Euler Lagrange equations in the weak form (integral form).
In my opinion the most simple correct proof is the Mathematico's proof: to di this it is sufficient to observe that $e^x \geq x+1$ for any $x \in RR$, and then $F(y) \geq F(1-|\cdot|)$ for each $y$ admissible. Notice that the solution of this problem is unique since $F$ is strictly convex.
The cirasa's solution, by means of the Jensen's inequality, is correct but it is a little bit complicated than the previous one: indeed first you have to show that $F(y) \geq 2$ for any $y$ admissibile and then you have to observe that $F(\bar y)=2$.
Finally the elgiovo's solution is not completely correct since you are working with non-differentiable (in general) functions. It seems that it works only because actually the minimum point for $F$ is absolutely continuous, and for this kind of extremals the classical methods of the Calculus of Variations work. But since we don't know a priori that $\bar y$ is a minimum point for $F$, we cannot assume this kind of regularity, and then we are able only to derive the Euler Lagrange equations in the weak form (integral form).
Well, ok, I have to show that:
[tex]\displaystyle \mathcal{I}[|y|]= \int_{-1}^1 e^{1-|x|-|y(x)|} dx \le \int_{-1}^1 e^{1-|x|-y(x)}dx= \mathcal{I}[y] \quad\forall y\in C_0([-1,1], \mathbb{R})[/tex].
Knowing that [tex]|y(x)|= y_+(x)+y_-(x)[/tex] then [tex]y(x)= y_+(x)-y_-(x)\le y_+(x)+y_-(x)= |y(x)|[/tex] hence [tex]1-|x|-|y(x)|\le 1-|x|-y(x)\quad\forall x\in [-1,1][/tex] (1.1). From the inequality it follows that:
[tex]\displaystyle \mathcal{I}[|y|]= \int_{-1}^1 e^{1-|x|-|y(x)|} dx \le \int_{-1}^1 e^{1-|x|-y(x)}dx= \mathcal{I}[y] \quad\forall y\in C_0([-1,1], \mathbb{R})[/tex] as we want. Moreover:
[tex]\displaystyle\mathcal{I}[|y|]=\mathcal{I}[y]\iff \int_{-1}^1 e^{1-|x|-y(x)}dx- \int_{-1}^1 e^{1-|x|-|y(x)|} dx = 0\implies \text{for linear property of integral operator}\implies[/tex]
[tex]\displaystyle\int_{-1}^1 e^{1-|x|-y(x)}- e^{1-|x|-|y(x)|}dx =0[/tex].
Now some notes:
1. The function [tex]e^{1-|x|-y(x)}- e^{1-|x|-|y(x)|}[/tex] is not negative: it follows from (1.1)
2. It's continue because it's sum of continue functions.
Then:
[tex]\displaystyle\int_{-1}^1 e^{1-|x|-y(x)}- e^{1-|x|-|y(x)|}dx =0\iff e^{1-|x|-y(x)}- e^{1-|x|-|y(x)|}=0[/tex] if and only if
[tex]|y(x)|=y_{+}(x)+y_{-}(x)=y_{+}(x)-y_{-}(x)= y(x) \iff y_-(x)= 0\ \forall x\in [-1,1][/tex].
mmm please check it, i hope there are no mistakes
.
[tex]\displaystyle \mathcal{I}[|y|]= \int_{-1}^1 e^{1-|x|-|y(x)|} dx \le \int_{-1}^1 e^{1-|x|-y(x)}dx= \mathcal{I}[y] \quad\forall y\in C_0([-1,1], \mathbb{R})[/tex].
Knowing that [tex]|y(x)|= y_+(x)+y_-(x)[/tex] then [tex]y(x)= y_+(x)-y_-(x)\le y_+(x)+y_-(x)= |y(x)|[/tex] hence [tex]1-|x|-|y(x)|\le 1-|x|-y(x)\quad\forall x\in [-1,1][/tex] (1.1). From the inequality it follows that:
[tex]\displaystyle \mathcal{I}[|y|]= \int_{-1}^1 e^{1-|x|-|y(x)|} dx \le \int_{-1}^1 e^{1-|x|-y(x)}dx= \mathcal{I}[y] \quad\forall y\in C_0([-1,1], \mathbb{R})[/tex] as we want. Moreover:
[tex]\displaystyle\mathcal{I}[|y|]=\mathcal{I}[y]\iff \int_{-1}^1 e^{1-|x|-y(x)}dx- \int_{-1}^1 e^{1-|x|-|y(x)|} dx = 0\implies \text{for linear property of integral operator}\implies[/tex]
[tex]\displaystyle\int_{-1}^1 e^{1-|x|-y(x)}- e^{1-|x|-|y(x)|}dx =0[/tex].
Now some notes:
1. The function [tex]e^{1-|x|-y(x)}- e^{1-|x|-|y(x)|}[/tex] is not negative: it follows from (1.1)
2. It's continue because it's sum of continue functions.
Then:
[tex]\displaystyle\int_{-1}^1 e^{1-|x|-y(x)}- e^{1-|x|-|y(x)|}dx =0\iff e^{1-|x|-y(x)}- e^{1-|x|-|y(x)|}=0[/tex] if and only if
[tex]|y(x)|=y_{+}(x)+y_{-}(x)=y_{+}(x)-y_{-}(x)= y(x) \iff y_-(x)= 0\ \forall x\in [-1,1][/tex].
mmm please check it, i hope there are no mistakes
.
"Mathematico":
Question:
Is it possible to prove the gugo's problem, assuming that y is a variable sign function ("funzione a segno variabile") and using [tex]y(x)= y_+(x)-y_-(x)[/tex] where [tex]y_+(x)= \max(0, y(x))[/tex] while [tex]y_-(x)= \max(0, -y(x))[/tex]?
My aim is to show that [tex]y_-(x) =0\quad\forall x\in [-1,1][/tex]... Maybe there is a better way
No, there isn't.
To prove [tex]$\overline{y}_- =0$[/tex] you have to use the "hint" given by gac.
"gac":
It is enough to observe that, if $y\in C_0([-1,1])$, then $|y|\in C_0([-1,1])$ (with the same $L^1$ norm).
Moreover
$I[|y|] \le I[y]$
with equality holding if and only if $|y| = y$, i.e. if and only if $y$ is non-negative.
This is my solution.
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