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
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
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
What if one imposes the constraint [tex]$||y||_1=\int_{-1}^1 |y(x)|\text{ d} x=1$[/tex]?
Can one get a priori information on the behaviour of the minima of the functional [tex]$\int_{-1}^1 e^{1-|x|-y(x)} \taxt{ d}x$[/tex]?
I'm going to make the last question more precise...
Suppose there exists [tex]$\overline{y} \in C^0([-1,1])$[/tex] which is a minimum of [tex]$\mathcal{I}[y]:=\int_{-1}^1 e^{1-|x|-y(x)} \taxt{ d}x$[/tex] under the constraint [tex]$||y||_1=1$[/tex].
Can we say a priori (i.e. without explicitly finding the minima) that [tex]$\overline{y} \geq 0$[/tex]?
Can one get a priori information on the behaviour of the minima of the functional [tex]$\int_{-1}^1 e^{1-|x|-y(x)} \taxt{ d}x$[/tex]?
I'm going to make the last question more precise...
Suppose there exists [tex]$\overline{y} \in C^0([-1,1])$[/tex] which is a minimum of [tex]$\mathcal{I}[y]:=\int_{-1}^1 e^{1-|x|-y(x)} \taxt{ d}x$[/tex] under the constraint [tex]$||y||_1=1$[/tex].
Can we say a priori (i.e. without explicitly finding the minima) that [tex]$\overline{y} \geq 0$[/tex]?
"cirasa":
Sorry, Mathematico, I didn't understand very well your post
Do not worry about that!
It's my fault, my english sucks! 
"Mathematico":
Hi cirasa, I found that inequality by monotony, in fact:
[tex]\displaystyle\int_{-1}^1 e^{1-|x|-y(x)} dx \ge \int_{-1}^1 2-|x|- y(x)dx = 2[/tex]. Can i say that the equality holds iff [tex]1-|x|-y(x) =0[/tex]?
Sorry, Mathematico, I didn't understand very well your post.
And I don't know what it means that "$\bar{y}$ solves the minimum problem".
I thought it meant that
$\int_{-1}^1e^{1-|x|-\bar{y}(x)dx}\le\int_{-1}^1e^{1-|x|-y(x)dx}$ for all $y\in C^0$ with $\int_{-1}^1y(x)dx=1$.
Instead I have to prove that equality holds iff $y(x)=1-|x|$, too.
I'm sorry...
Great solution elgiovo!
It comes out that one can use the Euler-Lagrange equation to minimize the functional, but a straightforward application leads to strange results, since the hypotheses of the theorem are not immediately satisfied. Some artifices are thus needed.
The function [tex]1-|x|-y(x)[/tex] is not differentiable, so we may consider the substitution [tex]f(x)=|x|+y(x)[/tex], so that [tex]L(x,f,f')=e^{1-f}[/tex] is differentiable in [tex]f[/tex]. Anyway, [tex]f[/tex] is not differentiable, so it's better to put [tex]f'(x)=|x|+y(x)[/tex]. This way, [tex]L(x,f,f')=e^{1-f'}[/tex] becomes differentiable in both [tex]f[/tex] and [tex]f'[/tex]. The partial derivatives are
[tex]\displaystyle \frac{\partial L(x,f,f')}{\partial f}=0[/tex], [tex]\displaystyle \frac{\partial L(x,f,f')}{\partial f'}=-e^{1-f'(x)}[/tex].
The Euler-Lagrange equation is thus [tex]\displaystyle \frac{\text{d}}{\text{d}x} e^{1-f'(x)}=0[/tex], or [tex]-e^{1-f'(x)}f''(x)=0[/tex], which leads to [tex]f''(x)=0[/tex].
Remembering the previous substitution, we must have [tex]\text{sgn}(x)+y'(x)=0[/tex], which is a differential equation whose solution is (intentionally forgetting some Dirac distribution arising somewhere ) [tex]y(x)=A-|x|[/tex], with [tex]A \in \mathbb{R}[/tex].
The constraint [tex]\displaystyle \int_{-1}^1 y(x)\text{d}x=1[/tex] finally gives [tex]A=1[/tex].
The function [tex]1-|x|-y(x)[/tex] is not differentiable, so we may consider the substitution [tex]f(x)=|x|+y(x)[/tex], so that [tex]L(x,f,f')=e^{1-f}[/tex] is differentiable in [tex]f[/tex]. Anyway, [tex]f[/tex] is not differentiable, so it's better to put [tex]f'(x)=|x|+y(x)[/tex]. This way, [tex]L(x,f,f')=e^{1-f'}[/tex] becomes differentiable in both [tex]f[/tex] and [tex]f'[/tex]. The partial derivatives are
[tex]\displaystyle \frac{\partial L(x,f,f')}{\partial f}=0[/tex], [tex]\displaystyle \frac{\partial L(x,f,f')}{\partial f'}=-e^{1-f'(x)}[/tex].
The Euler-Lagrange equation is thus [tex]\displaystyle \frac{\text{d}}{\text{d}x} e^{1-f'(x)}=0[/tex], or [tex]-e^{1-f'(x)}f''(x)=0[/tex], which leads to [tex]f''(x)=0[/tex].
Remembering the previous substitution, we must have [tex]\text{sgn}(x)+y'(x)=0[/tex], which is a differential equation whose solution is (intentionally forgetting some Dirac distribution arising somewhere
) [tex]y(x)=A-|x|[/tex], with [tex]A \in \mathbb{R}[/tex].The constraint [tex]\displaystyle \int_{-1}^1 y(x)\text{d}x=1[/tex] finally gives [tex]A=1[/tex].
Hi cirasa , I found that inequality by monotony, in fact:
[tex]\displaystyle\int_{-1}^1 e^{1-|x|-y(x)} dx \ge \int_{-1}^1 2-|x|- y(x)dx = 2[/tex]. Can i say that the equality holds iff [tex]1-|x|-y(x) =0[/tex]?
, I found that inequality by monotony, in fact:[tex]\displaystyle\int_{-1}^1 e^{1-|x|-y(x)} dx \ge \int_{-1}^1 2-|x|- y(x)dx = 2[/tex]. Can i say that the equality holds iff [tex]1-|x|-y(x) =0[/tex]?
To solve this exercise you can use the Jensen inequality:
If [tex]\displaymath \varphi[/tex] is a convex function, then
[tex]\displaymath \varphi\left(\frac{1}{2}\int_{-1}^1f(x)\,dx\right)\leq\frac{1}{2}\int_{-1}^1\varphi\circ f(x)dx[/tex] (*)
Put [tex]\displaymath f(x)=1-|x|-y(x)[/tex] and [tex]\displaymath \varphi(x)=e^x[/tex]. Using (*) you have
[tex]\displaymath 1=\varphi(0)=\varphi\left(\frac{1}{2}\int_{-1}^11-|x|-y(x)\,dx\right)\leq\frac{1}{2}\int_{-1}^1e^{1-|x|-y(x)}\,dx[/tex].
So for all [tex]\displaymath y\in C^0[/tex] you have
[tex]2\leq\int_{-1}^1e^{1-|x|-y(x)}\,dx[/tex].
If [tex]\displaymath \varphi[/tex] is a convex function, then
[tex]\displaymath \varphi\left(\frac{1}{2}\int_{-1}^1f(x)\,dx\right)\leq\frac{1}{2}\int_{-1}^1\varphi\circ f(x)dx[/tex] (*)
Put [tex]\displaymath f(x)=1-|x|-y(x)[/tex] and [tex]\displaymath \varphi(x)=e^x[/tex]. Using (*) you have
[tex]\displaymath 1=\varphi(0)=\varphi\left(\frac{1}{2}\int_{-1}^11-|x|-y(x)\,dx\right)\leq\frac{1}{2}\int_{-1}^1e^{1-|x|-y(x)}\,dx[/tex].
So for all [tex]\displaymath y\in C^0[/tex] you have
[tex]2\leq\int_{-1}^1e^{1-|x|-y(x)}\,dx[/tex].
"apatriarca":
There are several different meanings of resolve (http://www.merriam-webster.com/dictionary/resolve). One of the definitions in the dictionary is "to find a mathematical solution to" so I think it's correct to use resolve in this context even if I think solve is more common.
Ok,
. I' m happy to know that it's correct, but to tell the truth i used it without knowing its existence . I just anglicized the term "risolvere"
There are several different meanings of resolve (http://www.merriam-webster.com/dictionary/resolve). One of the definitions in the dictionary is "to find a mathematical solution to" so I think it's correct to use resolve in this context even if I think solve is more common.
"dissonance":
I think "solves" would fit better in there. In fact, I believe that "to resolve" is more like the Italian "decidere", as in "I can't resolve myself to go". I'm not sure, though, and my dictionary doesn't fully unveil this. Let's wait for the opinion of the experienced.
Thanks a ton!
"Mathematico":I think "solves" would fit better in there. In fact, I believe that "to resolve" is more like the Italian "decidere", as in "I can't resolve myself to go". I'm not sure, though, and my dictionary doesn't fully unveil this. Let's wait for the opinion of the experienced.
I'm not sure that this resolves the problem
This problem is driving me crazy 
, but at the same time I love it .
Some consideration about this:
Let [tex]f(z)= e^z-z-1[/tex], the function has a minimum value for [tex]z_0 = 0[/tex] in fact [tex]f(z)[/tex] decreases for [tex]z<0[/tex] while for [tex]z>0, f(x)[/tex] grows, so [tex](z_0, f(z_0)) = (0, 0)[/tex] is a global minimum point then
[tex]\forall z\in \mathbb{R}, f(z)\ge0\implies e^z\ge z+1[/tex], the equality holds if and only if [tex]z=0[/tex].
Let [tex]z= 1-|x|-y(x)[/tex], we have that: [tex]e^{1-|x|-y(x)}\ge 1-|x|-y(x)+1[/tex] the equality holds iff [tex]1-|x|-y(x)=0\implies y(x) = 1-|x|[/tex]
Now, my idea is to use the monotony of integral operator, but I'm not sure that this solves the problem
Any ideas?
, but at the same time I love it .Some consideration about this:
Let [tex]f(z)= e^z-z-1[/tex], the function has a minimum value for [tex]z_0 = 0[/tex] in fact [tex]f(z)[/tex] decreases for [tex]z<0[/tex] while for [tex]z>0, f(x)[/tex] grows, so [tex](z_0, f(z_0)) = (0, 0)[/tex] is a global minimum point then
[tex]\forall z\in \mathbb{R}, f(z)\ge0\implies e^z\ge z+1[/tex], the equality holds if and only if [tex]z=0[/tex].
Let [tex]z= 1-|x|-y(x)[/tex], we have that: [tex]e^{1-|x|-y(x)}\ge 1-|x|-y(x)+1[/tex] the equality holds iff [tex]1-|x|-y(x)=0\implies y(x) = 1-|x|[/tex]
Now, my idea is to use the monotony of integral operator, but I'm not sure that this solves the problem
Any ideas?
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