Maximizing the area enclosed between two graphs

Messaggioda gugo82 » 19/07/2019, 18:37

Problem:

Let $M$ be the space of monotone increasing functions of $[0,1]$ in itself and let $L$ be the space of linear functions over $[0,1]$.

1. For each $f in M$, find the projection of $f$ onto $L$, i.e. the function $f^** in L$ s.t. $int_0^1 (f(x) - f^**(x))^2 text(d) x = min_(phi in L) int_0^1 (f(x) - phi (x) )^2 text(d) x$.

Testo nascosto, fai click qui per vederlo
Hint: Let $phi (x) = a x + b$ and evaluate $a,b$ in such a way that $int_0^1 (f(x) - phi (x))^2 text(d) x$ attains its minimum value.


2. Prove that the functional\( A(f) := \int_0^1 | f(x) - f^*(x)| \text{d} x = \| f - f^* \|_{1,[0,1]}\), which gives the value of the area enclosed by the graphs of $f$ and $f^**$, attains its maximum value over the set $M$.

Testo nascosto, fai click qui per vederlo
Hint: $M$ is a convex cone and $A(f)$ is convex on $M$, therefore it attains its maximum on some extreme point of $M$.


3. Prove that $max_(f in M) A(f) = 1/2$ and try to explicitly find some function $y in M$ s.t. $A(y) = 1/2$.
Sono sempre stato, e mi ritengo ancora un dilettante. Cioè una persona che si diletta, che cerca sempre di provare piacere e di regalare il piacere agli altri, che scopre ogni volta quello che fa come se fosse la prima volta. (Freak Antoni)
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gugo82
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Re: Maximizing the area enclosed between two graphs

Messaggioda gugo82 » 12/08/2019, 03:54

gugo82 ha scritto:Let $M$ be the space of monotone increasing functions of $[0,1]$ in itself and let $L$ be the space of linear functions over $[0,1]$.

1. For each $f in M$, find the projection of $f$ onto $L$, i.e. the function $f^** in L$ s.t. $int_0^1 (f(x) - f^**(x))^2 text(d) x = min_(phi in L) int_0^1 (f(x) - phi (x) )^2 text(d) x$.

Testo nascosto, fai click qui per vederlo
Hint: Let $phi (x) = a x + b$ and evaluate $a,b$ in such a way that $int_0^1 (f(x) - phi (x))^2 text(d) x$ attains its minimum value.

Testo nascosto, fai click qui per vederlo
If we let $f in M$ and $phi(x) = a x + b in L$, then we have:
\[
\begin{split}
\int_0^1 \big( f(x) - \phi (x) \big)^2\ \text{d} x &= \int_0^1 f^2(x)\ \text{d} x - 2a\ \int_0^1 x f(x)\ \text{d} x - 2b\ \int_0^1 f(x)\ \text{d} x \\
&\phantom{=} + a^2\ \int_0^1 x^2\ \text{d} x + 2ab\ \int_0^1 x\ \text{d} x + b^2\ \int_0^1 1\ \text{d} x \\
&= \frac{1}{3}\ a^2 + ab + b^2 - 2Ea - 2Fb + G \; ,
\end{split}
\]
in which $E := int_0^1 x f(x) text(d) x, F := int_0^1 f(x) text(d) x, G := int_0^1 f^2(x) text(d) x$. Therefore the problem of finding $phi$ minimizing the integral $int_0^1 (f(x) - phi (x) )^2 text(d) x$ recasts as a simple free minimum problem for the two variables quadratic function $Q(a,b) := \frac{1}{3}\ a^2 + ab + b^2 - 2Ea - 2Fb + G$.

Using standard Calculus techniques, we find:
\[
\begin{split}
Q_a (a,b) &= \frac{2}{3}\ a + b - 2E \\
Q_b (a,b) &= a + 2b - 2F
\end{split}
\]
and a unique critical point of coordinates $a^** = 12E - 6F, b^** = 4F - 6E$, which is indeed a (absolute global strong) minimum for $Q$, because $Q$ is strictly convex over $RR^2$.
Therefore:
\[
\begin{split}
f^* (x) &= a^* x + b^* \\
&= 6 \left( 2 \int_0^1 t f(t) \text{d} t - \int_0^1 f(t) \text{d} t\right)\ x + 2\ \left( 2 \int_0^1 f(t) \text{d} t - 3 \int_0^1 t f(t) \text{d} t \right) \\
&= \left( 6\ \int_0^1 ( 2 t - 1)\ f(t)\ \text{d} t\right)\ x + \left( 2\ \int_0^1 (2 - 3 t)\ f(t)\ \text{d} t\right)
\end{split}
\]
is the projection we were looking for.
Sono sempre stato, e mi ritengo ancora un dilettante. Cioè una persona che si diletta, che cerca sempre di provare piacere e di regalare il piacere agli altri, che scopre ogni volta quello che fa come se fosse la prima volta. (Freak Antoni)
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gugo82
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