Two optimization problems

[gugo82]

Not so many time ago, I proposed (and solved) an optimization problem for the inequality $c*||u'||_oo<=||u||_(C^1)$, con $u \in C^1([0,1])$ e $||u||_(C^1):=||u||_oo+||u'||_oo$ (see here for further details).

Now I propose the following two optimization problems which are similar to the previous.

***

Let $E$ be a $RR$-vector space and $|\cdot |_1,|\cdot |_2:E\to [0,+oo [$ two (non trivial) seminorms on $E$.
Suppose that there exists a positive number $\bar(c)$ s.t. the inequality:

(a) $\quad \bar(c)*|u|_1<=|u|_2$

holds true for all $u \in E$.
Under this hypothesis, it make sense to ask what is the greatest constant $C>0$ s.t. inequality $C*|u|_1<=|u|_2$ holds for all $u \in E$: a problem like this is usually called optimization problem for the inequality (a) and its solution $C$ is said to be the best constant in (a); moreover the inequality:

(b) $\quad C*|u|_1<= |u|_2$

(which is (a) with the best constant) is said to be the optimal form of (a).

The hypothesis made above is equivalent to the fact that the set $\Gamma:=\{ c>=0 : AA u \in E, c*|u|_1<=|u|_2\}$ isn't empty: then the best constant is by definition $C="sup "Gamma$.
It's not difficult to see that another characterization of the best constant $C$ is the following:

(c) $\quad C= "inf "\{ |u|_2/|u|_1, " with " u \in E " s.t. " |u|_1!=0\}$.

Suppose, now, that $C>0$ is the best constant for (a). It makes sense to ask if there exists any vector $v \in E\setminus \{0\}$ which satisfies (b) with equality and, if the answer is positive, if it's possible to identify the set of all the vectors for which equality holds in (b): this problem is called characterization of the equality case in (b) (the optimal form of (a)).
Again, if the answer to the first question is positive, it's possible to write $min$ instead of $"inf"$ in (c).

***

Problem:

1. Let $C^1([0,1])$ be normed with the usual norm $||u||_(C^1):=||u||_oo+||u'||_oo$ (here $||\cdot ||_oo$ is the maximum norm on $C([0,1])$).
Solve the optimization problem for the inequality:

(*) $\quad AA u in C^1([0,1]), \quad c*||u||_oo <= ||u||_(C^1) \quad$;

characterize the equality case in (*) when it is written in optimal form.

2. Let $(X,M,\mu)$ be a bounded measure space (i.e. a measure space with $\mu(X)<+oo$) and $p\in [1,+oo[$.
Solve the optimization problem for the inequality:

(**) $\quad AA u in L^oo(\mu ), \quad c*||u||_p <= ||u||_oo \quad$;

put (**) in optimal form and find some functions that satisfy (**) in optimal form with equality.

Hints: 1. It's easy to guess the possible best constant.
2. Use an easy Hölder's inequality.

[gugo82]

"Gugo82":2. Let $(X,M,\mu)$ be a bounded measure space (i.e. a measure space with $\mu(X)<+oo$) and $p\in [1,+oo[$.
Solve the optimization problem for the inequality:

(**) $\quad AA u in L^oo(\mu ), \quad c*||u||_p <= ||u||_oo \quad$;

put (**) in optimal form and find some functions that satisfy (**) in optimal form with equality.

Hints: 2. Use an easy Hölder's inequality.

Fixed $u \in L^oo(mu)$, we have also $|u|^p \in L^oo(mu)$ with $|||u|^p||_oo =||u||_oo^p$ and:

$||u||_p^p<=mu(X)*||u||_oo^p$

by Hölder's inequality; then:

(2) $\quad mu^(-1/p)(X) ||u||_p<= ||u||_oo$

and (**) holds true for all $c in [0, mu^(-1/p)(X)]$. The best constant $C$ in (**) is $>= mu^(-1/p)(X)$ and, to show that infact $C=mu^(-1/p)(X)$, it suffices to find a function $u \in L^oo(mu)$ s.t. equality holds in (2).
It's trivial to see that $u=\chi_X$* actually satisfies (2) with equality sign, so $C=mu^(-1/p)(X)$ is the best constant in (**) and (2) is (**) in optimal form.

For any fixed number $a in RR$, the function $a*\chi_X \in L^oo(mu)$ satisfies (2) with equality sign; if together with $a \in RR$ we choose a sequence $\mathcal(E)=(E_n)$ of measurable sets in $X$ s.t.:

$\{ (mu(E_n\cap E_m)=0 " for all "n!=m in NN) , (mu(X\setminus \bigcup_(n=1)^(+oo) E_n)=0) :}$,

then $u_(a,\mathcal{E}):=a*\sum_(n=1)^(+oo)(-1)^(n-1)\chi_(E_n)$ also satisfies (2) with equality.
It seems reasonable that equality holds in (2) iff the function $u$ is of type $u_(a,\mathcal(E))$, but I'm not able to prove this statement.

__________
* Here $chi_E$ is the characteristic function of the subset $E\subseteq X$, i.e.:

$chi_E(x):=\{(1, ", if " x\in E),(0, ", otherwise"):} \quad$;

therefore $chi_X$ is the function which assumes the value $1$ over the whole of $X$.

[gugo82]

"Gugo82":1. Let $C^1([0,1])$ be normed with the usual norm $||u||_(C^1):=||u||_oo+||u'||_oo$ (here $||\cdot ||_oo$ is the maximum norm on $C([0,1])$).
Solve the optimization problem for the inequality:

(*) $\quad AA u in C^1([0,1]), \quad c*||u||_oo <= ||u||_(C^1) \quad$;

characterize the equality case in (*) when it is written in optimal form.

Hints: 1. It's easy to guess the possible best constant.

For all $u$ of class $C^1$ we have:

(1) $\quad ||u||_oo<=||u||_(C^1) \quad$,

then the inequality (*) holds true for each $c\in [0,1]$; therefore the best constant $C$ for (*) is $>=1$.
Our aim is to show that actually $C=1$: to achive this, it suffices to find a function $u \in C^1([0,1])$ s.t. equality holds in (1).
If we choose $a >0$ and put $u_a(x):=a$, we find $||u_a||_oo=a=||u_a||_(C^1)$, so that equality in (1) holds trivially for $u_a$.

Actually (1) is (*) in optimal form. To characterize the equality case in (1) it suffices to note that $||u||_oo=||u||_(C^1)$ iff $||u'||_oo=0$: therefore equality holds in (1) only for constant functions.