Basic linear functional analysis' exercise
Let $X=C^0([0,1])$ be the vector space of all continuous functions $u : [0,1] \to \RR$; let
$||u||_1=\int_0^1 |u(x)|dx$ and $||u||_\infty=$sup$|u(x)|$. Show that $(X,||.||_1)$ and $(X,||.||_2)$ are infinite dimensional normed spaces and find a non continuous linear map $L : (X,||.||_1) \to (X,||.||_\infty)$.
$||u||_1=\int_0^1 |u(x)|dx$ and $||u||_\infty=$sup$|u(x)|$. Show that $(X,||.||_1)$ and $(X,||.||_2)$ are infinite dimensional normed spaces and find a non continuous linear map $L : (X,||.||_1) \to (X,||.||_\infty)$.
Risposte
Scusate l'italiano ma faccio fatica a scrivere in inglese.
Forse possiamo aprire la sezione in inglese del forum, ci spostiamo i topic già scritti in inglese.
Quale potrebbe essere il nome del forum?
Quale potrebbe essere la descrizione?
Suggerirei che entrambe fossero in inglese.
Forse possiamo aprire la sezione in inglese del forum, ci spostiamo i topic già scritti in inglese.
Quale potrebbe essere il nome del forum?
Quale potrebbe essere la descrizione?
Suggerirei che entrambe fossero in inglese.
Yes I am italian as you are !
We tried last year to have a part of the Forum in English considering the importance English has in all technical and scientific applications all around the world; just to keep in exercise with the language.
We tried last year to have a part of the Forum in English considering the importance English has in all technical and scientific applications all around the world; just to keep in exercise with the language.
ok now clear me something.. 
Why are you talking English instead of talking Italian?
You are Italian, aren't you?
Mega-X
Why are you talking English instead of talking Italian?
You are Italian, aren't you?
Mega-X
Finally, taking advantage of the suggestions……
*It is easy to show that $||.||_(1) $ and $||.||_(oo) $ are norms on $X $ since both satisfy the following conditions :
-$||x|| >=0 ;||x||=0 $ iff $x=0 $(Positivity).
-$||lambda*x|| = lambda*||x||; AA lambda in RR;AA x in X $ (Homogenity).
-$||x+y|| <= ||x|| +||y || ; AAx,y in X $ (Triangular inequality).
*$X $ is an infinite dimensional space: infact all polinomial functions on $[0,1]$ are a subset of $X$ and have dimension = n and no limit is fixed for n .
*Consider the identity map L such that $L(u ) = u $ ,$ AA u in X $.
L is linear but is not continuous .
Infact let us consider the sequence $(u_n)$ in $ X$ given by :
$u_n(x)= x^n; x in [0,1]$
$||u_n||_(1) = int_0^1|x^n|*dx = 1/(n+1) $ $rarr 0 $ for$n rarr oo$
But $ L(x_n)=x^n$ does not converge to $ 0 = L(0) $in $(X,||.||_(2))$ since :
$||u_n||_(2)= $ $max_(0,1) |x^n| =1 $, $AA n in NN $.
That is the sequence $(x^n)$ converges to $0$ in $(X,||.||_(1))$ while in $(X,||.||_(2)$ the sequence $L(x_n ) = x^n $ is always equal to $1$ ,$AA n in NN $ and does not converge to $0=L(0)$.
*It is easy to show that $||.||_(1) $ and $||.||_(oo) $ are norms on $X $ since both satisfy the following conditions :
-$||x|| >=0 ;||x||=0 $ iff $x=0 $(Positivity).
-$||lambda*x|| = lambda*||x||; AA lambda in RR;AA x in X $ (Homogenity).
-$||x+y|| <= ||x|| +||y || ; AAx,y in X $ (Triangular inequality).
*$X $ is an infinite dimensional space: infact all polinomial functions on $[0,1]$ are a subset of $X$ and have dimension = n and no limit is fixed for n .
*Consider the identity map L such that $L(u ) = u $ ,$ AA u in X $.
L is linear but is not continuous .
Infact let us consider the sequence $(u_n)$ in $ X$ given by :
$u_n(x)= x^n; x in [0,1]$
$||u_n||_(1) = int_0^1|x^n|*dx = 1/(n+1) $ $rarr 0 $ for$n rarr oo$
But $ L(x_n)=x^n$ does not converge to $ 0 = L(0) $in $(X,||.||_(2))$ since :
$||u_n||_(2)= $ $max_(0,1) |x^n| =1 $, $AA n in NN $.
That is the sequence $(x^n)$ converges to $0$ in $(X,||.||_(1))$ while in $(X,||.||_(2)$ the sequence $L(x_n ) = x^n $ is always equal to $1$ ,$AA n in NN $ and does not converge to $0=L(0)$.
Hint1:It is easy to show that $||.||_1$ and $||.||_\infty$ are norms on $X$. Consider all polynomial functions on $[0,1]$... this is a subspace of $X$ with....
Hint2: Consider the identity map $L$...
Hint2: Consider the identity map $L$...
The exercise is not a "basic exercise", but an exercise of "basic linear functional analysis".
OK clear , Luca and Fioravante : the sup norm is more general !!
Now there is only to solve the "basic exercise "
Now there is only to solve the "basic exercise "
Right Fioravante... when I consider the "max-norm" I always prefer to use the supremum notation instead of the maximum, even if the sup is a maximum.
Don't worry Camillo, if $u \in C^0([0,1])$ you can read $||u||_\infty=$max...
Don't worry Camillo, if $u \in C^0([0,1])$ you can read $||u||_\infty=$max...
Camillo, I bet it is just a matter of habit.
I remember, from my "previous life", this standard usage, stemming, perhaps, from the use of the functional space $L^{oo}$, where the "ess sup" is used.
Of course, as you say, Luca could have used as well the "max norm"
Have a nice weekend
Bye
I remember, from my "previous life", this standard usage, stemming, perhaps, from the use of the functional space $L^{oo}$, where the "ess sup" is used.
Of course, as you say, Luca could have used as well the "max norm"
Have a nice weekend
Bye
Question :
Why has been considered the sup norm instead of the max norm ?
The Weierstrass theorem anyhow assures the existence of the maximun for a function on a compact .
Why has been considered the sup norm instead of the max norm ?
The Weierstrass theorem anyhow assures the existence of the maximun for a function on a compact .
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