Basic linear functional analysis' exercise

[Luca.Lussardi]

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 $L : (X,||.||_1) \to (X,||.||_\infty)$.

[Camillo]

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 .

[Fioravante Patrone]

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

[Luca.Lussardi]

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...

[Camillo]

OK clear , Luca and Fioravante : the sup norm is more general !!
Now there is only to solve the "basic exercise "

[Luca.Lussardi]

The exercise is not a "basic exercise", but an exercise of "basic linear functional analysis".

[Luca.Lussardi]

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 $L$...

[Camillo]

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 $0=L(0)$.

[Mega-X]

ok now clear me something..

Why are you talking English instead of talking Italian?

You are Italian, aren't you?

Mega-X

[Camillo]

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.