Unbounded Operator

[Camillo]

Show that the Derivative $ [D:(Df )(x) =f'(x)] $ is an unbounded linear operator on the vector space of smooth functions equipped with the sup norm.

[Principe2]

We define $f_n(x)=x^n$ Then $||f'_n(x)||=n||x^{n-1}||=n||x^n||$. Thus, doesn't exist a costant $M$ such that
$||Df_n(x)||\leM||f(x)||$, where $D$ is the canonical differential operator.

[Camillo]

Hint : to show that Derivative is an unbounded operator find a suitable sequence of functions and work on it.The sequence must be such that [size=150]cannot[/size] be found a positive number M.. (refer to David_e post).

[david_e1]

"Operatore lineare illimitato sullo spazio delle funzioni regolari ($C^1(a,b)$) dotato della norma del sup"

This question sounds familiar to me!
(already solved this once when the English corner was just a couple of post in "our forum").

Just an hint. Limited linear operator from $X$ to $Y$ (both normed space) means that there exists a positive constant $M$ such that:

$ || T f ||_Y \leq M || f ||_X $

for any element $f$ of the normed space $X$. Where $||\cdot||_X$ is the norm in $X$. In this case:

$ || f ||_X = \text{sup}_{(a,b)} | f(x) | $

[Mega-X]

hmm can you translate for me this sentence?

"unbounded linear operator on the vector space of smooth functions equipped with the sup norm."

i used to translate in this way: "Operatore lineare illimitato sullo spazio vettoriale di funzioni liscie (?!) equipaggiate (?!?!) con la norma superiore"