Linear functional

[Camillo]

Let $X=C^1 ([0,1]) $ be the vector space of all continuous functions with continuous prime derivative $v : [0,1] rarr RR $ and $ (X,||.||_oo) $ the normed space with maximum norm.

Show that the linear functional : $ v(x) rarr v '(0) $ is not continuous.

EDIT : Corrected to read $X = C^1([0,1]) etc $ .

[david_e]

Let $\alpha$ be a positive real number and $\alpha > 2$. Now we consider the family of functions $f_\alpha:$

$f_\alpha = {(\alpha x \qquad x\in (0,1/\alpha)),(1-\alpha (x-1/\alpha) \qquad x \in (1/\alpha,2/\alpha)),(0 \qquad x>2/\alpha):}$

we have $||f_\alpha||=1$ for every $\alpha$. But $f'_\alpha(0)=\alpha$. So the linear operator $f_\alpha(x) rarr f'_\alpha(0)$ is unbounded so it can't be continuous.

*** EDIT ***
Some corrections...

PS: Sorry my English is terrible...

[Luca.Lussardi]

Camillo.... the definition of $v$ is ill posed since $v \in C^0([0,1])$ has not, in general, the prime derivative in $0$...

[Camillo]

Camillo.... the definition of $v$ is ill posed since $v \in C^0([0,1])$ has not, in general, the prime derivative in $0$...

Right, Luca it was a typing mistake ; I correct it.

[Camillo]

@David_e : the initial text was wrong ; now it is correct.
If you want to find the solution again , you are welcome

[david_e]

@David_e : the initial text was wrong ; now it is correct.
If you want to find the solution again , you are welcome

In fact I noticed it was strange to think of a derivate over $C^0$...

I thought you ment the operator from $C^0([0,1])$ to $[-\infty,+\infty]$ (taking also the $\infty$ values) using the right incremental limit. (I am not sure this is the right way to say "limite incrementale", but I don't have time right now to check the dictionary sorry ). I think this is not ill posed. Right?

I will try to solve it again this afternoon, but I need to know: what do you mean with $||\cdot||_\infty$?

Is it:

$||f||_\infty = \text{max}_{x\in [0,1]} |f(x)| $

or:

$||f||_\infty = \text{max}_{x\in [0,1]} |f(x)| + \text{max}_{x\in [0,1]} |f'(x)| $

I think the first one as with the second one the operator will be continuous, but I usually use $||\cdot||_\infty$ for the second one...

If it is the first one my proof holds, but I have to smouth my functions in $1/\alpha$ and $2/\alpha$...

[Camillo]

It is the first one as you thought ; but you have to smooth indeed the function

[Thomas]

Ei... I've forgotten almost everything, but I want to try

If we consider the succession of functions:

$f_n(x)=(sen(n^2x))/n$

I guess that this sequence:

- tends uniformly (and so in the max norm) to $g(x)=0$
- its prime derivative is $ncos(n^2x)$ and so it is continuous for every n;
- the prime derivative calculated in 0 is $n$, so the succession $f_n'(0)$ does not converge in R;

dealing with metric-spaces, I guess this is sufficient for concluding the exercise... or not?

[david_e]

Yes I think this is sufficient....