Is this a good definition?...

[fu^2]

We can consider the set of function [tex]f\in L^{1}(\mathbb{R})[/tex].

Then [tex]C_c^{\infty}[/tex] (the set of all function infinitely differentiable with compact support) is dense in [tex]L^{1}(\mathbb{R})[/tex], i.e. fix [tex]f\in L^{1}(\mathbb{R})[/tex], for each [tex]\epsilon>0[/tex] there exists [tex]g_{\epsion}[/tex]s.t. [tex]||g_{\epsion}-f||<\epsilon[/tex], therefore we can buid a sequence [tex]g_n[/tex] s.t. [tex]g_n\to f[/tex] in norm.

Then is "natural" think a method to define a "derivate" of [tex]f[/tex].
Today i have had an idea : We can say that [tex]f[/tex] is differentiable in same sense if [tex]g_n'[/tex] converge to some [tex]h\in L^1(\mathbb{R})[/tex]. In this case we set [tex]f'=h[/tex],

the problem is that the sequence [tex]g_n[/tex] which converges to [tex]f[/tex] isn't unique! therefore the derivative depends on the sequence [tex]g_n'[/tex], i.e. it's not well defined!

So it is possible "to walk in this way" to succeed in defining a generalization of the derivative for all integrable function on the real axis (i.e. is it possible to save my definition )?

[fu^2]

"gugo82":Once De Giorgi said: "Chi cerca, trova; chi ricerca, ritrova"... Well, fu^2, you've found one good way to generalize the concept of derivative: in fact you've re-discovered the definition of weak derivative.
[...]
Note that, in general, if [tex]$(g_n)\subset C_c^\infty$[/tex] is s.t. [tex]$||g_n-f||_1\to 0$[/tex] with [tex]$f\in W^{1,1}$[/tex], then [tex]$(g_n^\prime)$[/tex] have not to converge in [tex]$L^1$[/tex] to the weak derivative [tex]$f^\prime$[/tex]: the previous example of dissonance is illuminating.

In conclusion at my reasoning, if $f$ satisfy my definition, i.e. $u_n\to f$ in $L^1(RR)$ and $u_n'\to g\in L^1(RR)$, then we can say that $u_n\to f$ in $W^{1,1}(RR)$.

So the space of function that satisfy this definition we can write as follow ${f\in L^1(RR)|\exists u_n\in C_c^{\infty}(RR)}$, that is $W_0^{1,1}(RR)$. Then the weak derivative is more general...

[gugo82]

Once De Giorgi said: "Chi cerca, trova; chi ricerca, ritrova"... Well, fu^2, you've found one good way to generalize the concept of derivative: in fact you've re-discovered the definition of weak derivative.

Let [tex]$f,h\in L^1 (\mathbb{R})$[/tex] and [tex]$(g_n)\subset C_c^\infty (\mathbb{R})$[/tex] s.t. [tex]$g_n\to f$[/tex] and [tex]$g_n^\prime \to h$[/tex] in [tex]$L^1$[/tex] (i.e. [tex]$||g_n-f||_1,||g_n^\prime -h||_1 \to 0$[/tex]).
We'd like to show that [tex]$h$[/tex] is the weak derivative of [tex]$f$[/tex], i.e. that:

[tex]$\forall \phi \in C_c^\infty (\mathbb{R}),\quad \int_{-\infty}^{+\infty} f\ \phi^\prime =\int_{-\infty}^{+\infty} h\ \phi$[/tex].

We have:

[tex]$\int_{-\infty}^{+\infty} f\ \phi^\prime =\lim_n \int_{-\infty}^{+\infty} g_n\ \phi^\prime =\lim_n \int_{-\infty}^{+\infty} g_n^\prime \ \phi =\int_{-\infty}^{+\infty} h\ \phi$[/tex]

by Lebesgue's dominate convergence theorem and integration by parts for [tex]$C_c^\infty$[/tex] functions; hence [tex]$f$[/tex] is differentiable in the weak sense (with [tex]$f^\prime =h$[/tex]) and therefore [tex]$f\in W^{1,1} (\mathbb{R})$[/tex].

Now, if [tex]$(u_n)\in C_c^\infty (\mathbb{R})$[/tex] is another sequence s.t. [tex]$||u_n-f||_1\to 0$[/tex] and if exists [tex]$k\in L^1(\mathbb (R))$[/tex] s.t. [tex]$||u_n^\prime -k||_1\to 0$[/tex], then [tex]$h=k$[/tex] in [tex]$L^1$[/tex]: in fact, applying again Lebesgue's theorem and integration by parts, we have:

[tex]$\int_{-\infty}^{+\infty} k\ \phi =\lim_n \int_{-\infty}^{+\infty} u_n^\prime\ \phi$[/tex]
[tex]$=\lim_n \int_{-\infty}^{+\infty} u_n\ \phi^\prime$[/tex]
[tex]$= \int_{-\infty}^{+\infty} f\ \phi^\prime$[/tex]
[tex]$=\lim_n \int_{-\infty}^{+\infty} g_n\ \phi^\prime$[/tex]
[tex]$=\lim_n \int_{-\infty}^{+\infty} g_n^\prime\ \phi$[/tex]
[tex]$=\int_{-\infty}^{+\infty} h\ \phi \quad \Rightarrow$[/tex]

[tex]$\Rightarrow \quad \int_{-\infty}^{+\infty}(k-h)\ \phi =0$[/tex]

for each test function [tex]$\phi \in C_c^\infty (\mathbb{R})$[/tex], hence (by a remarkable lemma) [tex]$k-h=0 \text{ a.e. in } \mathbb{R}$[/tex] therefore [tex]$h=k$[/tex] in the [tex]$L^1$[/tex] sense.
It then follows that the function [tex]$f^\prime$[/tex] doesn't depend on the choice of the approximating sequence [tex]$(g_n) \subset C_c^\infty (\mathbb{R})$[/tex] as soon as it is taken to be Cauchy's in [tex]$L^1$[/tex] together with the sequence [tex]$(g_n^\prime)$[/tex].

Note that, in general, if [tex]$(g_n)\subset C_c^\infty$[/tex] is s.t. [tex]$||g_n-f||_1\to 0$[/tex] with [tex]$f\in W^{1,1}$[/tex], then [tex]$(g_n^\prime)$[/tex] have not to converge in [tex]$L^1$[/tex] to the weak derivative [tex]$f^\prime$[/tex]: the previous example of dissonance is illuminating.

[dissonance]

I forgot to say, in my first post, that convergence should be intended in $W^{1, 1}$ sense, rather than just $L^1$. This way you have continuity for the weak derivative operator $D$ and so from the $g_n\to f$ you get, applying $D$ to both sides, $D(g_n)\toD(f)$. If $g_n$ are taken $C^infty$, then the weak derivative operator is just the ordinary derivative, so that $g'_n\toD(f)$, no matter which $g_n$ you choose.

If you take convergence only in the $L^1$ sense you're going to have problems, like I said abstractly (too much abstractly, on second thought ) in my last post. As a trivial example, take the $L^1(0, 1)$ function $0$. You can approximate it with the constant sequence $f_n(x)=0$ or with $g_n(x)=x^n$. The problem is that $g'_n(x)=nx^{n-1}$ doesn't converge $L^1$, as it is readily seen: $nx^{n-1}\to 0$ for all $x\in(0, 1)$, but $int_0^1nx^{n-1}"d"x=1$ identically.

OK, this doesn't really prove anything, since you required convergence for $g_n$ and also for $g'_n$ in your definition. This amounts to require convergence in $W^{1, 1}$ sense, if $f$ is $W^{1, 1}$. And so, in this case, we are back to our weak derivative operator $D$.

[edit]What's left behind is the case $f \notin W^{1, 1}$. Not at all: see Gugo's next post.[/edit] In that case I think we could say more or less the same things if we replace the weak derivative operator $D$ with the distributional derivative operator.

[fu^2]

"dissonance":This way you can speak of "weak derivative" for a $L^1(RR)$ function; I believe that taking your $f\in L^1(RR)$ as $f\inW^{1, 1}(RR)$ instead, your definition is well-posed and equivalent to that of weak derivative.

I don't think that my definition is equivalent to the weak derivative.

In fact in $f$ satisfy my definition, than $f$ admit the weak derivative (easy), but the vice versa is false... i think.


In fact let $f\in L^1(RR)$ and let $g$ its weak derivative (i.e. $\intf\phi'=-\intg\phi$ for each test function $\phi$, and $int=\int_{-\infty}^{+\infty}$), then we have to consider a sequence $u_n\in C_C^{\infty}$ s.t. $||u_n-f||_{L^1}\to 0$, when $n\to +\infty$. We must show that $||u_n'-g||_{L^1}\to 0$.

Integrate by part we obtain that $\intu_n\phi'=-\intu_n'\phi$ and by definition (since $||u_n\phi'-fphi'||_{L^1}=||phi'(u_n-f)||_{L^1}\to 0$) we have $\int u_nphi'\to\int f\phi'$, then $|\intf\phi'-\intu_n\phi'|=|intg\phi-\intu_n'\phi|$ so we can conclude that $\intu_n'\phi \to \intg\phi$, for each $phi\in C_c^{\infty}$, but this result don't involve that $\int|g-u_n'|\to 0$... or not?... So i don't see other way "to get the goal" and i think that my definition isn't well define

what do you think?...

... i'll think to your second topic, but the way is hard ...

[dissonance]

We can put your idea in these terms.

Call $D$ the derivative operator on $C_C^infty(RR)$. Since we know that $C_C^infty(RR)$ is a dense subspace of $L^1(RR)$, it is natural to think at a way to extend $D$ to the whoie $L^1(RR)$. This could be done automatically should $D$ happen to be continuous with respect to the natural norm of $L^1(RR)$, but that's not the case: the bounded sequence $f_n(x)={(sin(nx), x\in[-pi, pi]), (0, "otherwise"):}$ is mapped by $D$ into an unbounded one.

In cases like these one speaks of "closed" and "closable" operators. A linear operator $L: D(L)\subsetV \to W$ is said to be closed if its graph ($G(L)={(x, Lx)\ :\ x\inD(L)}$) is closed. Likewise, a linear operator is said to be closable if the closure of its graph ($bar{G(L)}$) is the graph of a linear operator. Operatively, we have the following criterion:

$L$ defined as above is closed $iff$ for every sequence $x_n$ converging to $x$ and such that $Lx_n$ converges, we have $x\inD(L)$ and $Lx_n\toLx$.

$L$ defined as above is closable $iff$ for every sequence $x_n\to0$ such that $Lx_n$ converges, we have $Lx_n\to0$.

The operator $D$ which we have previously defined is not closed in $L^1(RR)$. In fact there exist sequences of differentiable functions converging $L^1$ to non-differentiable ones, take ${(sqrt(x^2+1/n),\ x\in[-1, 1]), (0, "otherwise"):}$ as a crude example. The problem is that I don't think it is closable either. In fact there is no need for a sequence of differentiable functions, converging $L^1$ to $0$, to have their derivatives' $L^1$ norm becoming arbitrarily small. I think we could even work out an example of a sequence $f_n$ such that $||f_n||_1\to0$ and $||f'_n||_1\to +\infty$ without much trouble.

For this, I see little chance of success in trying to extend the operator $D$ to the whole $L^1$. The best you could get is a non-closed operator. On the contrary, restricting $L^1$ to $W^{1, 1}$ everything will flow painlessly. In fact $D$ is continuous with respect to the norm of $W^{1, 1}$.

[dissonance]

[The verb is "to define", whose past participle is "defined" (it's a regular one - I checked the dictionary). Take a look at this Wikipedia page, especially at the example with "The letter is written".

Speaking of verbal tenses, you wrote "do you added...": that's wrong; the correct one is "why did you add...". Look at this page.

A word of warning. You used the noun "argument" to translate the Italian "argomento"; you should have used the word "topic" instead. "Argument" in English is typically used to say "discussione, litigio", if I remember correctly.]

[fu^2]

thank you for your correction, my english is not good, but if i don't write in this lenguage, i can't get knowledge.
I learn from your correction too

[P.S. i understand all correction, except one: i have wrote "is well define" and you have changed in "is well defined". Why do you added the "final d"?...]

Came back to my question, i don't know the sobolev space, i'm going to see this argument... so i will understand if i can save my definition

If someone has another idea, i want lissen that!

[dissonance]

OK, now that we have assessed (or at least, tried to) the English side let's talk about the mathematical one. I think that what you say should be put into the framework of Sobolev spaces, precisely in the framework of the space $W^{1, 1}(RR)$. This way you can speak of "weak derivative" for a $L^1(RR)$ function; I believe that taking your $f\in L^1(RR)$ as $f\inW^{1, 1}(RR)$ instead, your definition is well-posed and equivalent to that of weak derivative.

[dissonance]

I hope you won't get mad at me if I try and fix some English errors in your text. Remember I'm no English teacher so don't trust me too much.

First of all, the title is wrong: you should say "Is this a good definition?"

"fu^2":We can considered the set of function [tex]f\in L^{1}(\mathbb{R})[/tex].

Cut that "ed" off otherwise that won't make sense. Also you repeatedly say "the set of function", while the correct form is "the set of all functions".

fix [tex]f\in L^{1}(\mathbb{R})[/tex], for each [tex]\epsilon>0[/tex] there exists [tex]g_{\epsion}[/tex]s.t. [tex]||g_{\epsion}-f||<\epsilon[/tex], therefore we can build a sequence [tex]g_n[/tex] s.t. [tex]g_n\to f[/tex] in norm.

Added a "there" and a final "s" in "there exists". Changed "we can built" into "we can build".

Then is "natural" to think at a method to define a "derivative" of [tex]f[/tex].

Added a "to" and a "at" in "to think at a method". Also "derivate" is wrong: the correct English word is "derivative".

Today i have had an idea : We can say that [tex]f[/tex] is differentiable in some sense if [tex]g_n'[/tex] converges to some [tex]h\in L^1(\mathbb{R})[/tex]. In this case we set [tex]f'=h[/tex],

You said "same" but you really meant "some". Also added a final "s" in "converges".

the problem is that the sequence [tex]g_n[/tex] which converges to [tex]f[/tex] isn't unique! therefore the derivative depends on the sequence [tex]g_n'[/tex], i.e. it's not well defined!

Self-explaining.

So is it possible to "walk in this way" to succeed in defining a generalization of the derivative for all integrable functions on the real axis (i.e. is it possible to save my definition )?

Added a "it" in "is it possible"; changed "to succed define" into "to succeed in defining", changed "generalization of the derivate to all integrable function" into "generalization of the derivative to all integrable functions", added a "it" in "is it possible to save...".

P.S.: As I said before I'm no English teacher. So there is no guarantee the above corrections are correct for their own sake! Should it be the case, I beg the other users reading this to try and correct me as well. Thanks.