Differentiation on banach spaces

[cliowa]

Let $$L(A;B)$$ be the space of linear maps $$l:A\rightarrow B$$.

My goal is to derive the Leibniz (Product) Rule using the chain rule. Let $$f_i:U\subset E\rightarrow F_i, i=1,2$$ be differentiable maps and let $$B\in L(F_1,F_2;G)$$. Then the mapping $$B(f_1,f_2)=B\circ (f_1\times f_2):U\subset E\rightarrow G$$ is differentiable.
Using the chain rule I find for $$u\in U$$: $$D \{B\circ (f_1\times f_2)\}(u)=DB \{(f_1\times f_2)(u)\}\circ D(f_1\times f_2)(u)=B\circ D(f_1\times f_2)(u)\in L(E;G)$$, as B is a linear map. I could now go on and take the derivative of the cartesian product: $$...=B\circ(Df_1\times Df_2)(u)$$, but I feel it's all wrong. To me, there's no product rule in sight. Could you point out my error(s)?
Thanks a lot. Best regards...Cliowa

 

[Hurkyl]

What do you mean by $U \subset E$?

I'm going to assume that by $f_1 \times f_2$, you mean the function defined by:

$$(f_1 \times f_2)(a) = (f_1(a), f_2(a))$$

(IMHO $(f_1, f_2)$ is a better notation for this, and using $f_1 \times f_2$ to mean the map $(f_1 \times f_2)(a, b) = (f_1(a), f_2(b))$)


There's a problem in this expression:

$$DB \{(f_1\times f_2)(u)\}\circ D(f_1\times f_2)(u)$$

I can't figure out any way to parse this expression so that the composition operator makes sense.

 

[cliowa]

What do you mean by $U \subset E$?

U subset of E, I should have written that I meant an open subset U of E. That's simply in order to avoid trouble with closedness, bounds.

I'm going to assume that by $f_1 \times f_2$, you mean the function defined by:

$$(f_1 \times f_2)(a) = (f_1(a), f_2(a))$$

(IMHO $(f_1, f_2)$ is a better notation for this, and using $f_1 \times f_2$ to mean the map $(f_1 \times f_2)(a, b) = (f_1(a), f_2(b))$)

I'm sorry, I thought that was the usual notation, but of course I meant the same thing as you wrote: the cartesian product.

There's a problem in this expression:

$$DB \{(f_1\times f_2)(u)\}\circ D(f_1\times f_2)(u)$$

I can't figure out any way to parse this expression so that the composition operator makes sense.

As $$B\in L(F_1,F_2;G)$$ it follows that $$DB(v,w)=B, \forall (v,w)\in F_1\times F_2$$, right?
As $$(f_1\times f_2)=(f_1,f_2):U\subset E\rightarrow F_1\times F_2$$ i get $$D(f_1\times f_2)=D(f_1,f_2):E\rightarrow L(E;F_1,F_2)$$ so that the composition is perfectly fine:
$$L(F_1,F_2;G)\circ L(E;F_1,F_2) "="L(E;G)$$. Please do correct me if I'm wrong.

If this stuff seems strange to you, how would you do it?

 

[Hurkyl]

U subset of E, I should have written that I meant an open subset U of E. That's simply in order to avoid trouble with closedness, bounds.

I mainly found it confusing where it was being used. I figured that's what you meant, it just looked odd and wanted to make sure.

I'm sorry, I thought that was the usual notation, but of course I meant the same thing as you wrote: the cartesian product.

It might be in this particular subject -- I'm just showing my bias towards category theory.

so that the composition is perfectly fine:

You're right there, my mistake.


Oh, I see the problem. If B is something resembling multiplication, then it wouldn't be a linear map $F_1 \times F_2 \rightarrow G$, but instead a bilinear map.

It helps to look at special cases -- here, I was thinking about ordinary calculus. Here, we'd want something like B(a, b) = ab to recover the product rule, and that is clearly not linear on R².

If you want to consider linear maps instead of bilinear maps, you're going to have to use the tensor product of your spaces.

 

[cliowa]

Oh, I see the problem. If B is something resembling multiplication, then it wouldn't be a linear map $F_1 \times F_2 \rightarrow G$, but instead a bilinear map.

It helps to look at special cases -- here, I was thinking about ordinary calculus. Here, we'd want something like B(a, b) = ab to recover the product rule, and that is clearly not linear on R².

That one is clear to me, that's also how I figured it would work.

Oh, I see the problem. If B is something resembling multiplication, then it wouldn't be a linear map $F_1 \times F_2 \rightarrow G$, but instead a bilinear map.

If you want to consider linear maps instead of bilinear maps, you're going to have to use the tensor product of your spaces.

I read all this in a book by J. Marsden and R. Abraham called "Manifolds, Tensor Analysis, and Applications", available at https://www.amazon.com/gp/product/0387967907/?tag=pfamazon01-20.
They introduce topology, spaces, then derivatives. I attached a picture with some extracts of their claims, so you can have a look at the important parts. However they are not talking about bilinear maps, nor have they (yet) introduced tensor products, so there must be some other way of doing it. But I can't see how!
(You should also have a look at the example they're doing, as this is how they apply their theory.)
Thanks a lot and best regards...Cliowa

 

[mathwonk]

i have explained all this in an earlier response to a question about product rules and derivatives of vector valued functions.

anyway your whole setup is flawed foundationally. the derivative of a map in a banach space is not just a linear map but a continuous linear map.

 

[cliowa]

anyway your whole setup is flawed foundationally. the derivative of a map in a banach space is not just a linear map but a continuous linear map.

I feel I never claimed that, but let me get this straight: If $f:E\rightarrow F$, where E, F are Banach spaces, then the derivative function $Df:E\rightarrow L(E,F)$ must not be continuous, right? However, the differential at a point $e\in E$ is $Df(e)\in L(E,F)$ and therefore is continuous. Or what did you mean, by "foundationally flawed"?

i have explained all this in an earlier response to a question about product rules and derivatives of vector valued functions.

I guess you're referring to the thread https://www.physicsforums.com/showthread.php?t=124010". I have read you're response there, but I still have some questions: I understand perfectly fine your example for the bilinear map "multiplication in R". However, I can't quite follow when you proceed to bilinear maps in general: The proof you presented for multiplication makes use of the multiplication, how can one generalize then?

In a different thread, https://www.physicsforums.com/showthread.php?t=67268" you wrote:

I.e. a differential is a linear function on the tangent space. Since the tangent space to a curve is one dimensional, the space of linear functions is also one dimensional.

thus any two linear functionals are scalar multiples of each other, so their quotient is a scalar. this is not true for differentials on higher dimensional tangent spaces.

I cannot explain why this pont of view is prohibited in elementary calculus. perhaps they do not wish to do the work necessary to justify it.

As I am traditionally taught (and therefore quite confused): Could you explain this point of view a little more?
Thanks a jolly lot in advance.
Best regards...Cliowa
@mathwonk: BTW, thanks a lot for your lengthy and detailled explanations and valuable links in the de thread. That was very useful.

 

[Hurkyl]

However they are not talking about bilinear maps, nor have they (yet) introduced tensor products, so there must be some other way of doing it. But I can't see how!

It just struck me what might work -- you can "factor" the bilinear map into two linear maps.

If B is bilinear, then notice for each a that the map defined by:

$$B_a(b) := B(a, b)$$

is a linear map. Also, the map $a \rightarrow B_a$ is linear. Can you work with this?

 

[cliowa]

It just struck me what might work -- you can "factor" the bilinear map into two linear maps.

If B is bilinear, then notice for each a that the map defined by:

$$B_a(b) := B(a, b)$$

is a linear map. Also, the map $a \rightarrow B_a$ is linear. Can you work with this?

Oh yes, that's a good idea. Basically this is using the natural isomorphism $L(F_1,F_2;G)\simeq L(F_1;L(F_2;G))$. But how would you work that out in detail (the idea is perfectly clear, but how do I write this whole thing down nicely?)?
Best regards...Cliowa

 

[mathwonk]

I am saying that, unles the years have clouded my miknd, that even a linear map is not differentiable unkless iot is continuous.

linearmaps are not always continuous on banach spaces. my apologies if this was not an issue for you.

similarly bilinear maps are not differentiable unless they are continuous and this is not always the case.

 

[cliowa]

I am saying that, unles the years have clouded my miknd, that even a linear map is not differentiable unkless iot is continuous.

linearmaps are not always continuous on banach spaces. my apologies if this was not an issue for you.

Thanks for pointing that out. I should have mentioned this, but it never seemed like a problem to me, because in my mind I was working in finite dimensions.
Could you give an example of a linear map between banach spaces which is not continuous?
I still don't get how you transfer the derivative stuff from linear to bilinear forms: Is it so simple you don't even think it's worth mentioning?
Best regards...Cliowa

 

[Hurkyl]

Your goal is impossible. You cannot derive the product rule using just the chain rule. You must exploit the fact that the derivative evaluated at a point is a linear transformation.

Proof: define the "fake derivative operator" Q by:

Q{f}(a) = the constant function whose value is f(a)

This clearly satisfies the chain rule, and Q is even a linear map! But Q does not satisfy the product rule.


Therefore, it is impossible to derive the product rule using only the chain rule.

 

[cliowa]

Your goal is impossible. You cannot derive the product rule using just the chain rule. You must exploit the fact that the derivative evaluated at a point is a linear transformation.

That's alright, fine with me. I am willing to exploit this fact. I have seen how the product rule then follows for the bilinear map of multiplication.

But if you have a look at the excerpts from the book I mentioned you'll see that what they get is the Leibniz Rule in a general setting, i.e. where B is not necessarily multiplication but just some bilinear map. Now, it is this, that I do not understand. I don't know how to take this derivative:
$$D \{B\circ (f_1\times f_2)\}(u)$$

In my first post I wrote down a few steps which don't lead to the right thing, as we have discussed. What I should find (if the book is correct here), is that
$$D \{B\circ (f_1\times f_2)\}(u)e=B(Df_1(u)e,f_2(u))+B(f_1(u),Df_2(u)e)$$. How do I find that?

(P.S.: Your idea about splitting the bilinear map into to linear ones seems good to me, but I can't write it down, I can't formalize it. How would you do that?)

 

[Hurkyl]

It might help to stop thinking about $f_1$ and $f_2$, and just look at the derivative of B. What you need to show is that:

DB(a, b) . (c, d) = B(c, b) + B(a, d)

(where I use the notation `S.v' to signify the application of the linear transformation S to v, so that the expressions are easier for me to parse)

That's alright, fine with me. I am willing to exploit this fact. I have seen how the product rule then follows for the bilinear map of multiplication.

You try mimicing the proof, but replacing multiplication with an arbitrary bilinear form?

 

[cliowa]

It might help to stop thinking about $f_1$ and $f_2$, and just look at the derivative of B. What you need to show is that:

DB(a, b) . (c, d) = B(c, b) + B(a, d)

(where I use the notation `S.v' to signify the application of the linear transformation S to v, so that the expressions are easier for me to parse)


You try mimicing the proof, but replacing multiplication with an arbitrary bilinear form?

Oh geez, I must have been blind! But how the hell could I be so... nevermind. Thankfully this spiritual nightmare has come to an end. Thanks a lot.
Is there a second way of doing it? Does this factoring the bilinear maps into two linear ones lead anywhere?
humble regards...Cliowa

 

[Hurkyl]

Does this factoring the bilinear maps into two linear ones lead anywhere?

Probably. But to get there, I think you need to find the derivative of the evaluation map:

S x L(S, T) --> T

or maybe the composition map

L(U, V) x L(V, W) --> L(U, W)

and I didn't see an easy way to find that derivative.

 

[cliowa]

and I didn't see an easy way to find that derivative.

as you already guessed: It's the same thing for me...