Conclusion about the dimension of C°(R)?

[Rodrigo Schmidt]

[mentor note: thread moved from Linear Algebra to here hence no homework template]

So, i was doing a Linear Algebra exercise on my book, and thought about this.

We have a linear map A:E→E, where E=C°(ℝ), the vector space of all continuous functions.
Let's suppose that Aƒ= ^x∫₀ ƒ(t)dt.

By the Calculus Fundamental Theorem, d/dx(Aƒ) = ƒ, so we have a left inverse, which implies that ker(A)={0}.

Supposing that dim(E) is finite, by the rank-nullity theorem we have that im(A)=E . As a result of that:
(ƒ(x)=|x|) ∈ E ⇒ ƒ ∈ im(A)
⇒dƒ/dx ∈ E

But we know that ƒ's derivative is not continuous. So, supposing that dim(E) is finite lead us to a contradition (dƒ/dx ∈ E ∧ dƒ/dx ∉ E) therefore dim(E) must be infinite.

Is this argument valid? If not, could you guys point where does it fail? Thank you!

 

[S.G. Janssens]

We have a linear map A:E→E, where E=C°(ℝ), the vector space of all continuous functions.
Let's suppose that Aƒ= ^x∫₀ ƒ(t)dt.

Note that the right-hand side is $(Af)(x)$, not $Af$.

But we know that ƒ's derivative is not continuous.

The derivative is not even defined at $x = 0$.

Is this argument valid?

Yes, it is valid, but it is quite convoluted. If you insist on doing it this way, I would prefer to say that you have found a linear operator $A$ that is injective but not surjective (the latter because $A$ maps into $C^1(\mathbb{R})$). This already implies infinite dimensionality of $E$, since on a finite dimensional space injectivity and surjectivity are equivalent for linear operators.

Of course, if you merely care to show that $E$ is infinite dimensional, it is more straightforward to identify an infinite linearly independent set.

 

[Rodrigo Schmidt]

Thanks for the knowledge shared!

I would prefer to say that you have found a linear operator AAA that is injective but not surjective (the latter because AAA maps into C1(R)C1(R)C^1(\mathbb{R}))

I hadn't tought that, that's, indeed, much simpler.

Of course, if you merely care to show that EEE is infinite dimensional, it is more straightforward to identify an infinite linearly independent set.

So the basis of the set of all polynomials would be enough?

 

[LCKurtz]

Thanks for the knowledge shared!

I hadn't tought that, that's, indeed, much simpler.

So the basis of the set of all polynomials would be enough?

The set of all polynomials is too large to be a basis for $C_0$, but $\{1,x,x^2,\dots\}$ would do.

 

[mathwonk]

this is very nice, and shows a creative grasp of what you are learning. keep it up and you will eventually do some new research!

 

[S.G. Janssens]

this is very nice, and shows a creative grasp of what you are learning. keep it up and you will eventually do some new research!

While I usually like the straightforward approach best (maybe this is one of the differences between a "pure" mathematician and an (aspiring) "applied" mathematician?), I gave it some thought and then came to the conclusion that I agree with you.

Just as an exercise, you could try to complete the direct approach as well. (You are almost there, anyway.)

 

[Rodrigo Schmidt]

The set of all polynomials is too large to be a basis for C0C0C_0, but {1,x,x2,…}{1,x,x2,…}\{1,x,x^2,\dots\} would do.

you could try to complete the direct approach as well.

Hm, i see, but isn't that the basis of the set of all polynomials? So, by being a basis, that must be a linearly independent set, and it's also infinite, so that implies in the infinite dimensionality of C°(ℝ)?

this is very nice, and shows a creative grasp of what you are learning. keep it up and you will eventually do some new research!

That's where i want to get someday. Thanks for the inspiration!

 

[S.G. Janssens]

Hm, i see, but isn't that the basis of the set of all polynomials? So, by being a basis, that must be a linearly independent set, and it's also infinite, so that implies in the infinite dimensionality of C°(ℝ)?

I think that I and post #4 misunderstood you when you wrote

So the basis of the set of all polynomials would be enough?

We (or I, at least) thought that here you asserted that the set of all polynomials itself is a basis for a linear subspace of $C^0(\mathbb{R})$, maybe because you wrote "the basis". However, from what you wrote afterwards (quoted at the top of this post), I get that you meant any (Hamel) basis for the linear space of all polyniomials. (For example, you can indeed use the canonical basis mentioned in post #4. By the way, I think there the $C_0$ was written by small mistake.)

Very good, as far as I can see, you are more than done.

 

[Rodrigo Schmidt]

I get that you meant any (Hamel) basis for the linear space of all polyniomials.

Yeah, that's what i meant. Sorry if i wasn't clear enough! English is not my mother language so, mostly when writing about math and science, my texts can get a little confusing. I will try to be more specific in the next time.

Very good, as far as I can see, you are more than done.

Okay! Thanks for the support and knowledge shared!