# Conclusion about the dimension of C°(R)?

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1. Jul 12, 2017

### Rodrigo Schmidt

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

We have a linear map A:E→E, where E=C°(ℝ), the vector space of all continuous functions.
Let's suppose that Aƒ= x0 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!

Last edited by a moderator: Jul 12, 2017
2. Jul 12, 2017

### Krylov

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

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

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.

3. Jul 12, 2017

### Rodrigo Schmidt

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?

4. Jul 12, 2017

### LCKurtz

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

5. Jul 12, 2017

### 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!

6. Jul 13, 2017

### Krylov

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.)

7. Jul 13, 2017

### Rodrigo Schmidt

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

8. Jul 13, 2017

### Krylov

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

Last edited: Jul 13, 2017
9. Jul 13, 2017

### Rodrigo Schmidt

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.

Okay! Thanks for the support and knowledge shared!