- #1
Rodrigo Schmidt
- 14
- 5
[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!
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!
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