# Understanding linear transformation

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• schniefen
In summary: So, in a situation where there is a non-differentiable ##u(t)## but a continuous ##F(u)##, we would still have a vector in ##C({R})##.In summary, the function ##F## can be a linear transformation from ##U## to ##V=C(\mathbf{R})## if and only if the term ##\mathbf{u}^{(n)}(t)## is not eliminable.

#### schniefen

TL;DR Summary
Understanding the linear transformation of ##F(\mathbf{u})(t)=\mathbf{u}^{(n)}(t)+a_1\mathbf{u}^{(n-1)}(t)+...+a_n\mathbf{u}(t)##.
How can the function ##F(\mathbf{u})(t)=\mathbf{u}^{(n)}(t)+a_1\mathbf{u}^{(n-1)}(t)+...+a_n\mathbf{u}(t)##, where ##\mathbf{u}\in U=C^n(\mathbf{R})## (i.e. the space of all ##n## times continuously differentiable functions on ##\mathbf{R}##) be a linear transformation (from ##U##) to ##V=C(\mathbf{R})##? In ##F##, isn't the term ##\mathbf{u}^{(n)}(t)## not eliminable and so one always gets a vector in ##C^n(\mathbf{R})##?

Last edited:
What do you mean by eliminable? ##u^{n} ## is only continuous after ##n## differentiations and so is ##F(u)##.

fresh_42 said:
What do you mean by eliminable? ##u^{n} ## is only continuous after ##n## differentiations and so is ##F(u)##.
Doesn't ##(\mathbf{u}^{(n)}(t),\mathbf{u}^{(n-1)}(t),...,\mathbf{u}(t))## represent a basis for ##U=C^n(\mathbf{R})##, and so in order to transform a vector from ##U## to ##V=C(\mathbf{R})## (with basis ##\mathbf{u}(t)##), one would have to be able to cancel the other basis vectors by choosing appropriate coefficients in ##F##?

What is the dimension of ##C^n(\mathbb{R})\,?## Compare it with the dimension of the space of all polynomials of a certain degree and lower: What is ##\dim_\mathbb{R} \{\,a_nx^n+\ldots+a_1x+a_0\,|\,a_i\in \mathbb{R}\,\}\,?##

schniefen
Is ##\dim C^n(\mathbb{R})=\dim_\mathbb{R} \{\,a_nx^n+\ldots+a_1x+a_0\,|\,a_i\in \mathbb{R}\,\}\,=n##?

We have ##n+1## independent directions ##a_i##, which is finite. But regardless the plus one, do we have ##\sin (t) ## in the vector space of polynomials? It is certainly in ##C^n(\mathbb{R})##.

What is the basis for ##C^n(\mathbb{R})##?

I don't know one. Maybe Legendre polynomials, but I'm not even sure whether if it's countable or not. I suspect it isn't.

But doesn't ##\dim C^n(\mathbb{R})=n+1## imply a basis of ##n+1## vectors?

##\dim C^n(\mathbb{R})=n+1## implies anything, because it is wrong. We have ##P^k(\mathbb{R}) \subseteq C^n(\mathbb{R})## for all ##k,n \in \mathbb{N}##, where ##P^k## is the algebra of (real) polynomials of degree at most ##k##, and we have ##\sin \notin P^k## for any ##k##.

This shows, that ##\dim C^n(\mathbb{R})## is at least countably infinite and that ##P^n(\mathbb{R})## is a proper subset.

member 587159 and schniefen
schniefen said:
In ##F##, isn't the term ##\mathbf{u}^{(n)}(t)## not eliminable and so one always gets a vector in ##C^n(\mathbf{R})##?

A vector in ##C^n({R})## is also vector in ##C({R})##.

I think what you mean to ask about is a situation where ##u(t)## is ##n## times continuously differentiable, but not ##n+1## times continuously differentiable. You are worried that the ##u^n(t)## term in ##F(u)## might cause ##F'(u) = u^{(n+1)} + ...## not to exist. I think there is a distinction between the notation ##C({R})## and ##C^1({R})##. For ##F(u)## to be in ##C({R})## , it need not be differentiable. It only has to be continuous. The function ##u^{(n)} ## is continuous even though it may not be differentiable.

schniefen

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