# Showing that a set of differentiable functions is a subspace of R

• I
Mayhem
TL;DR Summary
Show that the set of differentiable real-valued functions ##f## on the interval ##(-4,4)## such that ##f'(-1) = 3f(2)## is a subspace of ##\mathbb{R}^{(-4,4)}##
Problem:

Show that the set of differentiable real-valued functions ##f## on the interval ##(-4,4)## such that ##f'(-1) = 3f(2)## is a subspace of ##\mathbb{R}^{(-4,4)}##

This is my first bouts with rigorous mathematics and my brain is not at all wired for attacking problems like this (yet). I know that a subspace ##U## of vector space ##V## must satisfy the conditions that ##0 \in U##, and for ##u, w \in U \Rightarrow u + w \in U## and lastly ##a \in \mathbb{R}## and ##u \in U \rightarrow au \in U## (additive identity, closed under addition, closued under scalar multiplication respectively).

How do I attack this intelligently? What question do I need to ask myself?

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You have the right idea, I believe! To show that the subset contains a zero element, it's enough to just write it down, i.e. ##z(x) = 0##.

To show that it's closed under scalar multiplication, then for any ##a \in \mathbb{R}## and ##g \in U \subset \mathbb{R}^{(-4,4)}##, you want to show that ##ag \in U##, or in other words that ##(ag)'(-1) = 3(ag)(2)##, given that ##g'(-1) = 3g(2)##.

And to show that it's closed under addition, then for any ##g,h \in U \subset \mathbb{R}^{(-4,4)}##, you want to show that ##g+h \in U##, i.e. that ##(g+h)'(-1) = 3(g+h)(2)##, given that ##g'(-1) = 3g(2)## and ##h'(-1) = 3h(2)##.

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• member 587159 and Mayhem
Mentor
Summary:: Show that the set of differentiable real-valued functions ##f## on the interval ##(-4,4)## such that ##f'(-1) = 3f(2)## is a subspace of ##\mathbb{R}^{(-4,4)}##
There's a small point about this problem that bothers me, the part about the set of functions being a subspace of the interval [-4, 4] on the real line.

Differentiable, real-valued functions constitute a kind of vector space, that is usually called a function space. For the problem of this thread, it would make more sense to show that the given set is a subspace of the function space of differentiable, real-valued functions that are defined on the interval [-4, 4].

It doesn't make sense to say that this set is a subspace of part of the real line. Was the terminology that was used an error in translation?

Mayhem
There's a small point about this problem that bothers me, the part about the set of functions being a subspace of the interval [-4, 4] on the real line.

Differentiable, real-valued functions constitute a kind of vector space, that is usually called a function space. For the problem of this thread, it would make more sense to show that the given set is a subspace of the function space of differentiable, real-valued functions that are defined on the interval [-4, 4].

It doesn't make sense to say that this set is a subspace of part of the real line. Was the terminology that was used an error in translation?
It's taken directly from the book.

There's a small point about this problem that bothers me, the part about the set of functions being a subspace of the interval [-4, 4] on the real line.

Differentiable, real-valued functions constitute a kind of vector space, that is usually called a function space. For the problem of this thread, it would make more sense to show that the given set is a subspace of the function space of differentiable, real-valued functions that are defined on the interval [-4, 4].

It doesn't make sense to say that this set is a subspace of part of the real line. Was the terminology that was used an error in translation?

I think the problem statement is fine. The space ##\mathbb{R}^{(-4,4)}## is a function space, specifically the space of all functions ##f : (-4,4) \subset \mathbb{R} \rightarrow \mathbb{R}##, i.e. functions whose domains are the open interval ##(-4,4)##.

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Mayhem
I believe the problem statement is fine. The space ##\mathbb{R}^{(-4,4)}## is a space of functions, specifically the space of all functions ##f : (-4,4) \in \mathbb{R} \rightarrow \mathbb{R}##, i.e. functions whose domains are the open interval ##(-4,4)##.
That's also how I interpreted it.

• etotheipi
That's also how I interpreted it.

Yes, I believe Axler defines somewhere something along the lines of, ##A^B## is the space of maps from the set ##B## to the set ##A##. So the problem statement is absolutely fine

Mentor
I believe the problem statement is fine. The space R(−4,4) is a function space, specifically the space of all functions f:(−4,4)⊂R→R, i.e. functions whose domains are the open interval (−4,4).
I'll take your word for it, but it's not notation that is widely used. I looked at five of the lin. algebra textbooks I still have, and none of them had anything like this notation.

• etotheipi
Mayhem

Since the subspace spans from ##(-4,4)## in the reals, then ##0## must by definition be contained in the subspace.

Let ##g, h \in \mathbb{R}^{(-4,4)}## be differentiable functions and let ##f = g + h##. Thus $$(g+h)'(-1) = 3(g+h)(2)$$. Or is this circular reasoning or is more work just needed? I could expand the terms.

Closed under scalar multiplication

Let ##a \in \mathbb{R}##, then $$(af)'(-1) = 3(af)(2)$$ is still in ##\mathbb{R}^{(-4,4)}## ... Is the argument then that scalar multiplication does not affect the domain of a function or something?

Just some thoughts. I really am not used to thinking this rigorously about problems like this, but I feel like I need linear algebra to really understand differential equations in the future.

A few things; the subspace is also a space of functions, and the requirement of a zero vector in this context means that the subspace must contain a zero function ##z##, i.e. ##z: (-4,4) \rightarrow \mathbb{R}, x \mapsto z(x) = 0##. Since ##z(x) = 0##, then ##z'(x) = 0##, and this clearly then satisfies ##z'(-1) = 3z(2)##. So, by construction, you showed that the subspace contains a zero vector (zero function).

For showing the closure conditions, you need to make use of the definitions ##(f+g)(x) = f(x) + g(x)## and ##(\lambda f)(x) = \lambda f(x)##. For instance, this is how we'd show the subspace is closed under addition, for ##g,h \in U \subset \mathbb{R}^{(-4,4)}##,\begin{align*} (g+h)'(-1) &= (g' + h')(-1) \\ &= g'(-1) + h'(-1) \\ &= 3g(2) + 3h(2) \\ & = 3(g(2) + h(2)) \\ &= 3(g+h)(2) \end{align*}so we showed that, if ##g## and ##h## are elements of ##U##, then so is ##g+h##.

• Mayhem
Mayhem
A few things; the subspace is also a space of functions, and the requirement of a zero vector in this context means that the subspace must contain a zero function ##z##, i.e. ##z: (-4,4) \rightarrow \mathbb{R}, x \mapsto z(x) = 0##. Since ##z(x) = 0##, then ##z'(x) = 0##, and this clearly then satisfies ##z'(-1) = 3z(2)##. So, by construction, you showed that the subspace contains a zero vector (zero function).

For showing the closure conditions, you need to make use of the definitions ##(f+g)(x) = f(x) + g(x)## and ##(\lambda f)(x) = \lambda f(x)##. For instance, this is how we'd show the subspace is closed under addition, for ##g,h \in U \subset \mathbb{R}^{(-4,4)}##,\begin{align*} (g+h)'(-1) &= (g' + h')(-1) \\ &= g'(-1) + h'(-1) \\ &= 3g(2) + 3h(2) \\ & = 3(g(2) + h(2)) \\ &= 3(g+h)(2) \end{align*}so we showed that, if ##g## and ##h## are elements of ##U##, then so is ##g+h##.
Yes, since if ##z(x) = 0## then ##z'(x) = 0## means that ##z'(x) = 3z(x)## for all ##x \in \mathbb{R}##. Yes?

• etotheipi
Yes, since if ##z(x) = 0## then ##z'(x) = 0## means that ##z'(x) = 3z(x)## for all ##x \in \mathbb{R}##. Yes?

Exactly, yes • Mayhem
Mayhem
Exactly, yes Nice. I'm a little unsure how your demonstration of closed under addition rigorously shows that that ##g+h \in \mathbb{R}^{(-4,4)}## though. It's what I would have done, but I can't see how it's a valid "proof", probably due to me not understanding something vital.

We have a space of all functions ##\mathbb{R}^{(-4,4)}## from the interval ##(-4,4)## to the real numbers. We also have a subspace ##U## of this space, consisting of all of the functions that are both in ##\mathbb{R}^{(-4,4)}## but also satisfy the equation ##f'(-1) = 3f(2)## [think, Venn diagram...].

For instance, the function ##\chi## defined by ##\chi(x) = x^2 - 14/3## with domain ##(-4,4)## is an element of ##U## (and, also an element of ##\mathbb{R}^{(-4,4)}##, since that's just a larger set), however the function ##\xi## defined by ##\xi(x) = x^2## with domain ##(-4,4)## is not an element of ##U##, but it is still an element of ##\mathbb{R}^{(-4,4)}##.

We're asked to show that, if we pick any two functions, which I just called ##g## and ##h##, from ##U## [i.e. both ##g## and ##h## have domains ##(-4,4)##, and they satisfy ##g'(-1) = 3g(2)## and ##h'(-1) = 3h(2)##] then their sum is also an element of ##U##. That's just another way of saying that we must require that ##(g+h)'(-1) = 3(g+h)(2)##, which we proved above.

You could call ##g+h## another letter, e.g. ##m = g+h##. Since ##m## has a domain ##(-4,4)##, and satisfies the definition of this particular subspace, it's also an element of ##U##. Does that make more sense now?

Homework Helper
The notation T^S for the space of functions from S to T is very widely used. The motivation is based on the fact that the number of such functions in case S and T are finite of order n and m, is m^n. The same holds as well for infinite sets.

A common example is the notation 2^S for the set of subsets of S, since a function from S to {0,1} is equivalent to designating a subset of S as the inverse image of 1. This also reflects that the number of such subsets is 2^n where n = cardinality of S. it is indeed possible that this notation is not common in elementary linear algebra texts. But this notation is actually almost universal at only a slightly higher level.

Could not find an online copy of hausdorff's classic set theory book but the notation occurs on page 18 of this introductory text:

https://openscholarship.wustl.edu/cgi/viewcontent.cgi?article=1020&context=books

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• etotheipi
Gold Member
The notation T^S for the space of functions from S to T is very widely used. The motivation is based on the fact that the number of such functions in case S and T are finite of order n and m, is m^n. The same holds as well for infinite sets.

Another piece of motivation for this notation that I like very much is that ##A^{B\times C}## is in natural bijection with ##(A^B)^C,## and also ##A^B\times A^C## is identified with ##A^{B+C}## (where ##+## means disjoint union).

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• etotheipi
Mayhem
We have a space of all functions ##\mathbb{R}^{(-4,4)}## from the interval ##(-4,4)## to the real numbers. We also have a subspace ##U## of this space, consisting of all of the functions that are both in ##\mathbb{R}^{(-4,4)}## but also satisfy the equation ##f'(-1) = 3f(2)## [think, Venn diagram...].

For instance, the function ##\chi## defined by ##\chi(x) = x^2 - 14/3## with domain ##(-4,4)## is an element of ##U## (and, also an element of ##\mathbb{R}^{(-4,4)}##, since that's just a larger set), however the function ##\xi## defined by ##\xi(x) = x^2## with domain ##(-4,4)## is not an element of ##U##, but it is still an element of ##\mathbb{R}^{(-4,4)}##.

We're asked to show that, if we pick any two functions, which I just called ##g## and ##h##, from ##U## [i.e. both ##g## and ##h## have domains ##(-4,4)##, and they satisfy ##g'(-1) = 3g(2)## and ##h'(-1) = 3h(2)##] then their sum is also an element of ##U##. That's just another way of saying that we must require that ##(g+h)'(-1) = 3(g+h)(2)##, which we proved above.

You could call ##g+h## another letter, e.g. ##m = g+h##. Since ##m## has a domain ##(-4,4)##, and satisfies the definition of this particular subspace, it's also an element of ##U##. Does that make more sense now?
Yes! This makes perfect sense. I was thinking about it the wrong way. I couldn't really see how showing that ##3(g+h)(2) = 3g(2)+3h(2) ## really showed us anything useful, going from left to right, but I see the logic plays out nicely from right to left (if that made any sense at all).

Could you give me an example or two of where closure under addition doesn't work out? Something a counter example helps fortify understanding (as Axler himself said).

An easy one would be to consider the set ##A = \{ x : x = 2k - 1, k \in \mathbb{Z} \}##, i.e. the set of odd integers. This set isn't closed under addition, since ##(2m-1) + (2n-1) = 2(m+n - 1) \notin A##. So, the set of odd integers cannot be a vector space.

• Mayhem
Mayhem
An easy one would be to consider the set ##A = \{ x : x = 2k - 1, k \in \mathbb{Z} \}##, i.e. the set of odd integers. This set isn't closed under addition, since ##(2m-1) + (2n-1) = 2(m+n - 1) \notin A##. So, the set of odd integers isn't a vector space.
That's neat.

Just for closure, we can verify closure under scalar multiplication by letting ##g = af## where ##a \in \mathbb{R}## and ##g \in \mathbb{R}^{(-4,4)}## and see that it holds up by substituting it in.

Nearly, but we're aiming to check that the subset is closed, and not just the set ##\mathbb{R}^{(-4,4)}##. So we choose a vector ##f \in U \subset \mathbb{R}^{(-4,4)}##, and a number ##a \in \mathbb{R}##, and we show that ##g = af \in U##. Viz$$(af)'(-1) = af'(-1) = a \cdot 3f(2) = 3(af)(2)$$

• Mayhem