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In this question someone asks about showing that the set of all functions of the form $y(t) = c_1\cos\omega t + c_2\sin\omega t$ is a vector space. But doesn't literally any set of functions of the form $y(t) = c_1f(t) + c_2g(t) + \ldots$ form a vector space? After all, there will always be a zero element (coefficients = 0) and an additive inverse (coefficients of opposite sign), and trivially scaling or adding two $y(t)$ will yield another function of the same form.

So what is an example of a set of functions that do NOT form a vector space? The most common pedagogic example I've seen is unsatisfyingly contrived: the set of all polynomials of degree N. This is explained to not form a vector space because the zero element is not of degree N. However technically the zero element is still of the form $c_1 + c_2x + \ldots + c_Nx^N$, so I'm not sure how comfortable I am with this example.

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    $\begingroup$ The space of all constant functions except $f(x)=3$. The space of all functions such that $F(0)=1$. The space of all polynomials of degree exactly $2$. The space of all polynomials with lead coefficient $1$. And so on. $\endgroup$ – lulu Oct 11 '19 at 19:09
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    $\begingroup$ Yes, the set of all finite linear combinations of any set of complex- (or real-)valued functions on some domain forms a vector space. That's still a very special set of functions. How about the set of discontinuous real-valued functions on $[0,1]$? That's not a vector space. $\endgroup$ – saulspatz Oct 11 '19 at 19:18
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    $\begingroup$ Take any set of functions and remove zero (given it's in the set). $\endgroup$ – amsmath Oct 11 '19 at 19:34
  • $\begingroup$ @saulspatz, isn't not allowing functions to be continuous a more special set of functions than just the set of complex/real valued functions? $\endgroup$ – user1247 Oct 11 '19 at 20:00
  • $\begingroup$ Yes. Any set of functions is more special than the set of all functions. So what? $\endgroup$ – saulspatz Oct 11 '19 at 20:04
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Take your favourite differential equation, as long as it's not a homogeneous linear one. Then its solution set isn't a vector space under the usual definitions for functions of scaling and addition. Surely that's not a contrived example.

For the sake of physical examples, note that the linearity of quantum mechanics and electromagnetism is starkly at odds with the nonlinearity of general relativity and nuclear interactions. The Higgs field satisfies a nonlinear differential equation. While electromagnetism is linear, it's inhomogeneous with a source. Fluid mechanics is nonlinear too.

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  • $\begingroup$ Do you mean that the solution set considered as functions separately from the DE do not constitute a vector space, or do you mean that linear combinations of solutions do not anymore satisfy the DE. The latter is obvious to me, but the former is not. Can you provide an example set of several functions that are not a vector space? $\endgroup$ – user1247 Oct 11 '19 at 21:30
  • $\begingroup$ @user1247 The statements are equivalent. For example, consider $y^\prime=-y^2$; the solutions are $y=1/(x+c)$. $\endgroup$ – J.G. Oct 11 '19 at 21:33
  • $\begingroup$ The space of functions being considered would be indexed as $(c_1,c_2,\ldots)$ meaning $1/(x+c_1)$, $1/(x+c_2)$, ...? Just making sure I understand -- we are not talking about linear combinations where the $c_n$ are the coefficients in front of each solution? $\endgroup$ – user1247 Oct 11 '19 at 21:44
  • $\begingroup$ @user1247 Compare that to solutions of an inhomogeneous wave equation, which are far too general to give a similar treatment. $\endgroup$ – J.G. Oct 11 '19 at 22:10
  • $\begingroup$ Thanks -- but doesn't any situation in which you don't allow the multiplicative coefficient to vary, almost trivially (by definition) not form a vector space? Or am I not imaginative enough -- are there examples similar to your non-example? $\endgroup$ – user1247 Oct 12 '19 at 1:10

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