Is U a Subset of V?

[Theorem.]

Homework Statement

Determine in each case below whether U is a subspace V. If it is, verify all conditions for U to be a subspace, and if not, state a condition that fails and give a counter-example showing that the condition fails.

There are parts (a) through (f) to this question. I am only having trouble with (e), which reads:

V is the space of all polynomials with real coefficients, viewed as functions $$\Re\rightarrow\Re$$ and U is the set of all differentiable functions $$\Re\rightarrow\Re$$

Homework Equations

A subspace of a vector space V is a subset U of V that has three properties:

1. The zero vector of V is in U
2. U is closed under vector addition
3. U is closed under muliplication by scalars

The Attempt at a Solution

actually the only trouble I am having with this question is deciding whether or not U is even a subset of V. I know that if U is a subset of V, all the properties are satisfied, since the zero vector is an element of U, and from the basic properties of functions it is trivial that scalar multiplication and closure under vector addition holds.

I have been told that all differentiable functions can be written as polynomials via taylor series? if this is the case, the U is indeed a subset of V. Yet it is definitely true that all polynomials are differentiable, and thus one could say that V is in fact a subset of U. If the second statement I have suggested is true, then how could I use a counter example in this question? I suppose I'd need a function which is differentiable but can't be represented as a polynomial. Originally i thought of the exponential function and trig functions (cos and sine) as counter examples but they are analytic and thus can be represented by their respective taylor series which yield polynomials?

Any help would be much much appreciated
Thanks,
Curtis

 

[Unit]

In short, U is not a subset of V. Not all differentiable functions are polynomials, e.g. sin(x), e^x, ln(x).

For a differentiable function to be written as a polynomial via Taylor series, it http://en.wikipedia.org/wiki/Taylor_series#Definition". A function like $f(x) = \sqrt[3]{x}$ is once-differentiable, but because $f'(x) = \frac{1}{3x^{2/3}}$ is not continuous (at x = 0), f is not twice-differentiable.

They may have mixed up U and V.

 

[Theorem.]

In short, U is not a subset of V. Not all differentiable functions are polynomials, e.g. sin(x), e^x, ln(x).

For a differentiable function to be written as a polynomial via Taylor series, it needs to be infinitely differentiable. A function like $f(x) = \sqrt[3]{x}$ is once-differentiable, but because $f'(x) = \frac{1}{3x^{2/3}}$ is not continuous (at x = 0), f is not twice-differentiable.

They may have mixed up U and V.

Thanks Unit. I agree, perhaps I can go see him tomorrow to clarify whether he meant that the functions were infinitely differentiable or not. I was approaching the question with this assumption which was kind of stupid of me. If this is not the case, then it would suffice to list an example such as the one you provided (i.e. a function that is only once differientiable), and simply state that it is an element of U but is not an element of V. Thanks : )

 

[HallsofIvy]

Are you sure that the problems is whether or not U is a subspace of V, where U is the space of differentiable functions and V is the space of polynomials? If so, the answer is trivially, "no", because there exist differentiable functions that are not polynomials. Whether the functions were to be infinitely differentiable is not relevant since the fact that some infinitely differentiable functions are equal to their Taylor series (not all are) is NOT saying that they are polynomials.

Check to make sure the problem isn't to show that V is a subspace of U or that you don't have the definitions of U and V reversed. It would make much more sense to be asked to show that the space of all polynomials is a subspace of the space of all differentiable functions.

 

[Theorem.]

Are you sure that the problems is whether or not U is a subspace of V, where U is the space of differentiable functions and V is the space of polynomials? If so, the answer is trivially, "no", because there exist differentiable functions that are not polynomials. Whether the functions were to be infinitely differentiable is not relevant since the fact that some infinitely differentiable functions are equal to their Taylor series (not all are) is NOT saying that they are polynomials.

Check to make sure the problem isn't to show that V is a subspace of U or that you don't have the definitions of U and V reversed. It would make much more sense to be asked to show that the space of all polynomials is a subspace of the space of all differentiable functions.

Hey HallsofIvy thanks for your response. Yeah I am sure, and I agree it would make more sense. it was a multiple part question so it would make sense if he threw in one question like this. I suppose most people will just show that U contains the zero vector, and satisfies the two closure properties listed without directly thinking whether or not one is a subset of the other-which would be wrong.
Thanks for your help