"Given the vector space, C(−infty,infty) as the set of all continuous functions that are always continuous, is the set of all exponential functions, U={a^x∣a>=1}, a subspace of the given vector space? As far as I'm aware, proving a subspace of a given vector space only requires you to prove closure under addition and scalar multiplication, but I'm kind of at a loss as to how to do this with exponential functions (I'm sure it's way simpler than I'm making it). My argument so far is that the set U is a subset of the set of all differentiable functions, which itself is a subset of C(−infty,infty), but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure)."

apopiw83

apopiw83

Answered question

2022-11-04

Given the vector space, C ( , ) as the set of all continuous functions that are always continuous, is the set of all exponential functions, U = { a x a 1 }, a subspace of the given vector space?As far as I'm aware, proving a subspace of a given vector space only requires you to prove closure under addition and scalar multiplication, but I'm kind of at a loss as to how to do this with exponential functions (I'm sure it's way simpler than I'm making it).My argument so far is that the set U is a subset of the set of all differentiable functions, which itself is a subset of C ( , ), but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).

Answer & Explanation

Taniyah Lin

Taniyah Lin

Beginner2022-11-05Added 14 answers

As originally written, no, since 0 U . In addition, your U seems to be the set of monic monomials in x, not exponential functions, which would instead have the form a x for some positive constant a. Even then, this would not be a subspace, for exactly the same reason.
Edit: Another simple approach would be to show that U (in either version) is not closed under scalar multiplication.
Showing failure to be closed under addition is more difficult in both cases, but not too tricky for the original U.
reevelingw97

reevelingw97

Beginner2022-11-06Added 4 answers

Not, it is not a subspace. For instance the null function does not belong to U.

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