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For the brief explanation: $\mathcal{P}$ contains $0$ by choice $p(x) = 0$ and polynomial plus a polynomial is a polynomial, and a scalar times a polynomial is a polynomial. So $\mathcal{P}$ is a non-empty subset of $\mathcal{C}^{\infty}$ that's closed under addition and scalar multiplication, and hence a subspace of $\mathcal{C}^{\infty}$.
(i) Not closed under addition. $p(x) = x^2$ and $q(x) = -x^2+x+1$ for are polynomials of precisely degree $2$ but their sum $(p+q)(x) = x+1$, a polynomial of degree $1.$
(ii) The set of even polynomials is not empty because we can take $p(x) = 0$. The sum of two even functions is even, so it's closed under addition. Also constant times an even polynomial is still an even polynomial, so it's closed under scalar multiplication. Hence it's a subspace of $\mathcal{P}$.
(iii) Let $p(x) = x^2$ and $q(x) = x^3$ then $p'(1) = 1$ and $q'(1) = 1$ but $(p+q)'(1) = 2$, so it's closed under addition.
(iv) It's not empty since it contains $0$ by the choice $\lambda = \mu = 0$. Let $p(x) = \lambda x + \mu x^2+ (\lambda-\mu)x^3$ and $q(x) = \lambda' x + \mu' x^2+ (\lambda'-\mu')x^3$ then $(p+q)(x) = (\lambda+\lambda') x + (\mu+\mu') x^2+ (\lambda-\mu+\lambda'-\mu')x^3$ which is again in the set. Also multiplying by a constant, we're still in the set, so it's a subspace.
Is this alright, and how do I do part (b)?
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