Question: Is the zero vector a linear combination of any nonempty set of vectors ? Justify. Is the set of all differential real valued functions defined on R a subspace of c(R)? Justify your answer.

First part:
Yes,it is
\(\vec{0} = 0.\vec{v1} + 0.\vec{v2} + ...+0.\vec{vn}\)
Moreever, an empty sum, that is the sum of no vectors, is usually define to be zero, and with that defination 0 is the linear combination of set of vectors, empty or not.
Second Part:
To show differential function is subspace of real valued function define on R a subspace of c(R). Let \(\mathcal{D}(R)\) denote the set of all differentiable real valued function over R. If \(f : \mathcal {R} \Rightarrow \mathcal{R}\) is differential then f is continious and so \(\mathcal{D}(R)\) is a subset of \(\mathcal{C}(R)\)\

We can show that this subset is subspace by showing that this is closed under addition and scalar multiplication, we can proceed as follows.
If f and g are in \(\mathcal{D}(R)\). Then
\((f + g)(x)\) = \(f(x) + g(x)\) is define for any \(x \in R\)
\(V_{ss1}\) : Zero is a constant value.
We know that differentiation of constant is zero.
Thus, f(x) = 0
\(\forall x\in R\). so we have \(f \in \mathcal{c}(R)\) .
\(V_{ss2}\) :Sum of two differential function is also differential.
We also have differential function is Continuous.
Let \(f(x) \in R\ and\ f(y) \in R\)
Then, \(f(x) + f(y) \in R \Rightarrow f(x) + f(y) \in \mathcal{C}(R)\)
\(V_{ss3}\): Scalar multiplication to any differential function is also differential. This implies also Continuous
i.e \(\forall c \in R\ and\ f(x) \in R\)
\(\mathcal{C}f(x) \in (R) \Rightarrow \mathcal{C}f(x) \in \mathcal {C}(R)\).