I am wondering whether linearity with respect to scalar functions $f \in C^{\infty}(M, \mathbb{R})$ is part of the definition of a tensor?
Let me explain it by referring to the Riemann curvature tensor
$$R(U,V)(W) :=\nabla_U(W)\nabla_V(W)-\nabla_V(W) \nabla_U(W)-\nabla_{[U,V]}(W).$$
From the definition it is not immediate (at least to me) that we actually have for a smooth function $f \in C^{\infty}(M, \mathbb{R})$ that
$$R(fU,V)(W)|_p = f(p)R(U,V)(W)|_p.$$ But this seems to hold and similar linearity properties hold for the other components, too.
Now, we defined a tensor at a point $p \in M$ to be a multilinear map from the product of cotangent and tangent spaces. This of course only means that for scalars and sums of cotangent/tangent vectors, we have linearity in each component, but we did not talk about functions in the definition.
Therefore, I was wondering how the linearity with respect to scalar functions is related to the definition of a tensor? Cause my textbook says that tensor are characterized by the fact that they do not depend on things that happen in a neighborhood of $p \in M$ which explain this "scalar function linearity." But I don't see how this interpretation is captured in our tensor definition?
Let $V(M)$ denote the space of vector fields on $M$. There are two equivalent requirements for a function $V(M)\times\ldots\times V(M)\to C^\infty(M)$ in order for it to be a tensor. One can require the function to be linear over $C^\infty(M)$ or alternatively require the function to induce a multilinear form at every point, depending only on the value of every vector field at this point. As written above, the two requirements are equivalent to one another.
As for the curvature tensor, it is not so immediate that it is a tensor, but it is. The way I know to verify that, is indeed check linearity over $C^\infty(M)$.