How can I show that the function is smooth?

Question

I got an assignment which I just can't find the right way to solve.

It goes like this:

Let $\Omega \in R^n$ be a domain and $b_1,...,b_n:\Omega \to R$ smooth mappings (or functions, don't know the correct translation into english), so that for every $x\in\Omega$ the vectors $b_1(x),...,b_n(x)$ linear independent.

Let $c_1, ... , c_n:\Omega\to R$ be mappings (or functions).

Show that the function $F(x):=c_1(x)b_1(x)+...+c_n(x)b_n(x)$ is smooth when $c_1,...,c_n$ are smooth.

Answer 1

Since the $c_{i}$'s are smooth and $b_{i}$'s are also smooth, their product is smooth for every $i$(and the derivative canbe calculated by Leibniz rule). Then, your function is a sum of smooth functions, so it must be smooth.

Answer 2

Your question is not clear: since the $b_i$ are functions that take real values, how can the set $\{b_i(x)\}_i$ be linearly independent? Maybe you wanted to say that the set of functions $\{b_i\}_i$ is linearly independent?

Anyway, just use that the product and sum of two smooth functions are again smooth. This has nothing to do with linear independence.