Prove that Bernstein Transformation is linear The definition of the transformation is $$(B_nf)(x) = \sum_{k=0}^n f(\frac{k}{n}) {n \choose k}x^k(1-x)^{n-k}$$

How can I show this is a linear map? I know the sum of $B_n$ should be 1, but can I consider $f(\frac{k}{n})$ as a constant?
$(B_n(f+a g))(x) = \sum_{k=0}^n (f+a g)(k/n) \binom{n}{k} x^k (1-x)^{n-k} = \sum_{k=0}^n (f(k/n) + a g(k/n)) \binom{n}{k} x^k (1-x)^{n-k} = \sum_{k=0}^n f(k/n)  \binom{n}{k} x^k (1-x)^{n-k} +\sum_{k=0}^n a g(k/n)  \binom{n}{k} x^k (1-x)^{n-k} = sum_{k=0}^n f(k/n)  \binom{n}{k} x^k (1-x)^{n-k} +a \sum_{k=0}^n g(k/n)  \binom{n}{k} x^k (1-x)^{n-k} = (B_n f)(x) + (B_n g)(x)$.

So $B_n(f+a g) = B_n f +a B_n g$. So linear.
Hint: $B_n$ associates to each function f another function defined as you wrote. In order to show that $B_n$ is linear you have to prove that $B_n(f+g)= B_n(f)+ B_n(g)$, and $B_n(af)=aB_n(f)$, for any functions f and g, and any real number a. But two functions are equal if they have equal values for each argument x.