Why is this a linear transformation? In class the professor said that 
f(x)-->g(x)*f(x) is considered a linear transformation.  I don't understand why that is, can someone explain?  
 A: Your domain is the "vector space of functions", . You didn't specify what functions but it could be something like the space of continuous real valued functions defined on the real line. This is an infinite dimensional vector space. The functions themselves can be wildly non-linear but this doesn't make your professor's statement false.
The linear operator is multiplication by the function g, which again could be a continuous real valued functions defined on the same domain. Multiplication by g would just give you another real valued function. This result is linear in f, i.e. g(x)*(c1*f1(x) + c2*f2(x)) = c1*g(x)*f1(x) + c2*g(x)*f2(x). 
A: Let $V$ be the set of all functions $\mathbb{R}\to\mathbb{R}$.  Fix $g\in V$.  Then $f\mapsto g\cdot f$ is a linear transformation $V\to V$ because it satisfies both axioms of linearity: For all $c\in\mathbb{R}$ and $f,f'\in V$, we have:
$$g\cdot(f+f') = g\cdot f + g\cdot f'$$
$$g\cdot(c\cdot f) = c\cdot (g\cdot f)$$
A: I think your professor was trying to state that multiplication by a function $g$ is a linear operation in a function space (or between to function spaces).
For example, consider the space $C(\mathbb R)$ of continuous functions $\mathbb R\to\mathbb R$.
It is a vector space.
Now fix any continuous function $g$.
Then the map $f\mapsto gf$ (pointwise product) is a linear transformation $C(\mathbb R)\to C(\mathbb R)$.
A: I can only guess, but perhaps it refers to the left translation map by $g(x)$ which is a linear operator. More precisely, let $A$ be any $K$-algebra with underlying vector space $V$ and consider the map $L(x)\colon y\mapsto xy$. This is an endomorphism of $V$.
