prove that this operator is not compact

Question

Let $g\in C[0,1]$ be a continuous function and $g\ne 0$. Let $G:C[0,1]\to C[0,1]$ the operator defined by: $G(f)(x)=f(x)g(x)$. I proved that the operator is linear and continuous. I want to prove that $T$ is not a compact operator. But I don't know how. Please help me with this problem.

Answer 1

The space $C[0,1]$ is not complete if it is normed with $\lVert \rVert_2$. Thus if we take a bounded sequence $(f_n)\subset C[0,1]$ it may not have a subsequence that converges. Now $G$ is compact iff for every bounded sequence $(f_n)\subset C[0,1]$ the sequence $G(f_n)$ has a subsequence that converges<=>$(f_n)g$ has a subsequence that converges. if $(f_n)$ has not then we have what you desire.

Answer 2

I assume that by $C[0,1]$ you mean continuous functions on $[0,1]$ with the sup norm. Now take $g$ to be the constant function $1$. Then consider the sequence of functions $\{x^n\}$. This is a bounded sequence and so if $T$ is compact we have $T(x^n) = x^n$ having a convergent subsequence. But this is not the case because any convergent subsequence would have to converge pointwise to the function that is $0$ on $0 \leq x < 1$ and $1$ at $x = 1$. This function is not even continuous, so the convergence cannot be uniform (by the uniform limit theorem).