Exercise 2, § 1

Let $\mu$ the Lebesgue measure on $[0,1]$, H the Hilbert space $L^2_{\mathbf{C}}([0,1], \mu)$, A = $\mathscr{L}(H)$, and $x\in$ A the operator which transforms the function $f(t)$ into the function $tf(t)$. Prove that Sp$_A(x) = [0,1]$, but that $x$ admits no eigenvalue.