Surjectivity of operator $(Vf)(t)= \begin{cases} f(t-1) & \text{if $x\geq1 $ } \\ 0 & \text{if $x<1 $} \end{cases}$in $L_2(0, \infty)$

Question

Let $V:L_2(0, \infty)\to L_2(0, \infty)$ defined by

$(Vf)(t)= \begin{cases} f(t-1) & \text{if $x\geq1 $ } \\ 0 & \text{if $x<1 $} \end{cases}$

I want to show $V$ is isometry that is not surjective, I get stuck when to prove $V$ is not surjective

Any ideas or insight would be greatly appreciated

Answer 1

$g(x)=1$ for $x<1$ and $0$ for $x>1$ defines a function in $L^{2}$ and this is not in the range of $V$.