Operators with integrals defintion: $Q^*Q=Q?$

Question

Definite two operators $Q,V$ on $E=L^2(0,1)$ as follow:

for all $f\in E;$ $$(Qf)(x)=f(x)-\int_0^1 f(x)dx$$ $$(Vf)(x)=\int_0^x f(t)dt$$

Now denote $A=QV$ and $T=A^*A$.

I would like to prove that $T=V^*A.$

We have that $T=(QV)^*QV=V^*Q^*QV.$ Which means I need to prove that $Q^*Q=Q.$

Now, $\langle \phi,Qf\rangle_{E,E}=\langle Q^*\phi,f\rangle_{E,E}$ as $E^*=E.$

If I am correct I have to prove that $\langle Qf,Qf\rangle=(Qf)(x),$ but $$\langle Qf,Qf\rangle=(Qf)(x)-\int_0^1 (Qf)(x)dx=f(x)-2\int_0^1f(x)dx-\int_0^1\int_0^1f(x)dxdx.$$

What am I missing?

Answer 1

$$\int_0^1\int_0^1f(x)dxdx=\int_0^1f(x)dx $$ Also the last term in your last equation should be positive. $$\langle Qf,Qf\rangle=(Qf)(x)-\int_0^1 (Qf)(x)dx=f(x)-2\int_0^1f(x)dx+\int_0^1\int_0^1f(x)dxdx.$$