Properties of (Af)(x, y)= (x+iy)f(x,y) on compact domain

Question

Consider in $L^2(Q)$, where $Q=[0,1]\times[0,1],$ the linear operator: $$(Af)(x, y)= (x+iy)f(x,y)$$ 1. Show that A is bounded 2. Calculate $||A||$ 3. Calculate $A^\dagger$

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I'm unsure about my solution I would very much appreciate if someone could check it and give me some suggestions.

1. $$||A||=||x^2f(x,y) +iy^2f(x,y)|| \le (x^2+y^2)||f|| < \infty$$ Since $f \in L^2(Q)$, $||f||$ is limited in Q and Q is compact (we can be sure given Weierstrass that $x^2+y^2$ won't diverge).

2. $$||A||=\sup_{||f||=1}||f(x,y)(x^2 +iy^2)||=1$$

3. $$A^\dagger=(x-iy)f(x,y)$$

Answer 1

For 1 and 2, \begin{align} \|Af\|_2^2=\int_0^1\int_0^1 (x^2+y^2)|f(x,y)|^2\,dx\,dy \leq2\int_0^1\int_0^1 |f(x,y)|^2\,dx\,dy=2\|f\|^2. \end{align} So $A$ is bounded and $\|A\|\leq\sqrt2$. Now, for any $n$, let $f=1_{E}$, where $E$ is the rectangle $[1-1/n,1]\times [1-1/n,1]$. Then \begin{align} \|Af\|_2^2&=\int_0^1\int_0^1 (x^2+y^2)|f(x,y)|^2\,dx\,dy\\ \ \\ &\geq\left((1-\frac1n)^2+(1-\frac1n)^2\right) \,\|f\|^2. \end{align} So, for any $n$, $$\|A\|\geq\sqrt2\,(1-1/n).$$ It follows that $\|A\|=\sqrt2$.

For the adjoint, you have \begin{align} \langle A^*f,g\rangle &=\langle f,Ag\rangle =\int_0^1\int_0^1 f(x,y)\,\overline{(x+iy)g(x,y)}\,dx\,dy\\ \ \\ &=\int_0^1\int_0^1 (x-iy)\,f(x,y)\,\overline{g(x,y)}\,dx\,dy\\ \ \\ \end{align} As $g$ is arbitrary, it follows that $(A^*f)(x,y)=(x-iy)f(x,y)$.