1. ## Some functional analysis

Let $H$ be a real Hilbert space. Define the complexification of $H$ by $H_{\mathbb{C}}$= $\{x+iy: x,y \in H\}$.

Prove:

1. $H_{\mathbb{C}}$ is a Hilbert space

2. Define $B(x+iy) = A (x) + i A(y)$ where $A$ is a bounded operator on $H$. Prove that B is bounded on H_C.

3. Let $A:{L^2}(0,\pi)\longrightarrow{L^2}(0,\pi)$ be defined by:

$(Af)(x) = \int_0^\pi sin(x-y)f(y)dy$
find the point spectrum and spectrum of A.

For 1. I say, suppose $(z_n)=(x_n + iy_n)$ is a Cauchy sequence in $H_{\mathbb{C}}$. Then $(x_n), (y_n)$ are Cauchy in H and since H is complete converge to some x,y in H. Therefore, $z_n\longrightarrow{z}=x+iy$ which is in $H_{\mathbb{C}}$ so it's complete. Is this ok?

For 2. I can only show that $||B(w)||_{op} \leq M (||x|| + ||y|| )$ where $w = x + iy$ which is not good enough as it should be M (||x+iy||) to fit the definition of a bounded operator. So please help.

For 3. sadly I have no clue.

2. Originally Posted by davidmccormick
Let $H$ be a real Hilbert space. Define the complexification of $H$ by $H_{\mathbb{C}}$= $\{x+iy: x,y \in H\}$.

Prove:

1. $H_{\mathbb{C}}$ is a Hilbert space

2. Define $B(x+iy) = A (x) + i A(y)$ where $A$ is a bounded operator on $H$. Prove that B is bounded on H_C.

3. Let $A:{L^2}(0,\pi)\longrightarrow{L^2}(0,\pi)$ be defined by:

$(Af)(x) = \int_0^\pi sin(x-y)f(y)dy$
find the point spectrum and spectrum of A.

For 1. I say, suppose $(z_n)=(x_n + iy_n)$ is a Cauchy sequence in $H_{\mathbb{C}}$. Then $(x_n), (y_n)$ are Cauchy in H and since H is complete converge to some x,y in H. Therefore, $z_n\longrightarrow{z}=x+iy$ which is in $H_{\mathbb{C}}$ so it's complete. Is this ok?

For 2. I can only show that $||B(w)||_{op} \leq M (||x|| + ||y|| )$ where $w = x + iy$ which is not good enough as it should be M (||x+iy||) to fit the definition of a bounded operator. So please help.

For 3. sadly I have no clue.
For 1., you have dealt with the completeness requirement for a Hilbert space, but you have not explained the algebraic structure of the complexification. You need to define $\langle x+iy,z+iw\rangle = \langle x,z\rangle + \langle y,w\rangle + i\bigl(\langle y,z\rangle - \langle x,w\rangle\bigr)$, for $x+iy$ and $z+iw$ in $H_{\mathbb{C}}$, and show that this satisfies the axioms for a complex inner product.

Notice that that definition of the inner product implies that the norm on $H_{\mathbb{C}}$ is given by $\|x+iy\|^2 = \|x\|^2+\|y\|^2$. That should help you with the problem that you are having with 2.

I don't offhand know the answer to 3. I'll try to think about that when I have a bit more time.

3. Originally Posted by davidmccormick
3. Let $A:{L^2}(0,\pi)\longrightarrow{L^2}(0,\pi)$ be defined by:

$(Af)(x) = \int_0^\pi sin(x-y)f(y)dy$
find the point spectrum and spectrum of A.
Let $c,\,s$ denote the elements of $L^2(o,\pi)$ given by $c(x) = \cos x$, $s(x) = \sin x$. Use the relation $\sin(x-y) = \sin x\cos y-\cos x\sin y$ to see that $Af = \langle f,c\rangle s - \langle f,s\rangle c$. Thus A is a finite-rank operator, with two-dimensional range R spanned by $c$ and $s$, and A is zero on the orthogonal complement of R. So the spectrum of A is equal to the point spectrum, which consists of 0 together with the spectrum of the operator A restricted to R.

Also, A restricted to R has matrix $\frac\pi2\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ (with respect to the basis consisting of $c$ and $s$), from which you can easily find the other two points in the spectrum.

4. man you're a legend. I still can't show that B is bounded, but ur still a legend.

5. ## solved

The boundedness part also solved. thanks again.