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Thread: Some functional analysis

  1. #1
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    Some functional analysis

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

    Prove:

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

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

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

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

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

    For 2. I can only show that $\displaystyle ||B(w)||_{op} \leq M (||x|| + ||y|| )$ where $\displaystyle w = x + iy$ 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.
    thanks for your help.
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  2. #2
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    Quote Originally Posted by davidmccormick View Post
    Let $\displaystyle H$ be a real Hilbert space. Define the complexification of $\displaystyle H$ by $\displaystyle H_{\mathbb{C}}$=$\displaystyle \{x+iy: x,y \in H\}$.

    Prove:

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

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

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

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

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

    For 2. I can only show that $\displaystyle ||B(w)||_{op} \leq M (||x|| + ||y|| )$ where $\displaystyle w = x + iy$ 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 $\displaystyle \langle x+iy,z+iw\rangle = \langle x,z\rangle + \langle y,w\rangle + i\bigl(\langle y,z\rangle - \langle x,w\rangle\bigr)$, for $\displaystyle x+iy$ and $\displaystyle z+iw$ in $\displaystyle H_{\mathbb{C}}$, 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 $\displaystyle H_{\mathbb{C}}$ is given by $\displaystyle \|x+iy\|^2 = \|x\|^2+\|y\|^2$. 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.
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  3. #3
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    Quote Originally Posted by davidmccormick View Post
    3. Let $\displaystyle A:{L^2}(0,\pi)\longrightarrow{L^2}(0,\pi)$ be defined by:

    $\displaystyle (Af)(x) = \int_0^\pi sin(x-y)f(y)dy$
    find the point spectrum and spectrum of A.
    Let $\displaystyle c,\,s$ denote the elements of $\displaystyle L^2(o,\pi)$ given by $\displaystyle c(x) = \cos x$, $\displaystyle s(x) = \sin x$. Use the relation $\displaystyle \sin(x-y) = \sin x\cos y-\cos x\sin y$ to see that $\displaystyle Af = \langle f,c\rangle s - \langle f,s\rangle c$. Thus A is a finite-rank operator, with two-dimensional range R spanned by $\displaystyle c$ and $\displaystyle s$, 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 $\displaystyle \frac\pi2\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ (with respect to the basis consisting of $\displaystyle c$ and $\displaystyle s$), from which you can easily find the other two points in the spectrum.
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  4. #4
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    man you're a legend. I still can't show that B is bounded, but ur still a legend.
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  5. #5
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    solved

    The boundedness part also solved. thanks again.
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