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- #1

$(Ef)(x)=0.5(f(x)+f(-x)$ is self adoint.

It seems that in order for this to be true we have that f(-x) is the conjugate of f(x) but I don't know why this is true.

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- Thread starter
- #1

$(Ef)(x)=0.5(f(x)+f(-x)$ is self adoint.

It seems that in order for this to be true we have that f(-x) is the conjugate of f(x) but I don't know why this is true.

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- #2

- Jan 26, 2012

- 4,197

$$ \frac{1}{2} \int_{ \mathbb{R}}\left[ f(x)+f(-x) \right]^{*} g(x) \, d\mu

= \frac{1}{2} \int_{\mathbb{R}} f^{*}(x) \left[ g(x)+g(-x) \right] \, d\mu,$$

or

$$ \int_{ \mathbb{R}}\left[ f^{*}(x) g(x)+f^{*}(-x) g(x) \right]\, d\mu

= \int_{\mathbb{R}} \left[f^{*}(x) g(x)+f^{*}(x)g(-x) \right] \, d\mu,$$

or

$$\int_{ \mathbb{R}} f^{*}(-x) g(x)\, d\mu

= \int_{\mathbb{R}} f^{*}(x)g(-x) \, d\mu.$$

Can you think of a way to show this?

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- #3

I had thought of using parts and then using the fact that square integrable functions vanish at + and - infinity but sine there are no derivatives involved that looks to be a non-starter.

$$ \frac{1}{2} \int_{ \mathbb{R}}\left[ f(x)+f(-x) \right]^{*} g(x) \, d\mu

= \frac{1}{2} \int_{\mathbb{R}} f^{*}(x) \left[ g(x)+g(-x) \right] \, d\mu,$$

or

$$ \int_{ \mathbb{R}}\left[ f^{*}(x) g(x)+f^{*}(-x) g(x) \right]\, d\mu

= \int_{\mathbb{R}} \left[f^{*}(x) g(x)+f^{*}(x)g(-x) \right] \, d\mu,$$

or

$$\int_{ \mathbb{R}} f^{*}(-x) g(x)\, d\mu

= \int_{\mathbb{R}} f^{*}(x)g(-x) \, d\mu.$$

Can you think of a way to show this?

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- #4

- Jan 26, 2012

- 4,197

Well, think of the integral (temporarily) as a Riemann integral:I had thought of using parts and then using the fact that square integrable functions vanish at + and - infinity but sine there are no derivatives involved that looks to be a non-starter.

$$\int_{-\infty}^{\infty}f^{*}(x)g(-x) \, dx.$$

What happens when you let $y=-x$?

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- #5

you get the equality.Well, think of the integral (temporarily) as a Riemann integral:

$$\int_{-\infty}^{\infty}f^{*}(x)g(-x) \, dx.$$

What happens when you let $y=-x$?