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show self adjointness

Fermat

Active member
Nov 3, 2013
188
Conisider the space $L_{2}$ of square integrable functions on R with the usual integral inner product. Show that the operator E defined by, for f in $L_{2}$,
$(Ef)(x)=0.5(f(x)+f(-x)$ 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.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
So, we need to show that $\langle Ef | g \rangle = \langle f | Eg \rangle$ for all $f,g\in L_{2}$. This is tantamount to showing that
$$ \frac{1}{2} \int_{ \mathbb{R}}\left[ f(x)+f(-x) \right]^{*} g(x) \, d\mu
= \frac{1}{2} \int_{\mathbb{R}} f^{*}(x) \left[ g(x)+g(-x) \right] \, d\mu,$$
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?
 

Fermat

Active member
Nov 3, 2013
188
So, we need to show that $\langle Ef | g \rangle = \langle f | Eg \rangle$ for all $f,g\in L_{2}$. This is tantamount to showing that
$$ \frac{1}{2} \int_{ \mathbb{R}}\left[ f(x)+f(-x) \right]^{*} g(x) \, d\mu
= \frac{1}{2} \int_{\mathbb{R}} f^{*}(x) \left[ g(x)+g(-x) \right] \, d\mu,$$
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?
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.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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.
Well, think of the integral (temporarily) as a Riemann integral:
$$\int_{-\infty}^{\infty}f^{*}(x)g(-x) \, dx.$$
What happens when you let $y=-x$?
 

Fermat

Active member
Nov 3, 2013
188
Well, think of the integral (temporarily) as a Riemann integral:
$$\int_{-\infty}^{\infty}f^{*}(x)g(-x) \, dx.$$
What happens when you let $y=-x$?
you get the equality.