- #1
bjnartowt
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- 3
Homework Statement
Show that [itex]\int {{f^*}(x)g(x) \cdot dx} [/itex] is an inner product on the set of square-integrable complex functions.
Homework Equations
Schwarz inequality:
[tex]\left| {\int {{f^*}(x)g(x) \cdot dx} } \right| \le \sqrt {\int {{{\left| {f(x)} \right|}^2} \cdot dx} \int {{{\left| {g(x)} \right|}^2} \cdot dx} } [/tex]
The Attempt at a Solution
Rephrase what we want to prove: see if:
[itex]\int {{f^*}(x)g(x) \cdot dx} < \infty [/itex]
...is true.
Since we are considering the set of square-integrable-functions: “f” and “g” are in the set of square-integrable functions: they are elements of Hilbert space:
[tex]\left\{ {f(x),g(x)} \right\} \in {L^2}[/tex]
This means the following integrals exist:
[tex]\int_{ - \infty }^{ + \infty } {{f^*}(x)f(x) \cdot dx} = \int_{ - \infty }^{ + \infty } {{{\left| {f(x)} \right|}^2} \cdot dx} < \infty {\rm{ }}\int_{ - \infty }^{ + \infty } {{g^*}(x)g(x) \cdot dx} = \int_{ - \infty }^{ + \infty } {{{\left| {g(x)} \right|}^2} \cdot dx} < \infty [/tex]
In turn: this gaurantees:
[tex]\sqrt {\int {{{\left| {f(x)} \right|}^2} \cdot dx} \int {{{\left| {g(x)} \right|}^2} \cdot dx} } < \infty [/tex]
Schwarz inequality says:
[tex]\left| {\int {{f^*}(x)g(x) \cdot dx} } \right| \le \sqrt {\int {{{\left| {f(x)} \right|}^2} \cdot dx} \int {{{\left| {g(x)} \right|}^2} \cdot dx} } [/tex]
Together: the previous two equations require:
[tex]\left| {\int {{f^*}(x)g(x) \cdot dx} } \right| < \infty [/tex]
Another inequality is that:
[tex]\left| {\int {{f^*}(x)g(x) \cdot dx} } \right| < \int {\left| {{f^*}(x)g(x)} \right| \cdot dx} [/tex]
…which actually ruins what we’re trying to prove. Gah. Well, it doesn't counter-prove it, but I've obviously used the wrong inequality. Somehow it must be that:
[tex]\int {\left| {{f^*}(x)g(x)} \right| \cdot dx} \le \sqrt {\int {{{\left| {f(x)} \right|}^2} \cdot dx} \int {{{\left| {g(x)} \right|}^2} \cdot dx} } [/tex]
but I'm not sure how to get that. Triangle inequality? Perhaps I am staring too hard, but it doesn't seem so?