- #1
Lajka
- 68
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Hi,
Quick question here: I know that C-S inequality in general states that
[itex]|<x,y>| \leq \sqrt{<x,x>} \cdot \sqrt{<y,y>}[/itex]
and, in the case of [itex]L^2(a,b)[/itex]functions (or [itex]L^2(R)[/itex] functions, for that matter), this translates to
[itex]|\int^{b}_{a}f(x)g(x)dx| \leq \sqrt{\int^{b}_{a}|f(x)|^2dx} \cdot \sqrt{\int^{b}_{a}|g(x)|^2dx}[/itex]
What I don't understand is, in a book I read, it says
[itex]||fg||_1 \leq ||f||_2 \cdot ||g||_2[/itex]
which means
[itex]\int^{b}_{a}|f(x)g(x)|dx \leq \sqrt{\int^{b}_{a}|f(x)|^2dx} \cdot \sqrt{\int^{b}_{a}|g(x)|^2dx}[/itex]
I suppose that both of these correct, but I don't how to justify the transition from [itex]|\int^{b}_{a}f(x)g(x)dx|[/itex] to [itex]\int^{b}_{a}|f(x)g(x)|dx[/itex].
I suppose I should use the fact that
[itex]|\int^{b}_{a}f(x)g(x)dx| \leq \int^{b}_{a}|f(x)g(x)|dx[/itex]
but that can't be sufficient, e.g., if 2<5 and 2<17 doesn't mean that 5<17. Any thoughts?
Thanks.
EDIT: I'm just going to get greedy and pop-in another small question from the book I use
http://i.imgur.com/l4jD0.png
Can anybody explain to me why is this cleary true? (I hate it when they say it like that, I feel dumb)
Quick question here: I know that C-S inequality in general states that
[itex]|<x,y>| \leq \sqrt{<x,x>} \cdot \sqrt{<y,y>}[/itex]
and, in the case of [itex]L^2(a,b)[/itex]functions (or [itex]L^2(R)[/itex] functions, for that matter), this translates to
[itex]|\int^{b}_{a}f(x)g(x)dx| \leq \sqrt{\int^{b}_{a}|f(x)|^2dx} \cdot \sqrt{\int^{b}_{a}|g(x)|^2dx}[/itex]
What I don't understand is, in a book I read, it says
[itex]||fg||_1 \leq ||f||_2 \cdot ||g||_2[/itex]
which means
[itex]\int^{b}_{a}|f(x)g(x)|dx \leq \sqrt{\int^{b}_{a}|f(x)|^2dx} \cdot \sqrt{\int^{b}_{a}|g(x)|^2dx}[/itex]
I suppose that both of these correct, but I don't how to justify the transition from [itex]|\int^{b}_{a}f(x)g(x)dx|[/itex] to [itex]\int^{b}_{a}|f(x)g(x)|dx[/itex].
I suppose I should use the fact that
[itex]|\int^{b}_{a}f(x)g(x)dx| \leq \int^{b}_{a}|f(x)g(x)|dx[/itex]
but that can't be sufficient, e.g., if 2<5 and 2<17 doesn't mean that 5<17. Any thoughts?
Thanks.
EDIT: I'm just going to get greedy and pop-in another small question from the book I use
http://i.imgur.com/l4jD0.png
Can anybody explain to me why is this cleary true? (I hate it when they say it like that, I feel dumb)
Last edited: