- #1
Alesak
- 111
- 0
Hi,
in the book "Applied analysis by Hilbert space method", I have no idea what is happening in the last step here:
[itex]<L(f), g>=\int^{1}_{0} L(f)\overline{g}dx=-\int^{1}_{0} f''\overline{g}dx=-\int^{1}_{0}\overline{g} d(f')[/itex]
where L(f) = -f'' and both f and g are functions of x.
It seems like some kind of substitution is going on. But how can the author disappear f'', change dx and not change g(x) at the same time?
He is showing here that l(y)=-y'' is hermitian operator and this is first step.
in the book "Applied analysis by Hilbert space method", I have no idea what is happening in the last step here:
[itex]<L(f), g>=\int^{1}_{0} L(f)\overline{g}dx=-\int^{1}_{0} f''\overline{g}dx=-\int^{1}_{0}\overline{g} d(f')[/itex]
where L(f) = -f'' and both f and g are functions of x.
It seems like some kind of substitution is going on. But how can the author disappear f'', change dx and not change g(x) at the same time?
He is showing here that l(y)=-y'' is hermitian operator and this is first step.