Processing math: 100%
With a B(x), u∉V, but ∑jcjψj∈V
- {ψi}i∈Is is a basis for V
- ∑j∈Iscjψj(x)∈V
- But u∉V!
- Reason: say u(0)=C and u∈V; any v∈V has v(0)=C, then
2u∉V because 2u(0)=2C (wrong value)
- When u(x)=B(x)+∑j∈Iscjψj(x),
B∉V (in general) and
u∉V, but (u−B)∈V since ∑jcjψj∈V