Thanks for the explanation, actually only one addition for 06:17. If assume there is a point where z(t)=f(t)-g(t) e 0. Let's say a point T. And consider limit of z(t) as t approaches T. Due to that z(t) is continuous and continuous functions preserved sign of its value in the limit point then there is exists interval [T-\sigma, T+\sigma] where Z(T) is e 0. And this implies that the integral of the norm of Z(t) over [a,b] will be definitely not zero. Continuity is crucial, because for two function R->R -- x(t)=0 and y(t)=0 for all t, except t=0 where y(0)=1 in fact have that inegral |y(t)-x(t)| will be zero,....But y(t) is not equal to x(t).
5:57 to make it properly rigorous, this argument needs a little more. If the space was all functions instead of just the continuous ones, it wouldnt work. We can have an argument like this: Assume we have some (x,y) s.t. y=f(x)=/=0. Let u be in [a,b] s.t. it's the point closest to x satisfying f(u)=0. If no such u exists, then clearly the integral is non-zero because the function is always non-zero. Hence this assumption is legit. Then by the Intermediate value thm, every L on (0,L) can be obtained by a t on (u,x) (or (x,u) depends on the situation. WLOG, lets assume the former). So then the integral is greater or equal to just integrating over (u,x) which is integrating over a positive function, which yields a positive answer, which then is a contradiction. So the assumption that there is (x,y) s.t. y=f(x)=/=0 is false and f is identically 0.
![C[a,b] with the Supremum Metric is Complete Part 1](http://i.ytimg.com/vi/g5UjOR1w5NE/mqdefault.jpg)
Thanks for the explanation, actually only one addition for 06:17.
If assume there is a point where z(t)=f(t)-g(t)
e 0. Let's say a point T. And consider limit of z(t) as t approaches T.
Due to that z(t) is continuous and continuous functions preserved sign of its value in the limit point then there is exists interval [T-\sigma, T+\sigma] where Z(T) is
e 0. And this implies that the integral of the norm of Z(t) over [a,b] will be definitely not zero.
Continuity is crucial, because for two function R->R -- x(t)=0 and y(t)=0 for all t, except t=0 where y(0)=1 in fact have that inegral |y(t)-x(t)| will be zero,....But y(t) is not equal to x(t).
Great explanations. Thanks!
At 6.08, shall it be f=g a.e. ?
Beautiful.
Im using function spaces to calculate the height of my house
haha
why is S ?
5:57 to make it properly rigorous, this argument needs a little more. If the space was all functions instead of just the continuous ones, it wouldnt work. We can have an argument like this: Assume we have some (x,y) s.t. y=f(x)=/=0. Let u be in [a,b] s.t. it's the point closest to x satisfying f(u)=0. If no such u exists, then clearly the integral is non-zero because the function is always non-zero. Hence this assumption is legit. Then by the Intermediate value thm, every L on (0,L) can be obtained by a t on (u,x) (or (x,u) depends on the situation. WLOG, lets assume the former). So then the integral is greater or equal to just integrating over (u,x) which is integrating over a positive function, which yields a positive answer, which then is a contradiction. So the assumption that there is (x,y) s.t. y=f(x)=/=0 is false and f is identically 0.