$\displaystyle f_n(x)=x^n$ [0,1]
on how to find the value to put for the nirm convergence
??
my prof put zero as the limit function although the function converges to 1 in x=1
||f_n -0||
i dont know why
?
$\displaystyle f_n(x)=x^n$ [0,1]
on how to find the value to put for the nirm convergence
??
my prof put zero as the limit function although the function converges to 1 in x=1
||f_n -0||
i dont know why
?
A sequence of functions, $\displaystyle {f_n}$ converges to a function, g, "in norm", if and only if, given $\displaystyle \epsilon> 0$, there exist N such that if n> N, then ||f_n(x)- g(x)||< \epsilon[/tex]. That is, of course, just the "usual" definition of convergence of a sequence with absolute value replaced by "norm". Now, the question is, what norm are you using? Here, you appear to be using either $\displaystyle ||f- g||= \int |f(x)-g(x)|dx$ or $\displaystyle ||f-g||= \sqrt{\int (f(x)- g(x))^2 dx$. In any case, the point is that individual values will not affect those. You can change the values as you please at individual points and not change the values of the integrals.
(Where you say "equally convergence" in the title, I thing the correct English term is "uniformly convergent" or "uniform convergence".