math8 said:I think the maximum of |fn(x)-f(x)| on [0,1] is 1/1+n. So can I conclude that the convergence is uniform because there is an "N" that doesn't depend on x such that for all n> or eq. to N, 1/1+n gets arbitrarily small?
math8 said:that N should be > than 1-e/e right?
Uniform convergence in [0,1] refers to the convergence of a sequence of functions to a limit function on the interval [0,1]. This means that the functions in the sequence approach the limit function at the same rate on the entire interval, rather than just at certain points.
Pointwise convergence only requires that the functions in a sequence approach the limit function at each individual point, whereas uniform convergence requires that the functions approach the limit function at the same rate on the entire interval.
Uniform convergence is important because it allows us to guarantee that the limit function of a sequence of functions is continuous, even if the individual functions in the sequence may not be. This makes it a useful tool in proving the convergence of integrals and series.
Yes, it is possible for a sequence of functions to converge uniformly without converging pointwise. This means that the functions in the sequence approach the limit function at the same rate on the entire interval, but may not approach the limit function at each individual point.
One way to test for uniform convergence on an interval [a,b] is to use the Weierstrass M-test. This test involves finding an upper bound for the absolute value of each function in the sequence and then checking if the series formed by these upper bounds converges. If the series converges, then the sequence of functions converges uniformly on the interval [a,b].