Uniform Convergence Examples James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University October 13, 2017
Outline 1
Example Let (x n ) be the sequence of functions on [0, 1] defined by x n (t) = 2nt e nt2. Discuss pointwise and uniform convergence on [0, 1]. Solution First, it is easy to see the pointwise limit function is x(t) = 0 on [0, 1]. Is the convergence uniform? Let s begin by graphing some of these functions in Octave. The code is straightforward. 1 T = l i n s p a c e ( 0, 1, 5 1 ) ; f = @( t, n ) 2 n t. / exp ( n t. ˆ 2 ) ; p l o t (T, f (T, 5 ),T, f (T, 1 0 ),T, f (T, 2 0 ),T, f (T, 3 0 ) ) ; x l a b e l ( t ) ; y l a b e l ( y ) ; t i t l e ( x_n ( t ) = 2 nt / e ^{ nt ^2} for n = 5,10,20 and 30 ) ; 6 l e g e n d ( x5, x10, x20, x30, location, north ) ;
Here you see a plot of x 5, x 10, x 20 and x 30 on the interval [0, 1]. You can clearly see the peaks of the functions are increasing with the maximums occuring closer and closer to 0.
Solution The derivative here, after some simplification, is x n(t) = 2n(1 2nt2 ) which e nt2 is zero at t = 0 and t = ±1/ 2n. The critical point at t = 0 is uninteresting and the maximum occurs at t n = 1/ 2n and has value 2n/e. Let s calculate x n x here. We have x n x = sup 2nt/e nt2 0 = 2n/e 0 t 1 For the convergence to be uniform, given ɛ > 0, we would have to be able to find N ɛ so that x n x < ɛ when n > N ɛ. Here that means we want 2n/e < ɛ when n > N ɛ. But for n > N ɛ, 2n/e. So this cannot be satisfied and the convergence is not uniform.
Now let s look at this same sequence on a new interval. Example Examine the convergence of the sequence x n (t) = 2nt e nt2 on [ 2, 2]. Solution We can graph some of the functions in this sequence on this new interval using the code below. T = l i n s p a c e ( 2,2,101) ; f = @( t, n ) 2 n t. / exp ( n t. ˆ 2 ) ; p l o t (T, f (T, 5 ),T, f (T, 1 0 ),T, f (T, 2 0 ),T, f (T, 3 0 ),T, f (T, 5 0 ) ) ; 4 x l a b e l ( t ) ; y l a b e l ( y ) ; t i t l e ( x_n ( t ) = 2 nt / e ^{ nt ^2} on [ -2,2] ) ;
Here you see a plot of some of these functions on the interval [ 2, 2]. You can clearly see the peaks of the functions are increasing with the minimums and maximums occuring closer and closer to 0. The minimum occur at 1/ 2n with value 2n/e, while the maximums occur at 1/ 2n with value 2n/e.
Solution Now take a small positive number a and mentally imagine drawing a vertical line through previous picture at that point. The maximum s occur at 1/ 2n. If 1/ 2n <.1 or n > 50, the maximum values all occur before the value x =.1. We generate a plot for this as follows: (here we do not add the code for the axis labels etc.) T = l i n s p a c e ( 0. 0 5, 0. 5, 1 0 1 ) ; f = @( t, n ) 2 n t. / exp ( n t. ˆ 2 ) ; p l o t (T, f (T, 5 0 0 ),T, f (T, 7 0 0 ),T, f (T, 9 0 0 ) ) ; On the interval [0.1, 0.5], the functions have their maximum value at x n (0.1) =.2n/e.01n. Since.2n/e.01n 0 as n, we see given ɛ > 0, there is N ɛ so that.2n/e.01n < ɛ when n > N ɛ. You can see this behavior clearly in the next figure.
Solution There we graph x 500, x 700 and x 900 and you can easily see the value of these functions at 0.1 is decreasing. Hence, for n > N ɛ, x n 0 < ɛ unif and so we can say x n 0 on [.1, r] for any r > 0.1. A similar analysis works for any a > 0. In fact, if we looked at the other side, we would show x n unif 0 on any interval of the form [ r, a] with a > 0. Indeed, this convergence is uniform on any interval [c, d] as long as 0 [c, d].
Now let s look at this same sequence on [.1, 2] analytically. We know the minimum of x n (t) occur at 1/ 2n with value 2n/e, while the maximum of x n (t) occurs at 1/ 2n with value 2n/e. There is an N so that n > N implies < 1/ 2n <.1; i.e. the maximum value before x =.1. It occurs before the interval [.1, 2]. Since on the right of the maximum, x n (t) decreases, this tells us the maximum of x n (t) on [.1, 2] is given by x n (.1). So sup x n (t) 0 = x n (.1) =.2n t [.1.2] e.01n and we see this goes to zero with n. So convergence is uniform on [.1, 2].
Example Examine the pointwise and uniform convergence of (x n ) where x n (t) = 3nt/e 4nt2 on intervals of R. Solution Work this out in class following the example just done. Use MatLab and blackboard sketches to see what is going on.
Example Examine the pointwise and uniform convergence of (x n ) where x n (t) = 2nt/e 3nt2 on intervals of R. Solution Work this out in class following the example just done. Use MatLab and blackboard sketches to see what is going on.
Homework 21 21.1 Let x n be defined by x n (t) = { 0, 0 t 1/n n, 1/n < t < 2/n 0, 2/n t 1 Determine if (x n ) converges uniformly to its limit function. 21.2 Examine the pointwise and uniform convergence of (x n ) where x n (t) = 6nt/e 2nt2 on intervals of R. Do a careful analysis just like we have done for the other examples. Sketches are required!