Problem 3. Give an example of a sequence of continuous functions on a compact domain converging pointwise but not uniformly to a continuous function

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1 Problem 3. Give an example of a sequence of continuous functions on a compact domain converging pointwise but not uniformly to a continuous function Solution. If we does not need the pointwise limit of f n (x) to be continuous, then this problem would be a lot easier. For example we know that f n (x) = x n, x [0, ] will serve as a perfect example. However the pointwise limit of this sequence is not a continuous function on [0, ] so we need some more efforts. Now if U is a compact set in the problem we need to find a sequence of functions f n (x) whose pointwise limit is f(x) but we have lim sup f n (x) f(x) 0. n x U Inspired by this motivation, we can find the following modifications of the earlier sequence: n x, x [0, ] n f n (x) = 2 n x, x [ n, 2 ] n 0, x [ 2, ]. n It is not hard to verify that the pointwise lmit of f n (x) is 0. In fact, it is obvious that f n (0) 0 at first. Now for any x (0, ] we will have a sufficiently large N such that f n (x) = 0 if n N. However, f n (x) does not converge uniformly to 0 due to the fact that sup f n (x) f(x) =. x [0,] Problem 5. If lim n f n = f and the functions f n are all monotone increasing, must f be monotone increasing? What happens if f n are all strictly increasing? Solution. We recall that a function f is called monotone(strictly) increasing if for x < y in its domain we will have f(x) (<)f(y). Now assume f n (x) converge to f(x) uniformly and f n (x) are all monotone increasing. We apparently have the following inequality for x < y: f(x) f(y) = [f(x) f n (x)] + [f n (x) f n (y)] + [f n (y) f(y)] f n (x) f(x) + [f n (y) f n (x)] + f n (y) f(y) in which n is an arbitrary natural number. Now fix x and y, so we have two sequence of numbers on both sides of the inequality. So take n in () we will see that f(x) f(y). The previous argument surely fails if we replace monotone increasing by strictly increasing, as the limit now might be monotone increasing only insteading of being strictly increasing. For example consider f n (x) = x n for x (/2, 2/3). It is easy to see that f n are all strictly increasing, but their limit f(x) = 0 is not. Here one should notice that a constant function is still regarded as a monotone increasing function, which is not quite the same as our common sense. Problem 7. uniformly. If f n (x) < a n for all x and n= a n converges, prov that n= f(x) converges Solution. If f N (x) = N n= f n(x) converge uniformly to certain function f(x), f(x) must be the pointwise limit, whose existence is guaranteed because a n converges. Therefore we ()

2 must have, for a fixed x, that f(x) = n= f(x). Now we have f N (x) f(x) n=n+ for some x R. If we take sup norm of both sides of this inequality as they are both functions in x we will have sup f N (x) f(x) x R a n n=n+ whose right hand side converges to 0 when N. Therefore we proved that f n converges to f(x), its pointwise limit, uniformly. Problem 8. Give an example of a sequence of continuous functions on a non-compact domain D that does not converge uniformly, yet lim n f n (x n ) = f(x) for every sequence {x n } converge to x in D Solution. We use the example in Problem 3 again here, with the only modification that we consider their restriction instead of function f n themself on (0, ). Notice that this is critical to our problem or else lim n f n (/n) = will contradict with the convergence requirement in the current problem. Nevertheless, we now need to prove that lim a n = f n (x n ) = 0 for an arbitrary sequence x n x, x (0, ). Again this isn t very hard to prove. Assume x > 2/n x for some natural number n x. Because x n x, for sufficiently large n we will have x n > /n x, which conseqently means a n = 0 for such large n. Therefore we completed the proof. Problem 7. Compute the power-series expansion of the following functions about x = 0 a. f(x) = x 2 /( x 2 ), b. f(x) = /( x) 2, c. f(x) = + x. a n Solution. Before we compute the expansions of these functions, let us clarify a few notions which might be helpful for us to understand what a power-series expansion is and how to compute it. 2

3 For any smooth function, there is an associated formal Taylor series f (n) (x 0 )(x x 0 ) n. n! k=0 Here we should notice that this is just a formal series, we do not require it to be convergent or not. If there is a positive number r > 0 such that when x x 0 < r the above series of functions converges pointwisely to a function g(x), then we say the radius of convergence of the above power series is at least r. You can take the supremum of all such r as R and call it the convergence radius of the Taylor series. If g(x) = f(x) when x x 0 < r, for some 0 < r < R, then we say f has a power series expansion near x 0. In general, R might not exsit at all, which means the Taylor series of a function might not be convergent at all except at the trivial point x = x 0. The example can be given by the following function: f(x) = e n cos n 2 x. and its Taylor series at point 0. In general, r might not exist at all. If you consider the following function: {e x f(x) = 2, x 0 0, x = 0 n= then you will realize its derivatives at 0 will be f (n) (0) = 0, which means the Taylor series of f(x) will be constantly 0! But surely f(x) = 0 only when x = 0. Notice that the above example also shows that r may not be equal to R, as R = in this case but r does not exsit, or you can say r = 0 heuristically. Uniqueness of power-series expansion. This is a critically important but also trivial property of power-series expansion. By this property, you have two ways to compute the power-series expansion of a function f(x), either directly finding such power-series expansion, or computing the Taylor series, then verify it converges back to f(x). a. Because we have when x < we realize that the power-series x 2 = + x2 + x x 2n +..., x 2 + x x 2n

4 will be convegent to x 2 x, 2 which is naturally the power-series expansion of it. Notice that there is an alternative way to compute the expansion, which is based on the following algebraic manipulations f(x) = x 2 = 2 ( + x + x ). Now given thhis formula, it will be not that hard to compute the derivatives d n dx n x 2 x 2. Not surprisingly the Taylor series you finally obtain will be identical to the power series obtained in the first way, say x 2 + x x 2n However, there is another issue you need to address, to verify this power series converges back to x 2 /( x 2 ) when x <. Thankfully this is not very hard because it is a geometric series, but we still have to do the arguments in the first solution again! Notice that this is necessary, because as the counterexample I showed before, Taylor series might not converge back to f at all, which will certainly result in the conclusion that f does not have a powerseries expansion near 0(why?). b. This function is easier to handle than the previous one because we should have no difficulty in computing its derivatives f (n) (0). More specifically, we have Therefore we have the following Taylor series: d n (n + )! f(x) =. (2) dxn ( x) n+ g(x) = (n + )x n. (3) n=0 But we are not done yet. We need to show that this series converges back to /( x) 2 when x <. The easiest way to to this, is to estimate the error term in the Taylor theorem. Recall that such error term if we compute the Taylor series up to order n, will be f (n+) (x 0 )x n+ (n + )! for some x 0 [0, x]. By (2) we know that this is ( x 0 ) n+ x n+ 0 4

5 which converges to 0 as n because we know x 0 <. Therefore we proved that when x <, (3) not only converges, but also converges back to f(x) = /( x) 2. Notice that there is another tricky solution to this problem: ( x) = 2 x x = ( + x + x )( + x + x ) = (n + )x n. n=0 The last equity is because if you combine all the x n terms together, you will realizz there are x n, xx n, x 2 x n 2,..., x n, n + many such terms. c. It is easy to calculate d n dx f(0) = n 2 ( 2 ) ( 3 2 ) ( 5 2 ) ( 2n 3 2 if n 2. So we will have the Taylor series ( )n n k=2 (2k 3) x n n n! ) = ( )n n k=2 (2k 3) (4) 2 n Now we still need to prove that this series converges back to + x when x <. If we observe the general terms carefully in (4) we notice that if we have x 0 x <, then f (n+) (x 0 ) = n! 2 Therefore we completed the proof n 2n + 2 x 0 n+ x 0 n+. Problem 8. Compute the radius of convergence of n 2 2 n x n. Solution. We need to compute the limit lim n (n2 2 n ) n = 2(n n ) 2. Now for the limit of n n, the best way to compute it is to consider that log(n n ) = log n/n 0, which means n n. So we will have the radius of convergence as /2. 5