HW 8: Challenge 3(a,b)

Consider the sequence of functions g_n:[0,\infty]\to R defined by g_n(x):=\frac{x^n}{n}

a) Prove that (g_n) converges uniformly on [0,1] and find g=\lim g_n.

(g_n)=(\frac{x^n}{n})=\{x, \frac{x^2}{2},\frac{x^3}{3},\hdots\frac{x^n}{n}\}

In order to better examine this, look at x=.5, then (g_n)=(\frac{.5^n}{n})=\{.5, \frac{.5^2}{2},\frac{.5^3}{3},\hdots\frac{.5^n}{n}\}. The limit of \frac{.5^n}{n} as n\to\infty is 0. This holds similarly for x\in(0,1)

For x=0, 0^n/n=0\;\forall n and x=1, 1^n/n=1/n\to 0, so (g_n) converges pointwise to g(x)=0=\lim g_n

Let \epsilon>0 be arbitrary and choose \frac{1}{N}<\epsilon (which exists by Archimedean Property). Let n\geq N, x\in[0,1] both be arbitrary.

Examine |g_n(x)-g(x)|=|g_n(x)-0|=|\frac{x^n}{n}|. Notice that |\frac{x^n}{n}|\leq \frac{1}{n}\leq \frac{1}{N} <\epsilon, since n\geq N. Thus, since \epsilon, n, \text{ and } x were arbitrary (within the stated constraints), for all \epsilon>0,\exists N>\frac{1}{\epsilon}\in N such that g_n(x)-g(x)|<\epsilon whenever n \geq N and x\in [0,1]. Thus by definition of unform converges, (g_n) converges uniformly on [0,1].

b) Show that g=\lim g_n is differentiable on [0,1] and compute g'(x).

As shown in part a) g=\lim g_n=0.

Let c\in[0,1] be arbitrary and consider \lim_{x\to c} \frac{g(x)-g(c)}{x-c}. Since g is constant and defined for c, this is the same as \lim_{x\to c} \frac{0-0}{x-c}=0. This however, if the definition of the derivative of g at c, which therefore exists. Since c was arbitrary, the derivative g'(x)=0 exists for all points in [0,1], and thus by definition of differentiable,g'(x)=0 is differentiable on [0,1].

Posted on by Elise Favia
Categories: Challenges, Peer Problems

2 Responses

  1. Jeremy LeCrone says:
    November 14, 2017 at 3:10 pm

    https://richmond.qualtrics.com/jfe/form/SV_4SeHycnUpf89yex

  2. Gregory Bischoff says:
    November 19, 2017 at 8:17 pm

    All around very good! Structure is nice and clear. I’d only say to be a bit careful with typesetting. Also, when showing the existence of the limit in part b, I think you’d need to account for the fact that the denominator is 0 when x = c. You could do this by invoking L’Hopital’s rule. (though I feel I may not be right there as it is ALWAYS zero otherwise…)

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