$$$$

Assignment 10

We have been doing a lot of proofs in class recently, and its important not just to watch a proof be done, but to really rewrite it by yourself. So, most of this homework is asking you to prove things extremely similar to things we’ve proven in class, as motivation to write out those proofs yourself. You should be able to follow the book / notes rather closely, for hints!

Exercise 1 (Convergence at the Endpoints) We saw in class that proving in general that a power series is convergent at its endpoints is rather complicated. But in certain situations its easy to show. In this problem, assume $f(x)=\sum _n{a}_n{x}^n$ has radius of convergence 1, and also that $f(1)=\sum _n{a}_n$ converges absolutely.

Use this to prove that $f(x)$ is continuous at $x=1$.

Hint: follow the same proof technique we used for continuity on the interior, with Dominated convergence, adapted to this situation

Solution. Let $x_k\in (-1,1)$ be a sequence converging to $1$, we wish to show that $limf(x_n)=f(1)$. Writing out $f=\sum _n{a}_n{x}^n$ as a power series, we see that we wish to switch the limit and sum:

$$\underset{k}{lim}f(x_k)=\underset{k}{lim}\sum _n{a}_n{x}_k^n$$ $$\sum _n\underset{k}{lim}{a}_n{x}_k^n=\sum _n{a}_n\underset{k}{lim}{x}_k^n=\sum _n{a}_n(\underset{k}{lim}{x}_k{)}^n=\sum _n{a}_n(1{)}^n=f(1)$$

So, we must show that dominated convergence applies, which means that we need to find $M_n$ with $|a_n{x}_k^n|<M_n$ for all $k$. Because of the assumption that $\sum a_n$ converges absolutely, we can set $M_n=|a_n|$ and note $\sum M_n$ converges. Now for every $k$, $x_k\in (-1,1)⟹|x_k|<1$ so $|x_k{|}^n<1$ and $|a_n{x}_k^n|<|a_n|=M_n$. This is exactly what we needed! The terms of $f(x_k)$ are bounded above by a convergent sum, so dominated convergence applies, and the two lines above (where we switched the order of the limits and sum) are equal: $f$ is continuous at 1.

Exercise 2 (Continuity and Dense Sets I) Let $X\subset \mathbb{R}$ be a subset of the real line, and $f,g:X\to \mathbb{R}$ two continuous functions with domain $X$. Let $D\subset X$ be a dense subset of $X$, and prove that if $f(d)=g(d)$ for all $d\in D$, then actually $f$ and $g$ are the same function.

*Hint: follow closely the proof we did for $\mathbb{Q}\subset \mathbb{R}$ in class.

Solution. This argument is very similar to the argument in the textbook.

Exercise 3 (Continuity and Dense Sets II) Let $f\mathbb{R}\to \mathbb{R}$ be a continuous and $D$ a dense subst of $\mathbb{R}$ on which $f$ takes only positive values.

Prove that $f$ never takes any negative values, on all of $\mathbb{R}$.

Solution. Let $x\in \mathbb{R}$ be arbitrary. For each $n$, by density, we can find a point $d_n\in D$ with $x<d_n<x+1/n$. Using the squeeze theorem, we see $d_n\to x$, and so by continuity of $f$ we know $$f(x)=f(lim{d}_n)=limf(d_n)$$

But on $D$ $f$ is positive, so for each $n$ we know $f(d_n)>0$. Thus by the limit inequalities, we see $$limf(d_n)\ge 0$$

So putting these two together, we have $f(x)\ge 0$, as required.

Exercise 4 (Extreme Value Theorem) Show that if $f$ is a continuous function on a closed interval $[a,b]$, that $f$ achieves a minimum value: that is, there exists a $q\in [a,b]$ such that $f(a)\le f(x)$ for all $x\in [a,b]$.

Hint: follow closely the proof we did in class, for the existence of a maximum.

Solution. This argument is very similar to the argument in the textbook.

Exercise 5 (Intermediate Value Theorem) Let $f$ be continuous on $[a,b]$ and assume $f(a)>0$ and $f(b)<0$. Show there exists a $z\in [a,b]$ with $f(z)=0$.

Hint: follow closely the proof we did in class, where we had $f(a)<0$ and $f(b)>0$

Solution. This argument is very similar to the argument in the textbook.

Exercise 6 (Using the IVT) Prove that there is a real number which solves the equation $$x^3=10x+10,000$$

Solution. Consider the polynomial $f(x)=x^3-10x-10,000$, and observe that a zero of $f$ is a solution to the equation we seek. Since polynomials are continuous functions we aim to use the Intermediate Value Theorem, and prove the existence of a zero by finding a point where $f$ is negative and another point where $f$ is positive. At zero, $f(0)=-10,000$ is certainly negative. But at $x=100$ $$f(100)=(100{)}^3-10\cdot 100-10,000=1,000,000-(11,000)=9,989,000$$ which is very positive. Thus, there must exist some $z\in [0,100]$ where $f(z)=0$.

Exercise 7 (Uniform Continuity) Recall function is called uniformly continuous if for $ϵ>0$ you can pick a single $\delta$ that works on the entire domain. (You do not need the optional chapter uniform continuity to do this problem: just that definition).

Let $f,g$ be two uniformly continuous functions defined on the same domain $X\subset \mathbb{R}$. Prove that $f+g$ is uniformly continuous on $X$.

Solution. Choose $ϵ>0$ arbitrary. Then by the uniform continuity of $g,f$ we can find a ${\delta }_f$ such that $|x-y|<{\delta }_f⟹|f(x)-f(y)|<ϵ/2$, and similarly a ${\delta }_g$ such that $|x-y|<{\delta }_g⟹|g(x)-g(y)|<ϵ/2$. Set $\delta =min\{{\delta }_f,{\delta }_g\}$, and let $x,y$ be two points int eh domain such that $|x-y|<\delta$. Then for $h=f+g$,

$$|h(x)-h(y)|=|(f(x)+g(x))-(f(y)+g(y))|=|(f(x)-f(y))+(g(x)-g(y))|$$ By the triangle inequality, $$\le |f(x)-f(y)|+|g(x)-g(y)|$$ And as $|x-y|<\delta$, its $\le$ both ${\delta }_f$ and ${\delta }_g$ by design. So each of these quantities is less than $ϵ/2$, and the entire thing is less than $ϵ$.

Since $x,y$ were arbitrary in the domain subject only to the constraint of being within $\delta$ of one another, this same uniform $\delta$ works for all such pairs, and $f+g$ is uniformly continuous.