$$$$

Assignment 9

Exercise 1 Prove, directly from the $ϵ-\delta$ definition that the function $f(x)=1/x$ is continuous at $x=2$.

Solution. Scratch Work

Let $ϵ>0$ be given. We want to find $\delta >0$ such that whenever $|x-2|<\delta$, it follows that $$|\frac{1}{x}-\frac{1}{2}|<ϵ.$$

To simplify this expression, observe: $$|\frac{1}{x}-\frac{1}{2}|=\left|\frac{2-x}{2x}\right|=\frac{|x-2|}{2|x|}.$$

We want to make this less than $ϵ$, which happens by makin the numerator small, while making sure to not make the denominator too small.
To keep ourselves from having to worry about a zero in the denominator, lets look for a $\delta$ which is less than $1$, so that $x$ can only stray that far from 2. That is, if $|x-2|<\delta \le 1$, we have $1<x<3$, so $|x|<3$ and $|x|>1$. Therefore, $$\frac{|x-2|}{2|x|}<\frac{|x-2|}{2\cdot 1}=\frac{|x-2|}{2}.$$

To ensure this is less than $ϵ$, it suffices to require $$\frac{|x-2|}{2}<ϵ\text{or}|x-2|<2ϵ.$$

So letting $\delta =2ϵ$ would do it, but remembering we earlier restricted $\delta <1$ we should really take $\delta =min\{1,2ϵ\}$.

Rigorous Proof

Fix an $ϵ>0$, and set $\delta =min\{1,2ϵ\}$. Then for any $x$ with $|x-2|<\delta$, we have $$|\frac{1}{x}-\frac{1}{2}|=\frac{|x-2|}{2|x|}<\frac{|x-2|}{2}<ϵ.$$

Therefore, by the $ϵ$–$\delta$ definition, $f(x)=\frac{1}{x}$ is continuous at $x=2$.

Exercise 2 Show that the following function is continuous at $0$ and discontinuous everywhere else:

$$g(x)=\{\begin{array}{ll}x& x\in \mathbb{Q}\\ 0& x\notin \mathbb{Q}\end{array}$$

Solution. First, we prove that $x$ is continuous at $0$. Choose $ϵ>0$ and set $\delta =ϵ$. Then if $|x-0|<\delta$ (so $x\in (-\delta ,\delta )$) we must show that $|g(x)-g(0)|<ϵ$ (which, as $g(0)=0$ means we must show that $|g(x)|<ϵ|$).

There are two cases: either $x$ is rational or $x$ is irrational. If $x$ is irrational, then $g(x)=0$ so $|g(x)|=0$, which is certainly less than $ϵ$. If $x$ is rational, then $g(x)=x$, and we already know $|x|<\delta =ϵ$, so $|g(x)|=|x|<ϵ$. Thus $g$ is continuous at $0$.

Now we show $g$ is discontinuous elsewhere, by showing its discontinuous at $a$ for arbitrary $a$.
This means for some specific epsilon we must show that no delta works (if we go with the $ϵ-\delta$ approach) or we have to show there exists a sequence $x_n\to a$ where $g(x_n)\nrightarrow g(a)$ (if we take the sequences approach). We did $ϵ-\delta$ for part I, so for the sake of practicing multiple techniques I’ll use sequences here.

We break into two cases, if $a$ is rational or irrational. First, if $a$ is rational, then $g(a)=a\ne 0$. Because the irrationals are dense, we can construct a sequence of irrational numbers $x_n\to a$ (if we forgot how to do this, use density to get an irrational $x_n\in (a-1/n,a+1/n)$ and then use the squeeze theorem to confirm $x_n\to a$). Now $g(x_n)=0$ for all $n$ as $x_n$ is irrational, so $limg(x_n)=lim0=0$ as the limit of a constant sequence. But $g(a)\ne 0$ so $g(x_n)$ does not converge to $g(a)$.

Similarly, if $a$ is irrational, now $g(a)=0$ but we can use the density of the rationals to produce a sequence $x_n\to a$ of rational numbers. Here $g(x_n)=x_n$, and so $limg(x_n)=lim{x}_n=a$. Since $a\ne 0$ we see $g(x_n)$ does not converge to $g(a)$, so again $g$ is not continuous at $a$.

Exercise 3 Prove the remaining “continuity theorems”: if $f,g$ are continuous at a point $a$, then so is $f(x)g(x)$, and if $g(a)\ne 0$ then so is $f(x)/g(x)$.

Solution. These follow immediately from the corresponding limit laws, and the ability to use sequences to detect continuity. Choose an arbitrary sequence $x_n\to a$. Then $f(x_n)\to a$ as $f$ is continuous at $a$, and $g(x_n)\to g(a)$ as $g$ is continuous at $a$. By the limit law for convergent sequences, $f(x_n)g(x_n)$ is convergent, wtih limit $f(a)g(a)$. Thus, for an arbitrary $x_n\to a$ we see $f(x_n)g(x_n)\to f(a)g(a)$ so the function $f(x)g(x)$ is continuous at $a$.

If furthermore we assume that $g(a)\ne 0$, we can attempt the same idea, computing the limit of $f(x_n)/g(x_n)$ with the limit laws. But there’s a slight hold - up. The limit law for quotients says that we need $g(a)\ne 0$ (which we have assumed) and also $g(x_n)\ne 0$ for all $n$ (or at least after some finite point we can truncate: otherwise the sequence of quotients wouldn’t even be defined!) This we don’t know yet, so we have to do a little work.

Since $g$ is continuous at $a$, for any $ϵ$ theres a $\delta$ where $|x-a|<\delta$ implies $|f(x)-f(a)|<ϵ$. Taking $ϵ=|g(a)|/2$, we get a $\delta$ where for $x\in (a-\delta ,a+\delta )$, $g(x)$ is also nonzero (its at most half as close to zero as $g(a)$ was). Now our sequence $x_n\to a$ is convergent, so for this $\delta$ we can find an $N$ for which $n>N$ implies $|x_n-a|<\delta$. That is, past this $N$, we see $g(x_n)\ne 0$ so we can truncate the first $N$ terms of our sequence: this doesn’t change the limit, but now we have a sequence of nonzero terms converging to a nonzero limit, and can apply the limit theorem for quotients:

$$lim\frac{f(x_n)}{g(x_n)}=\frac{limf(x_n)}{limg(x_n)}=\frac{f(a)}{g(a)}$$

Since our sequence was arbitrary this holds for all such sequences, so $f/g$ is continuous at $a$.

Exercise 4 (Continuity of Polynomials) Prove that every polynomial is a continuous function on the entire real line. Hint: prove $x^n$ is continuous for each $n$ by induction. Then prove the result for polynomials by induction on their degree!

Solution. We know that $f(x)=x$ is continuous from class (or we can rapidly prove this by setting $\delta =ϵ$ in the definition). So, for induction assume $x^n$ is continuous for some $n$ and we seek to prove $x^{n+1}$ is continuous. Noting that $x^{n+1}=x{x}^n$, we use the previous problem for products of continuous functions: if $a\in \mathbb{R}$ is arbitrary both $x$ and $x^n$ are continuous at $a$, so $x{x}^n=x^{n+1}$ is as well. Since $a$ was arbitrary, this is continuous on the entire real line.

For polynomials we proceed by a similar induction. In class we proved that constant functions are continuous, so again using the previous homework problem we see $ax$ is continuous, and then using continuity of sums $ax+b$ is continuous for all $a,b\in \mathbb{R}$. Thus all linear polynomials are continuous. We proceed by induction on degree. Assume all polynomials of degree $n$ or less are continuous, and let $f(x)$ be a polynomial of degree $n+1$. This means $f(x)$ contains a term of degree $n+1$, plus lower order terms, and so we can write $$f(x)=a{x}^n+g(x)\mathrm{deg}(g)\le n$$

Since $k\in \mathbb{R}$ is constant, we apply the product theorem to see $k{x}^n$ is continuous, and noting $g$ is continuous by our induction hypothesis, the sum theorem to see $k{x}^n+g$ is continuous, at each $a\in \mathbb{R}$. We’ve completed the induction, so this holds for all polynomials.

Exercise 5 (The Pasting Lemma) Let $f,g$ be two continuous functions and $a\in \mathbb{R}$ is a point such that $f(a)=g(a)$. Prove that the piecewise function below is continuous at $a$. $$h(x)=\{\begin{array}{ll}f(x)& x\le a\\ g(x)& x>a\end{array}$$

Solution. We will prove that $h$ is continuous at $x=a$ using the $\epsilon$–$\delta$ definition of continuity.

Let $\epsilon >0$ be given.

Since $f$ is continuous at $a$, there exists ${\delta }_1>0$ such that for all $x$ with $|x-a|<{\delta }_1$, we have $$|f(x)-f(a)|<\epsilon .$$

Since $g$ is also continuous at $a$, there exists ${\delta }_2>0$ such that for all $x$ with $|x-a|<{\delta }_2$, we have $$|g(x)-g(a)|<\epsilon .$$

Let $\delta =min\{{\delta }_1,{\delta }_2\}$. Now suppose $|x-a|<\delta$.

We consider two cases:

If $x\le a$, then $h(x)=f(x)$ and $h(a)=f(a)$. Since $|x-a|<\delta \le {\delta }_1$, we have $$|h(x)-h(a)|=|f(x)-f(a)|<\epsilon .$$

If $x>a$, then $h(x)=g(x)$ and $h(a)=f(a)=g(a)$. Since $|x-a|<\delta \le {\delta }_2$, we have $$|h(x)-h(a)|=|g(x)-g(a)|<\epsilon .$$

In both cases, $|h(x)-h(a)|<\epsilon$ whenever $|x-a|<\delta$. Thus $h$ is continuous at $a$ as required.

We can also do this with the sequence condition, so give a second proof, for practice

:::{.solution} We will prove that $h$ is continuous at $x=a$ using the sequential criterion for continuity. Let $(x_n)$ be any sequence such that $x_n\to a$. We want to show that $h(x_n)\to h(a)=f(a)=g(a)$.

We divide our sequence $x_n$ into two pieces: those that are greater than $a$ and those that are less than or equal to $a$. We argue in two cases:

Case 1: One of these two sets is finite. Then the sequence $(x_n)$ lies entirely on one side of $a$ after some point. That is, either $x_n\le a$ for all sufficiently large $n$, or $x_n>a$ for all sufficiently large $n$.

• If $x_n\le a$ eventually, then for large $n$, $h(x_n)=f(x_n)$, and since $f$ is continuous, $f(x_n)\to f(a)$.
• If $x_n>a$ eventually, then for large $n$, $h(x_n)=g(x_n)$, and since $g$ is continuous, $g(x_n)\to g(a)$.

Since $h(a)=f(a)=g(a)$, both of these cases end with $h(x_n)$ tending to $h(a)$.

Case 2: The sequence $(x_n)$ has infinitely many terms both less than or equal to $a$, and greater than $a$.

Define two subsequences: - Let $(x_n^{\prime })$ be the subsequence of $(x_n)$ with $x_n^{\prime }\le a$. - Let $(x_n^{″})$ be the subsequence of $(x_n)$ with $x_n^{″}>a$.

Since $x_n\to a$, both subsequences also converge to $a$. Now: - $h(x_n^{\prime })=f(x_n^{\prime })\to f(a)=h(a)$ by continuity of $f$. - $h(x_n^{″})=g(x_n^{″})\to g(a)=h(a)$ by continuity of $g$.

By assumption, $f(a)=g(a)$, so both subsequences converge to the same value. Now we use a previous result, that if a sequence decomposes as the union of two subsequences converging to the same limit, the overall sequence converges. Thus, our sequence $h(x_n)$ converges, to the limit $h(a)$.

In every case, when $x_n\to a$ we see $h(x_n)\to h(a)$, so $h$ is continuous at $a$. :::

Exercise 6 Using the ‘Continuity Laws’ on sums, products, quotients, inverses, together with the continuity of the two “basic functions” $1$ and $x$, prove the following function is continuous on the entire real line

$$f(x)=\frac{x-\sqrt{x^2+|x-1{|}^3}}{1+|x|+x^2}$$

Solution. We work ‘from the inside out’. This means we will first prove the numerator and denominator are continuous by building up from basic pieces, and use their continuity to justify the continuity of the quotient.

The Numerator To prove the continuity of $x-\sqrt{x+|x-1{|}^3},$ we proceed step by step:

• $x$ and $1$ are continuous, as proven in class
• $x-1$ is continuous as the difference of continuous functions
• So $|x-1|$ is continuous, as the composition with the (continuous) absolute value.
• $|x-1{|}^3$ is continuous as a composition of this with $x\mapsto x^3$.
• $x^2+|x-1{|}^3$ is continuous as a sum of continuous functions.
• Since $|x-1{|}^3\ge 0$, and $x^2>0$, their sum is also positive, so the square root exists. Since $x\mapsto \sqrt{x}$ is continuous on its domain, $\sqrt{x+|x-1{|}^3}$ is continuous as a composition.
• Therefore, $x-\sqrt{x+|x-1{|}^3}$ is continuous.

The Denominator The denominator is $1+|x|+x^2.$ Each piece is continuous:

• 1 is continuous.
• $|x|$ is continuous.
• $x^2$ is continuous.
• The sum $1+|x|+x^2$ is continuous.
• Moreover, $1+|x|+x^2>0$ for all $x$ (as its greater than 1), so the denominator never vanishes.

The Quotient Finally, by the quotient theorem for continuity, since the numerator and denominator are continuous for all $x$ and the denominator is nonzero, their quotient is defined and continuous for all $x\in \mathbb{R}$.

Exercise 7 (Continuity of Contraction Maps) Recall that a function $f$ is a contraction map if there exists a $k\in (0,1)$ with $|f(x)-f(y)|<k|x-y|$ for all $x,y$. Prove that contraction maps are continuous.

Solution. Assume $f$ is a contraction map with constant $k\in (0,1)$. Let $a$ be an arbitrary point in the domain of $f$; we will show $f$ is continuous at $a$.

Choose an arbitrary $ϵ>0$, and set $\delta =ϵ/k$. Then for any $x$ with $|x-a|<\delta$, we know that $$|f(x)-f(a)|<k|x-a|=k\delta =\frac{ϵ}{k}k=ϵ$$

But this is precisely the definition of $f$ being continuous at $a$. Since $a$ was arbitrary, $f$ is continuous at each point of its domain.

Exercise 8 Let $\sum a_n{x}^n$ be a power series with radius of convergence $r$, and assume that the ratio test works for computing this.

• Use the Ratio Test to prove that integrating term-by-term gives a power series with the same radius of converenge: $$\sum _{n\ge 0}\frac{a_n}{n+1}{x}^{n+1}$$

• While the radius of convergence is the same for both series, show the behavior at the endpoints need not be, by studying the example $\sum a_n{x}^n=\sum _{n\ge 0}(-1{)}^n{x}^{2n}$.
  

Solution. Let $\sum a_n{x}^n$ be a power series with radius of convergence $r>0$, and suppose the ratio test applies to compute this radius. That is, assume $$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=L,$$ so that the radius of convergence is $r=\frac{1}{L}$ (with the usual conventions if $L=0$ or $\mathrm{\infty }$). We consider the power series obtained by integrating term by term: $$\sum _{n\ge 0}\frac{a_n}{n+1}{x}^{n+1}.$$

Let us apply the ratio test to this new series, whose coefficients are $ b_n = .$ Then the ratio of successive terms is: $$\left|\frac{b_{n+1}}{b_n}\right|=|\frac{a_{n+1}}{a_n}\cdot \frac{n+1}{n+2}|.$$

Since $\frac{n+1}{n+2}\to 1$ as $n\to \mathrm{\infty }$, we have: $$\underset{n\to \mathrm{\infty }}{lim}\left|\frac{b_{n+1}}{b_n}\right|=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=L.$$

Therefore, by the ratio test, the radius of convergence of the integrated series is also $r=1/L$.

At the Boundary: While the radius of convergence is the same for both the original and the integrated series, the behavior at the endpoints of the interval may differ. The geometric series is already an example: $\sum a_n{x}^n=\sum _{n\ge 0}{x}^n.$ converges at the points $(-1,1)$ and diverges everywhere else.

Now consider the term-by-term integral: $$\sum _{n\ge 0}\frac{1}{n+1}{x}^{n+1}.$$

This series still has radius of convergence $1$, but now also converges at $x=-1$, where it is the alternating harmonic series.