$$$$

Assignment 11

In the mid 1800’s Cauchy came up with the idea of using functional equations to describe the important functions of mathematics. The first case was linearity:

Definition 1 (Cauchy’s Functional Equation for Linearity) A function $f:\mathbb{R}\to \mathbb{R}$ satisfies Cauchy’s functional equation if for all $x,y\in \mathbb{R}$, $$f(x+y)=f(x)+f(y)$$

Note it follows for any finite sum $x_1+x_2+\cdots +x_n$, one can ‘distribute’ a solution to Cauchy’s functional equation, by induction: $f(x_1+x_2+\cdots +x_n)=f(x_1)+f(x_2)+\cdots +f(x_n)$.

Exercise 1 Prove that if $f$ is a continuous solution to Cauchy’s functional equation, then $f(x)=kx$ for some $k\in \mathbb{R}$. That is, the continuous solutions to Cauchy’s Functional equation are precisely the linear functions.

Hint: follow the strategy below:

Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function where $f(x+y)=f(x)+f(y)$.

• Prove that $f(n)=nf(1)$ for all $n\in \mathbb{N}$.
• Extend this to negative integers.
• Show that $f(1/n)=\frac{1}{n}f(1)$ for $n\in \mathbb{N}$ *Hint: use that $\frac{1}{n}+\frac{1}{n}+\cdots +\frac{1}{n}=1$
• From the above, deduce that for rational $r=p/q$, $p\in \mathbb{Z}$ $q\in \mathbb{N}$ that $f(r)=rf(1)$.
• Now use continuity! If $k=f(1)$, then $f(r)=kr$ on $\mathbb{Q}$…

Solution (Solution). Let $k=f(1)$. We will show that for all $x\in \mathbb{R}$, $$f(x)=kx.$$

Natural numbers

We prove by induction that for every $n\in \mathbb{N}$, $$f(n)=nf(1)=nk.$$

• Base case ($n=1$): $f(1)=k$ by definition.
  
• Inductive step: Assume $f(n)=nk$. Then $$f(n+1)=f(n)+f(1)=nk+k=(n+1)k.$$

2. Integers

First, $$f(0)=f(0+0)=f(0)+f(0),$$ so $f(0)=0$. For $n\in \mathbb{N}$, $$0=f(0)=f(n+(-n))=f(n)+f(-n),$$ hence $$f(-n)=-f(n)=-nk.$$ Thus $f(m)=mk$ for all $m\in \mathbb{Z}$.

3. Reciprocals

Let $n\in \mathbb{N}$. Since $$\underset{n\text{ times}}{\underbrace{\frac{1}{n}+\frac{1}{n}+\cdots +\frac{1}{n}}}=1,$$ applying $f$ gives $$nf(\frac{1}{n})=f(1)=k,$$ so $$f(\frac{1}{n})=\frac{k}{n}.$$

4. Rationals

Any rational $r=\frac{p}{q}$ with $p\in \mathbb{Z},q\in \mathbb{N}$ satisfies $$f(\frac{p}{q})=pf(\frac{1}{q})=p\frac{k}{q}=\frac{p}{q}k=kr.$$ Hence $f(r)=kr$ for all $r\in \mathbb{Q}$.

5. All real numbers

Let $x\in \mathbb{R}$. By density of $\mathbb{Q}$, choose a sequence $(r_n)\subset \mathbb{Q}$ with $r_n\to x$. Continuity of $f$ gives $$f(x)=f(\underset{n\to \mathrm{\infty }}{lim}{r}_n)=\underset{n\to \mathrm{\infty }}{lim}f(r_n)=\underset{n\to \mathrm{\infty }}{lim}(k{r}_n)=kx.$$

Therefore, the only continuous solutions of Cauchy’s functional equation are the linear functions $$\boxed{{\displaystyle f(x)=kx,k\in \mathbb{R}.}}$$

Exercise 2 Prove the exponential function $a^x$ for $a>1$ is monotone increasing: that is, if $x<y$, $a^x\le a^y$.

Hint: we know its monotone on rational inputs, so the interesting cases are again when at least one is irrational (and, the argument for both irrational can be generalized to include the other case). Write down monotone increasing sequences, truncate the sequences until you can insure $x_n<y_n$ for all $n$, and then apply the limit laws.

Solution (Solution). Let $a>1$ and consider the function $x\mapsto a^x$. We will show that whenever $x<y$ in $\mathbb{R}$, one has $$a^x\le a^y.$$

1. Monotonicity on rationals

First observe that if $r<s$ are rational numbers, then writing $r=\frac{m}{n}$ and $s=\frac{p}{q}$ with $m,n,p,q\in \mathbb{Z}$, $n,q>0$, we have $$a^r=(a^{1/n}{)}^m\text{and}{a}^s=(a^{1/q}{)}^p.$$ Since $a^{1/n}>1$ and exponentiation by an integer preserves order, it follows that $r<s$ implies $a^r<a^s$. Hence $a^x$ is strictly increasing on $\mathbb{Q})$.

2. Approximating by rationals

Now let $x<y$ be arbitrary real numbers. Choose two sequences of rationals $(r_n),(s_n)\subset \mathbb{Q}$ such that $r_1<r_2<\cdots <x$ and $y<\cdots <s_2<s_1$. One can form these by using density of rationals to pick $r_n\in (x-1/n,x)$ and $s_n\in (y,y+1/n)$ repsectively. Then by the squeeze theorem $$\underset{n\to \mathrm{\infty }}{lim}{r}_n=x,\underset{n\to \mathrm{\infty }}{lim}{s}_n=y.$$ Because $r_n<x<y<s_n$, we see that $r_n<s_n$ for all $n$. (Alternatively, we could have chosen $r_n$ and $s_n$ both monotone increasing, and then found an $N$ where $s_n>x$ (by setting $ϵ=y-x$ and applying the definition of convergence). Truncating the sequences at $N$ then gives the same result, that $r_n<s_n$ for all $n$).

3. Taking limits

By the rational‐case monotonicity, $$a^{r_n}<a^{s_n}\text{for all }n.$$ Both sequences $(a^{r_n})$ and $(a^{s_n})$ converge (by the continity of $a^x$) to $a^x$ and $a^y$ respectively. Hence by the limit laws for inequalities $$a^x=\underset{n\to \mathrm{\infty }}{lim}{a}^{r_n}\le \underset{n\to \mathrm{\infty }}{lim}{a}^{s_n}=a^y.$$

Thus for any real $x<y$, $$a^x\le a^y,$$ showing that $x\mapsto a^x$ is monotone increasing on $\mathbb{R}$.

Exercise 3 Prove range of the exponential function $a^x$ is all positive real numbers: that is, for any positive $y$, show there is some $x$ where $a^x=y$.

Hint: can you find some $n\in \mathbb{N}$ where $a^n>y$? If so, can you modify the idea to get an $m$ with $a^{-m}<y$? Once you have these two values, can you apply a theorem about continuity?

Solution. Let $a>1$ and fix an arbitrary $y>0$. We will show there exists $x\in \mathbb{R}$ such that $a^x=y.$

1. Finding an exponent above $y$

Write $a=1+b$ with $b>0$. By Bernoulli’s inequality, $$a^n=(1+b{)}^n\ge 1+nb\text{for every }n\in \mathbb{N}.$$ Since $nb\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$, the Archimedean property guarantees we can choose $N\in \mathbb{N}$ so large that $1+Nb>y,$ hence $a^N\ge 1+Nb>y.$

2. Finding an exponent below $y$

For any positive integer $M$, $$a^{-M}=\frac{1}{a^M}<\frac{1}{1+Mb}.$$ Again by the Archimedean property we pick $M$ so large that $$1+Mb>\frac{1}{y}⟹\frac{1}{1+Mb}<y,$$ thus $ a^{-M} < y.$

3. Applying the Intermediate Value Theorem

The function $a^x$ is continuous, and we have exhibited the existence of an $N$ with $a^N>y$ and an $M$ with $a^M<y$. Thus on the closed interval $[M,N]$ there must be some $x$ where $a^x=y$. Since $y>0$ was arbitrary, the range of $x\mapsto a^x$ is all of $(0,\mathrm{\infty })$.

Exercise 4 Let $L(x)$ be a logarithm and $r\in \mathbb{Q}$ a rational number. Prove directly from the functional equation that $L(x^r)=rL(x)$.

Solution. Assume $L:(0,\mathrm{\infty })\to \mathbb{R}$ satisfies the logarithm functional equation $L(xy)=L(x)+L(y)\text{for all }x,y>0.$

1. Value at $1$

Set $x=y=1$ in the functional equation to get $$L(1)=L(1\cdot 1)=L(1)+L(1),$$ hence $L(1)=0$.

2. Positive integer powers

For any $n\in \mathbb{N}$, note $$x^n=\underset{n\text{ times}}{\underbrace{x\cdot x\cdots x}},$$ so by applying $L$ repeatedly, $$L(x^n)=L(x\cdot x^{n-1})=L(x)+L(x^{n-1})=\cdots =nL(x).$$

3. Negative Integer Powers

Since $L(1)=0$ and $1=x\cdot x^{-1}$, we have $$0=L(1)=L(x)+L(x^{-1}),$$ so $$L(x^{-1})=-L(x).$$ By replacing $x$ with $x^n$, it follows that for every $n\in \mathbb{N}$, $$L(x^{-n})=L((x^n{)}^{-1})=-L(x^n)=-nL(x).$$

4. Rational exponents

Let $r=\frac{p}{q}$ with $p\in \mathbb{Z}$ and $q\in \mathbb{N}$. By the definition of rational exponents, $$(x^{p/q}{)}^q=x^p.$$ Applying $L$ gives $$L((x^{p/q}{)}^q)=L(x^p).$$ But by the functional equation, $$L((x^{p/q}{)}^q)=qL(x^{p/q}),L(x^p)=pL(x),$$ so $$qL(x^{p/q})=pL(x),$$ and hence $$L(x^{p/q})=\frac{p}{q}L(x).$$

Thus for every rational $r=p/q$ the identity holds.

Exercise 5 Recall that an even function satisfies $f(-x)=f(x)$ for all $x$ and an odd function satisfies $f(-x)=-f(x)$.

Show that for differentiable functions, the derivative of an even function is an odd function, and the derivative of an odd function is an even function.

Solution. Let $f$ be an even function, so that $f(-x)=f(x)$ for all $x\in \mathbb{R}$. Taking the derivative of each side of this equality gives $f^{\prime }(x)=(f(-x){)}^{\prime }$, and evaluating the right hand side with the chain rule: $$f^{\prime }(x)=[f(-x){]}^{\prime }=-f^{\prime }(-x)$$ Thus, $f^{\prime }(-x)=-f^{\prime }(x)$ so $f^{\prime }$ is odd.

The same style of argument succeeds when $f$ starts out odd: now, differentiating the equality $f(-x)=-f(x)$ we see $$-f^{\prime }(x)=[-f(x){]}^{\prime }=[f(-x){]}^{\prime }=-f^{\prime }(-x)$$ So, $f^{\prime }$ satisfies $f^{\prime }(-x)=f^{\prime }(x)$ and the derivative is even.

Exercise 6 (Differentiating Sums) Let $f,g$ be functions which are both differentiable at a point $a\in \mathbb{R}$. Prove that $f+g$ is also differentiable at $a$, and $$(f+g{)}^{\prime }(a)=f^{\prime }(a)+g^{\prime }(a)$$

Solution. This calculation is directly analogous to derivative of $kf(x)$ done in the book: we use the limit laws to distribute the limit over addition, and note the result is the sum of two limit definitions of the derivative.

Exercise 7 Let $f$ be a function and $a\in \mathbb{R}$ be a point such that $f(a)\ne 0$ and $f$ is differentiable at $a$. Prove that $1/f$ is also differentiable at $a$ and $${\left(\frac{1}{f}\right)}^{\prime }(a)=\frac{-f^{\prime }(a)}{f(a{)}^2}$$

Solution. Let $f(a)\ne 0$, and let $f$ be differentiable at $a$, lets start with $p(x)=\frac{1}{f(x)}$. Then we can see $$\begin{array}{rl}{\left(\frac{1}{f}\right)}^{\prime }(a)& =p^{\prime }(a)\\ & =\underset{x\to a}{lim}{\displaystyle \frac{p(x)-p(a)}{x-a}}\\ & =\underset{x\to a}{lim}{\displaystyle \frac{\frac{1}{f(x)}-\frac{1}{f(a)}}{x-a}}\\ & =\underset{x\to a}{lim}{\displaystyle \frac{\frac{f(a)-f(x)}{f(x)f(a)}}{x-a}}\\ & =\underset{x\to a}{lim}{\displaystyle \frac{f(x)-f(a)}{(-f(x)f(a))(x-a)}}\\ & =\underset{x\to a}{lim}{\displaystyle \frac{-1}{f(x)f(a)}}\cdot {\displaystyle \frac{f(x)-f(a)}{x-a}}\end{array}$$ Applying the limit here, we know that the second fraction is equal to $f^{\prime }(a)$, so we get, $$\begin{array}{rl}& ={\displaystyle \frac{-1}{f(a{)}^2}}{f}^{\prime }(a)\\ {\left(\frac{1}{f}\right)}^{\prime }(a)& ={\displaystyle \frac{-f^{\prime }(a)}{f(a{)}^2}}\end{array}$$

Exercise 8 Prove the power rule, $$(x^n{)}^{\prime }=n{x}^{n-1}$$ where $n\ne 0$ is an integer

Hint: first do the positive $n$ case by induction; then use this to get the negative $n$ case.

Proof. We already know $(x{)}^{\prime }=1=1x^0$, so we can use this as our base case. To see how to proceed however we do the next case up as well. Using the product rule we see $$(x^2{)}^{\prime }=(x\cdot x{)}^{\prime }=(x{)}^{\prime }\cdot x+x\cdot (x{)}^{\prime }=x+x=2x$$

Proceeding by induction, assume that for some $n>1$ this holds, and consider $x^{n+1}$. Then $$(x^{n+1}{)}^{\prime }=(x\cdot x^n{)}^{\prime }=(x{)}^{\prime }\cdot x^n+x\cdot (x^n{)}^{\prime }$$ $$=1x^n+x(n{x}^{n-1})=x^n+n{x}^n=(n+1)x^n$$

Hence the claim is true for all $n>0$. For $n<0$, note that $x^{-n}{x}^n=1$ by definition, and we can differentiate this identity using the power rule:

$$0=(1{)}^{\prime }=(x^n{x}^{-n}{)}^{\prime }=(x^n{)}^{\prime }{x}^{-n}+x^n(x^{-n}{)}^{\prime }$$ Simplifying: $$0=n{x}^{n-1}{x}^{-n}+x^n(x^{-n}{)}^{\prime }=n{x}^{-1}+x^n(x^{-n}{)}^{\prime }$$ Solving this for $(x^{-n}{)}^{\prime }$ yields

$$(x^{-n}{)}^{\prime }=-n\frac{x^{-1}}{x^n}=-n{x}^{-n-1}$$

Thus the result holds for negative powers a well.