# Continuous functions of one variable

We propose exercises on the continuous functions of one variable. Also, we give applications of intermediate value theorem, Heine-Borel theorem real analysis. Moreover, we investigate some classes of continuous functions like uniformly continuous functions, Lipschitzian functions, and Holderian functions.

### Exercises of Continuous functions

Exercise: Determine if the following functions are continuous or not: \begin{align*} f(x)=\begin{cases}\frac{\sin(2x)-\sin(5x)}{\sin(3x)},& x\neq 0,\cr -1,& x=0.\end{cases}\qquad g(x)=\begin{cases} x^x,& x>0,\cr 2,&x=0.\end{cases} \end{align*}

Solution: 1) The function $f$ is continuous on $\mathbb{R}\backslash\{0\}$ as the sum and the quotient of continuous functions. Now let us verify the continuity of $f$ at $0$. To that purpose we shall calculate the limit of $f$ as $0$ and check if it equal to $f(0)$. Before doing so we recall the following standard result \begin{align*} \lim_{x\to 0,\;x\neq 0} \frac{\sin(x)}{x}=1. \end{align*} Now for any $x\neq 0,$ we have \begin{align*} f(x)&=\frac{x}{\sin(3x)}\times \left(\frac{\sin(2x)}{x}-\frac{\sin(5x)}{x}\right)\cr &= \frac{1}{3}\frac{3x}{\sin(3x)}\times \left(2\frac{\sin(2x)}{2x}-5\frac{\sin(5x)}{5x}\right). \end{align*} Hence \begin{align*} \lim_{x\to 0,\;x\neq 0}f(x)=\frac{1}{3}(2-5)=-1=f(0). \end{align*} It follows that $f$ is continuous on $\mathbb{R}$.

2) For any $x>0$ we can write \begin{align*} g(x)=e^{\ln(x^x)}=e^{x \ln(x)}. \end{align*} This shows that $g$ is continuous on $(0,\infty)$. On the other hand we now that \begin{align*} \lim_{x\to 0^+}x\ln(x)=0. \end{align*} This implies that \begin{align*} \lim_{x\to 0^+} g(x)=1\neq g(0). \end{align*} Then $g$ is not continuous at $0$.

Exercise: Consider the function \begin{align*} f(t)=\begin{cases}\frac{\ln(1+2t^2)}{\sin^2(t)} ,& t > 0,\cr 0,& x=2. \end{cases} \end{align*} Prove that $f$ is continuous on $[0,+\infty)$.

Solution: The function $f$ is continuous on $(0,\infty)$, as the quotient of continuous functions. We now show what happen if $t$ is very closed to $0$. For $t\neq 0,$ we have \begin{align*} f(t)=2\frac{\ln(1+2t^2)}{2t^2}\left(\frac{t}{\sin(t)}\right)^2 \end{align*} As \begin{align*} \lim_{t\to 0,t\neq 0} \frac{\ln(1+2t^2)}{2t^2}=1, \end{align*} then \begin{align*} \lim_{t\to 0,t\neq 0}f(x)=2=f(0). \end{align*} This shows that $f$ is also continuous at $0$.

Exercise: Let $f$ be defined by \begin{align*} f(x)=\begin{cases} 1,& x\in \mathbb{Q},\cr 0,& x\notin \mathbb{Q}.\end{cases} \end{align*} Prove that $f$ is not continuous at any point of $\mathbb{R}$.

Solution: We recall that $g:A\subset \mathbb{R}\to \mathbb{R}$ is continuous at $x\in A$ if and only if for any sequence $(u_n)_n\subset A$ such that $u_n\to x$ as $n\to\infty,$ we have $g(u_n)\to g(x)$.

Using this result, to prove that $f$ is not continuous at $x\in \mathbb{R}$ it suffices to find two sequences $(u_n)_n$ and $(v_n)_n$ both converging to $x$ as $n\to \infty$ but the limits of $f(u_n)$ and $f(v_n)$ as $n\to\infty$ are different. In fact, let $x\in \mathbb{R}$. By density of $\mathbb{Q}$ in $\mathbb{R}$ there exists $(u_n)_n\subset \mathbb{Q}$ such that $u_n\to x$ as $n\to \infty$. By definition of $f$ we have $f(u_n)=1$, so the limit of $f(u_n)$ at infinity is $1$. On the other hand, by density of $\mathbb{R}\backslash\mathbb{Q}$ in $\mathbb{R}$ there exists $(v_n)_n\subset \mathbb{R}\backslash\mathbb{Q}$ such that $v_n\to x$ as $n\to \infty$. By definition of $f,$ we have $f(v_n)=0,$ so the limit of $f(v_n)$ at infinity is $0$. We then conclude that $f$ is not continuous at $x$.

### Continuous functions of one variable and sequences

Exercise: Let $a,b\in\mathbb{R}$ and consider a continuous function $f:[a,b]\to [a,b]$ such that \begin{align*} |f(t)-f(s)|<|t-s|,\qquad \forall t,s\in [a,b],\;t\neq s. \end{align*}

• Show that there exists a unique $\lambda\in [a,b]$ such that $f(\lambda)=\lambda$.
• Define the sequence \begin{align*} v_0\in [a,b],\quad\text{and}\; v_{n+1}=f(v_n),\quad\forall n\in\mathbb{N}. \end{align*} Prove that there exists $\mu\in\mathbb{R}$ such that \begin{align*} \lim_{n\to+\infty}|v_n-\lambda|=\mu. \end{align*}
• Justify that there exists a sub-sequence $(v_{\varphi(n)})_{n\ge 0}$ of $(v_n)_{n\ge 0}$ such that \begin{align*} \lim_{n\to+\infty}v_{\varphi(n)}=\lambda\pm\mu. \end{align*}
• In this question, we assume that $(v_{\varphi(n)})_{n\ge 0}$ converges to $\mu+\lambda$. Show that $\mu=0$ and $v_n\to \lambda$ as $n\to+\infty$.

Solution: 1) Let $g:[a,b]\to \mathbb{R}$ the continuous function defined by $g(t)=f(t)-t$. The fact that $g([a,b])\subset [a,b]$ implies that \begin{align*} g(b)=f(b)-b\le 0\le f(a)-a=g(a). \end{align*} Using the intermediate value theorem, there exist at least $\lambda\in [a,b]$ such that $g(\lambda=0),$ which means that $f(\lambda)=\lambda$. On the other hand, assume that there exists another $\lambda’\in [a,b]$ tel que $\lambda’\neq\lambda$ et $f(\lambda’)=\lambda’$. Then \begin{align*} |\lambda-\lambda’|=|f(\lambda)-f(\lambda’)|< |\lambda-\lambda’|.\end{align*} Absurd. Thus we have uniqueness.

2) We will prove that the sequence $(|v_n-\lambda|)_{n\ge 0}$ is decreasing. In fact, as $f(\lambda)=\lambda,$ for any $n\in\mathbb{N},$ \begin{align*} |v_{n+1}-\lambda|=|f(v_n)-f(\lambda)|<|v_n-\lambda|. \end{align*} The claim then follows. On the other hand, as $|v_n-\lambda|\ge 0$ for any $n,$ then the sequence $(|v_n-\lambda|)_{n\ge 0}$ converges and then there exist a real $\mu\in\mathbb{R}^+$ tel que \begin{align*}\tag{$L_\mu$}\lim_{n\to +\infty} |v_n-\lambda|=\mu. \end{align*}

3) Here we shall use the important Bolzano-Weierstrass theorem which says that for any bounded sequence we can extract a convergent subsequence. Let us prove that the sequence. Observe that for any $n\ge 1,$ we have $v_n=f(v_{n-1})\in [a,b]$. Thus $(v_n)_n$ is bounded, so there is a increasing function $\varphi:\mathbb{N}\to \mathbb{N}$ and a real $m\in \mathbb{R}$ such that \begin{align*} \lim_{n\to+\infty}v_{\varphi(n)}=\alpha. \end{align*} In particular, we have \begin{align*} \lim_{n\to+\infty}(v_{\varphi(n)}-\lambda)=\alpha-\lambda. \end{align*} As the function $t\mapsto |t|$ is continuous then \begin{align*} \lim_{n\to+\infty}|v_{\varphi(n)}-\lambda|=|\alpha-\lambda|. \end{align*} By the uniqueness of the limit, we obtain \begin{align*} |\alpha-\lambda|=\mu. \end{align*} This means that \begin{align*} \alpha=\lambda\pm\mu. \end{align*}

4) Let $\alpha=\mu+\lambda,$ the limit of the subsequence $(v_{\varphi(n)})_n$. By continuity of $f,$ and the fact that $v_{\varphi(n)+1}=g(v_{\varphi(n)}),$ we have \begin{align*} \lim_{n\to+\infty}v_{\varphi(n)+1}=g(\alpha). \end{align*} Now by using ($L_\mu$), we obtain \begin{align*} |g(\alpha)-\lambda |=\mu. \end{align*} Moreover \begin{align*} \mu=|g(\alpha)-\lambda|=|g(\alpha)-g(\lambda) |< |\alpha-\lambda|=|\mu|. \end{align*} But $|\mu|=\max\{\mu,-\mu\}$, so $|\mu|=-\mu$, which implies that $\mu\le 0$. On the other hand, we know that $\mu\ge 0$. This shows that $\mu=0$ and then $\alpha=\lambda$.