1 Let us first prove (i) $\Rightarrow$ (ii). Suppose that $\left{x_{n}\right}$ is a sequence with $\lim {n \rightarrow \infty} x{n}=c$. We need to show that $\lim {n \rightarrow \infty} f\left(x{n}\right)=f(c) .$ Let $\varepsilon>0$. We must find an integer $N$ so that $n \geq N$ implies $\left|f\left(x_{n}\right)-f(c)\right|<\varepsilon .$ To do this, choose $\delta>0$ so that $|x-c|<\delta$ implies $|f(x)-f(c)|<\varepsilon$. The existence of such a $\delta$ is guaranteed by the continuity of $f .$ Since $\lim {n \rightarrow \infty} x{n}=c$, there exists $N \geq 1$ such that $n \geq N$ implies $\left|x_{n}-c\right|<\delta$. Hence for any $n \geq N$, we have $\left|f\left(x_{n}\right)-f(c)\right|<\varepsilon$ This yields the desired conclusion.
2 Next we show (ii) $\Rightarrow$ (i). Suppose, to the contrary, that $f$ is sequentially continuous but discontinuous at $c$. Then there is a neighborhood $V$ of $f(c)$ such that no neighborhood $U$ of $c$ satisfies $f(U) \subseteq V .$ Set $U_{n}=B(c, 1 / n)={x \in D:|c-x|<1 / n}$, for $n \geq 1 .$ In particular, we have that $f\left(U_{n}\right) \cap V^{c} \neq \emptyset$, for any $n \geq 1$. Hence for $n \geq 1$, choose some $x_{n} \in D$ with $|x-c|<1 / n$ and $f\left(x_{n}\right) \notin V .$ By construction, $\left{x_{n}\right}$ converges to $c$, but $\left{f\left(x_{n}\right)\right}$ does not converge to $f(c)$. This contradicts the sequential continuity of $f$.

we showed that $\lim {x \rightarrow a} f(x)=0$ for any $a \in(0,1)$. Therefore (a) if $a \in \mathbb{R} \backslash \mathbb{Q}$, then $f(a)=0$ which implies $\lim {x \rightarrow a} f(x)=f(a)$, i.e., $f(x)$ is continuous at $a$;
(b) if $a \in \mathbb{Q} \cap(0,1)$, then $f(a) \neq 0$ which implies $\lim _{x \rightarrow a} f(x) \neq f(a)$, i.e., $f(x)$ is not continuous at $a .$

The equation satisfied by $f(x)$ is usually known as a functional equation. First note that $f(n x)=$ $n f(x)$ for any $n \in \mathbb{N}$. Indeed, we have $f(1 \cdot x)=1 \cdot f(x)$. Assume that $f(n x)=n f(x)$. Then
$$
f((n+1) x)=f(n x+x)=f(n x)+f(x)=n f(x)+f(x)=(n+1) f(x) .
$$
By induction we get the desired identity. Let $r=\frac{p}{q} \in \mathbb{Q}$. Then we have $f(q r x)=f(p x)=p f(x)$, and since $f(q r x)=q f(r x)$, we get $f(r x)=\frac{p}{q} f(x)=r f(x) .$ In particular, we have $f(r)=r f(1)$. Let $x \in \mathbb{R}$. Then there exists a sequence of rational numbers $\left{r_{n}\right}$ such that $\lim r_{n}=x$. Since $f(x)=f\left(x-r_{n}+r_{n}\right)=f\left(x-r_{n}\right)+f\left(r_{n}\right)=f\left(x-r_{n}\right)+r_{n} f(1)$, we have $f(x)=f\left(x-r_{n}\right)+r_{n} f(1)$. Now
if we take the limit as $n \rightarrow \infty$, by the existence of $\lim _{x \rightarrow 0} f(x)=f(0)$, we have that $f(x)=f(0)+x f(1)$. Note that $f(x)=f(x+0)=f(x)+f(0)$ for any $x \in \mathbb{R}$, so $f(0)=0$. Therefore, we have
$$
f

a) Given $c \in \mathbb{R}$ and $\varepsilon>0$, set $\delta=\frac{\varepsilon}{k}$. Then
$|f(x)-f(y)|<\varepsilon$ for all $x \in \mathbb{R}$ such that $|x-c|<\delta$ Hence if $f$ is a contraction, then $f$ is continuous. b) To show that $\left(x_{n}\right)$ is Cauchy, observe: $$ \left|x_{n+1}-x_{n}\right|=\left|f\left(x_{n}\right)-f\left(x_{n-1}\right)\right| \leq k\left|x_{n}-x_{n-1}\right|=\left|f\left(x_{n-1}\right)-f\left(x_{n-2}\right)\right| \leq k^{2}\left|\left(x_{n-1}-x_{n-2}\right)\right| $$ we can repeat the above process to obtain $$ \left|x_{n+1}-x_{n}\right| \leq k^{n}\left|x_{1}-x_{0}\right| . $$ Thus if $n>m$,
$$
\begin{gathered}
\left|x_{n}-x_{m}\right| \leq\left|x_{n}-x_{n-1}\right|+\left|x_{n-1}-x_{n-2}\right|+\cdots+\left|x_{m+1}-x_{m}\right| \
\left|x_{n+1}-x_{n}\right| \leq\left(k^{n-1}+k^{n-2}+\cdots+k^{m}\right)\left|x_{1}-x_{0}\right| \leq \frac{k^{m}}{1-k}\left|x_{0}-x_{1}\right|
\end{gathered}
$$
using the limiting sum of a geometric series, which we may do since $00$ whenever $m$ and $n$ are sufficiently large. Hence $\left(x_{n}\right)$ is a Cauchy sequence. Since $\mathbb{R}$ is a complete space, every Cauchy sequence converges in it, and the existence of $\lim {n \rightarrow \infty} x{n}$ is assured. We set $\lim {n \rightarrow \infty} x{n}=x^{*}$.

c) To show $x^{}$ is a fixed point of $f$, we note that, for any positive integer $n$, $$ 0 \leq\left|f\left(x^{}\right)-x^{}\right| \leq\left|f\left(x^{}\right)-x_{n}\right|+\left|x_{n}-x^{}\right|=\left|f\left(x^{}\right)-f\left(x_{n-1}\right)\right|+\left|x_{n}-x^{}\right| \leq k\left|x^{}-x_{n-1}\right|+\left|x_{n}-x^{}\right| $$ and so $\left|f\left(x^{}\right)-x^{}\right|=0$ since $\left|x_{n}-x^{}\right| \rightarrow 0\left(\right.$ and $\left.\left|x^{}-x_{n-1}\right| \rightarrow 0\right)$. Thus $f\left(x^{}\right)=x^{}$. d) Suppose there are two fixed points say $x^{}$ and $\tilde{x}$, so also $f(\tilde{x})=\tilde{x}$. Then
$$
\left|x^{}-\tilde{x}\right|=\left|f\left(x^{}\right)-f(\tilde{x})\right| \leq k\left|x^{}-\tilde{x}\right| $$ which since $k<1$ can only be true when $\left|x^{}-\tilde{x}\right|=0 ;$ that is, $x^{*}=\tilde{x}$. Hence there is a unique fixed point of $f$.