There are two interpretations of the question possible:
- Is there a norm $\|\cdot\|$ on $S$ such that the topology defined by $\|\cdot\|$ agrees with the topology defined by $d$, and $(E, \|\cdot\|)$ is a Banach space?
- Is there a norm $\| \cdot \|$ on the vector space $S$ such that $(S, \|\cdot\|)$ is a Banach space?
I will show, with references, that the answer to 1 is no, even if we do not require that $(E, \|\cdot\|)$ be complete, and the answer to 2 is yes, using the axiom of choice in the form of "every vector space has a basis".
Answer to number 1:
The first thing to observe is that the topology defined by $d$ is the same as the product topology on $S$ (equivalently, the topology of pointwise convergence). Continuity of the identity map from $(S,d)$ to $\mathbb{R}^{\mathbb{N}}$ is most easily proved by showing that the projection mappings $\pi_n : S \rightarrow \mathbb{R}$ defined by $\pi_n((x_i)) = x_n$ are continuous. Proving that the identity mapping is continuous in the other direction is most easily done using the observation that $0 \leq \frac{|x|}{1 + |x|} < 1$ for all real $x$, which means $d(0,(x_i))$ can be bounded by bounding only finitely many coordinates.
Then we need the fact that every closed bounded subset of $S$ is compact. Let's recall the definitions. Recall that in a topological vector space $E$, a 0-neighbourhood $N \subseteq E$ is a set such that there exists an open set $U$ such that $0 \in U \subseteq N$. A set $B \subseteq E$ is bounded if for all 0-neighbourhoods $N$ there exists a real number $\alpha > 0$ such that $B \subseteq \alpha N$. If $f : E \rightarrow F$ is a continuous linear map between topological vector spaces, and $B \subseteq E$ is bounded, then $f(B)$ is bounded (see e.g. Schaefer's Topological Vector Spaces I.5.4, or prove it directly using the definition of continuity of linear maps in terms of 0-neighbourhoods). So if $B \subseteq S$ is bounded, for all $n \in \mathbb{N}$, $\pi_n(S) \subseteq \mathbb{R}$ is bounded. Let $B$ be a closed, bounded set. Since $[-1,1]$ is a 0-neighbourhood in $\mathbb{R}$, for each $n \in \mathbb{N}$, the set $\pi_n(B) \subseteq [-\alpha_n,\alpha_n]$ for some real $\alpha_n > 0$. It follows that $B \subseteq \prod_{i=1}^\infty [-\alpha_n,\alpha_n]$, which is compact in $S$ by Tychonoff's theorem (because $(S,d)$ has the product topology). As $B$ is closed, it is also compact.
Now, in a normed space $(E, \|\cdot\|)$, the closed unit ball $U$ is bounded (every 0-neighbourhood $N$ contains an open ball of radius $\epsilon > 0$, so $U \subseteq (\epsilon^{-1} + 1)N$). By Riesz's lemma, if $E$ is infinite-dimensional, then $U$ contains a sequence for which the distance between the elements is bounded below by some number, which therefore has no Cauchy subsequence, and therefore no convergent subsequence (even in an incomplete metric space, every convergent sequence is Cauchy). In a compact metric space, every sequence has a convergent subsequence, so this proves $U$ is not compact. Therefore there is no norm defining the topology of $(S,d)$.
Answer to number 2:
Since we are disregarding the original topology of $S$ in this section, we will be using the non-topological notions of basis and dimension. If $E$ is a vector space over $\mathbb{R}$, we say that a family $(x_i)_{i \in I}$ is a (Hamel) basis if:
- $(x_i)_{i \in I}$ spans $E$, i.e. for each $x \in E$ there is a finite set $K \subseteq I$ and a family $(\alpha_i)_{i \in K}$ with each $\alpha_i \in \mathbb{R}$ such that $x = \sum_{i \in K} \alpha_i x_i$.
- $(x_i)_{i \in I}$ is linearly independent, i.e. for all finite sets $K \subseteq I$ and families $(\alpha_i)_{i \in K}$ with $\alpha_i \in \mathbb{R}$ such that $\sum_{i \in K}\alpha_i x_i = 0$, we have $\alpha_i = 0$ for all $i \in K$.
The key facts are that every linearly independent set can be extended to a basis, all bases of a vector space $E$ have the same cardinality (called the dimension of $E$), and vector spaces are linearly isomorphic iff they have the same dimension. Unfortunately, standard references on linear algebra usually only prove these facts for the finite-dimensional case, but the general case is treated in Chapter IX of volume II of Jacobson's Lectures in Abstract Algebra.
It is clear that the dimension of $E$ is $\leq |E|$, because a linearly independent set in $E$ is a subset of $E$. We have $$
|S| = |\mathbb{R}|^{|\mathbb{N}|} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0 \times \aleph_0} = 2^{\aleph_0},
$$
so the dimension of $S$ is $\leq 2^{\aleph_0}$. For any separable Banach space $E$, such as $\ell^2$, every element is the limit of a sequence from a countable dense subset, so $\mathbb{N}^{\mathbb{N}}$ maps surjectively onto $E$, so $E$ also has cardinality and therefore dimension $\leq 2^{\aleph_0}$.
In his paper On Infinite-dimensional Linear Spaces, Mackey proves in Theorem I.1 that every infinite-dimensional Banach space is of dimension $\geq 2^{\aleph_0}$. He does this in two steps. The first is to show that for every infinite-dimensional Banach space $E$, there exists an injective linear map $f : \ell^\infty \rightarrow E$. The second is the observation that the $(0,1)$-indexed family $((\alpha^i)_{i \in \mathbb{N}})_{\alpha \in (0,1)}$ in $\ell^\infty$ is linearly independent, making the dimension of $E$ greater than or equal to $|(0,1)| = 2^{\aleph_0}$. Therefore any separable infinite-dimensional Banach space has dimension exactly $2^{\aleph_0}$. Since $\ell^\infty$ is a linear subspace of $S$, this also proves that $S$ has dimension exactly $2^{\aleph_0}$.
So by mapping a basis of $S$ onto a basis of $\ell^2$ and extending by linearity, we can define a linear isomorphism $f : S \rightarrow \ell^2$. Let's use $\| \cdot \|_2$ for the norm on $\ell^2$ and define, for $x \in S$, $\|x\|_1 = \|f(x)\|_2$. The linearity of $f$ ensures that $\|\cdot\|_1$ is a norm. If $(x_i)_{i \in \mathbb{N}}$ is $\|\cdot\|_1$-Cauchy, define $y_i = f(x_i)$, and expanding the definitions shows that $(y_i)_{i \in \mathbb{N}}$ is $\|\cdot\|_2$-Cauchy, so converges to some $y \in \ell^2$. Since $f$ is an isomorphism, there exists $x \in S$ such that $f(x) = y$, and expanding the definitions and using linearity of $f$ shows that $(x_i)_{i \in \mathbb{N}}$ converges to $x$ in $\|\cdot\|_1$. Therefore $(S,\|\cdot\|_1)$ is a Banach space.
Some examples in mathematics are mainly used to show why things are defined the way they are, rather than for practical use. This is one of those examples.
Best Answer
I'm changing the domain to $[-1, 1]$. Deal with it.
For a given $\varepsilon > 0$, consider the function $$f(x) = |x|^{1 + \varepsilon}.$$ This function is continuously differentiable, with derivative $$f'(x) = (1 + \varepsilon)|x|^\varepsilon.$$ So, with this in mind, define $$f_n(x) = |x|^{1 + 1/n}.$$ I claim that $f_n(x) \to |x|$ uniformly over the given interval. We have, $$\left||x|^{1 + 1/n} - |x|\right| = |x| \cdot \left||x|^{1/n} - 1\right|.$$ When $|x| \le 1$, we have $1 \ge |x|^{1/n} \ge |x|$, so we get \begin{align*} \left||x|^{1 + 1/n} - |x|\right| &\le |x|\left(1 - |x|^{1/n}\right) \\ &= \frac{|x|(1 - |x|)}{1 + |x|^{1/n} + |x|^{2/n} + \ldots + |x|^{(n-1)/n}} \\ &\le \frac{|x|(1 - |x|)}{1 + (n - 1)|x|}. \end{align*} I claim that $$\frac{|x|(1 - |x|)}{1 + (n - 1)|x|} \le \frac{1}{n},$$ or equivalently, \begin{align*} n|x|(1 - |x|) \le 1 + (n - 1)|x| &\iff n|x| - nx^2 \le 1 + n|x| - |x| \\ &\iff n|x| + nx^2 + 1 \ge |x|, \end{align*} which is pretty evidently true, since $1 \ge |x|$, and the other terms are positive. Thus, $$\Big|f_n(x) - |x|\Big| \le \frac{1}{n},$$ which is to say, $f_n$ converges to the absolute value function uniformly, even though it is not continuously differentiable.
Fun fact: the Stone-Weierstrass theorem says that every continuous function on a compact interval can be approximated uniformly by polynomial functions. So, it would be possible (though harder) to simply choose $f_n$ from polynomials, which are infinitely-differentiable.