Baire's Category Theorem and Banach–Steinhaus

Proof. (a) Let $N \subseteq X$ be nowhere dense and suppose, for contradiction, that $B(x, \varepsilon) \subseteq N$ for some $x \in X$ and $\varepsilon > 0$. Every point of $B(x,\varepsilon)$ then lies in $N$, so $N$ is dense in the nonempty open set $B(x,\varepsilon)$ — trivially, since each point is its own constant sequence. That contradicts $N$ being nowhere dense.

(b) Let $N$ be closed and suppose it is dense in some nonempty open $U$. Then $U \subseteq \operatorname{cl}(N) = N$, and $U$, being open and nonempty, contains an open ball. So $N$ contains an open ball. Contrapositively, a closed set with no open ball inside it is nowhere dense. □

Proof. Let $(X, \lVert \cdot \rVert)$ be a complete normed space. Suppose, for contradiction, that there is a sequence of closed nowhere dense sets $(C_i)_{i \geq 1} \subseteq X$ whose union $Y = \bigcup_{i \geq 1} C_i$ contains an open ball, say $B(x_1, \varepsilon_1) \subseteq Y$.

The plan is to burrow: to build a nested sequence of balls, the $(n+1)$st of which avoids $C_n$ entirely, and then to catch the point they close down on. Concretely, I claim there are points $x_n$ and radii $\varepsilon_n > 0$ such that for every $n \geq 1$,

$$B(x_{n+1}, \varepsilon_{n+1}) \;\subseteq\; B\!\left(x_n, \tfrac{\varepsilon_n}{2}\right) \setminus C_n, \qquad \varepsilon_{n+1} \leq \tfrac{\varepsilon_n}{2}.$$

Base case. $B(x_1, \varepsilon_1/2)$ is an open ball, so by Lemma 1.2(a) it is not contained in the nowhere dense set $C_1$. Choose $x_2 \in B(x_1, \varepsilon_1/2) \setminus C_1$. Since $C_1$ is closed, $X \setminus C_1$ is open, and its intersection with the open ball $B(x_1, \varepsilon_1/2)$ is open and nonempty — nonempty because it contains $x_2$. Every point of an open set is the centre of some ball contained in that set, so there is $\varepsilon_2 > 0$ with $B(x_2, \varepsilon_2) \subseteq B(x_1, \varepsilon_1/2) \setminus C_1$, and we may shrink $\varepsilon_2$ so that $\varepsilon_2 \leq \varepsilon_1/2$.

Inductive step. Given $x_n$ and $\varepsilon_n > 0$, the same three moves apply verbatim with $n$ in place of $1$: the open ball $B(x_n, \varepsilon_n/2)$ is not contained in the nowhere dense $C_n$, so pick $x_{n+1}$ in $B(x_n, \varepsilon_n/2) \setminus C_n$; that set is open because $C_n$ is closed; so some $B(x_{n+1}, \varepsilon_{n+1})$ with $\varepsilon_{n+1} \leq \varepsilon_n/2$ sits inside it. This proves the claim.

Two things follow. First, the radii collapse: $\varepsilon_{n+1} \leq \varepsilon_n / 2$ gives $\varepsilon_n \leq \varepsilon_1 2^{-(n-1)} \to 0$. Second, the balls nest, since $B(x_{n+1}, \varepsilon_{n+1}) \subseteq B(x_n, \varepsilon_n/2) \subseteq B(x_n, \varepsilon_n)$.

So $(x_n)_{n \geq 1}$ is Cauchy. Given $\varepsilon > 0$, pick $k$ with $\varepsilon_k < \varepsilon$, which is possible since $\varepsilon_n \to 0$. For every $n \geq k$ nesting puts $x_n$ inside $B(x_k, \varepsilon_k)$, so $\lVert x_n - x_k \rVert < \varepsilon_k < \varepsilon$. Since $X$ is complete, $x_n \to x$ for some $x \in X$.

Now locate the limit. Fix $n \geq 1$. For every $m > n$ we have $x_m \in B(x_{n+1}, \varepsilon_{n+1}) \subseteq B(x_n, \varepsilon_n / 2)$, so $\lVert x_m - x_n \rVert < \varepsilon_n/2$, and letting $m \to \infty$ gives $\lVert x - x_n \rVert \leq \varepsilon_n / 2 < \varepsilon_n$. This is exactly what the halved radius bought us: passing to the limit costs a non-strict inequality, and the factor of two absorbs it, so $x$ lands strictly inside $B(x_n, \varepsilon_n)$ rather than merely on its closure.

The same estimate at $n+1$ gives $x \in B(x_{n+1}, \varepsilon_{n+1}) \subseteq X \setminus C_n$ for every $n \geq 1$, so $x$ belongs to no $C_n$ and therefore $x \notin Y$. But $n = 1$ gives $x \in B(x_1, \varepsilon_1) \subseteq Y$. Contradiction. □

Proof. A normed vector space contains at least the zero vector, so $B(0, 1)$ is an open ball, and it is contained in $X$. If $X$ were a countable union of closed nowhere dense subsets, that union would contain an open ball, contradicting Theorem 2.1. □

Proof. ($\Rightarrow$) Let $M$ be dense and suppose $B(m, \varepsilon) \subseteq X \setminus M$ for some $m \in X$ and $\varepsilon > 0$. By density there is a sequence $(m_i)_{i \geq 1} \subseteq M$ with $m_i \to m$, so there is $N \geq 1$ with $\lVert m - m_N \rVert < \varepsilon$. Then $m_N \in B(m, \varepsilon) \subseteq X \setminus M$, while $m_N \in M$. These sets are disjoint, so this is a contradiction.

($\Leftarrow$) Suppose $X \setminus M$ contains no open ball, and let $x \in X$. For each $n \geq 1$ the set $B(x, 1/n)$ is an open ball, hence not contained in $X \setminus M$, so we may choose $m_n \in B(x, 1/n) \cap M$. Then $\lVert x - m_n \rVert < 1/n \to 0$, so $(m_n)_{n \geq 1} \subseteq M$ converges to $x$. As $x$ was arbitrary, $M$ is dense. □

Proof. Write $U = \bigcap_{n \geq 1} O_n$. By De Morgan,

$$X \setminus U \;=\; \bigcup_{n \geq 1}\, (X \setminus O_n).$$

Each $O_n$ is open, so each $X \setminus O_n$ is closed; each $O_n$ is dense, so by Lemma 2.3 each $X \setminus O_n$ contains no open ball; and a closed set containing no open ball is nowhere dense by Lemma 1.2(b). So $X \setminus U$ is a countable union of closed nowhere dense sets, and Theorem 2.1 says it contains no open ball. Applying Lemma 2.3 once more, this time to $U$, gives that $U$ is dense. □

Proof. Work in $C([0,1])$ with the supremum norm $\lVert f \rVert_\infty = \sup_{x \in [0,1]} \lvert f(x) \rvert$, under which it is complete. For $f$ to be differentiable somewhere there must be an $x \in [0,1]$ at which $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ exists. Inspired by that, define for each $n, m \geq 1$

$$\begin{aligned} A_{n,m} \;=\; \Bigl\{\, f \in C([0,1]) \;:\;\ &\exists\, x \in [0,1] \ \text{ with } \ \Bigl\lvert \tfrac{f(x+h) - f(x)}{h} \Bigr\rvert \leq n \\[2pt] &\text{for all } h \text{ with } 0 < \lvert h \rvert < \tfrac{1}{m} \ \text{ and } \ x + h \in [0,1] \,\Bigr\}. \end{aligned}$$

Every somewhere-differentiable function lands in one of these. Suppose $f$ has derivative $y$ at $x$. Choose $n > \lvert y \rvert$. By the definition of the limit there is $\varepsilon > 0$ such that $\lvert (f(x+h) - f(x))/h \rvert \leq n$ whenever $0 < \lvert h \rvert < \varepsilon$, and then choose $m$ with $1/m < \varepsilon$, which is possible since $1/m \to 0$. So $f \in A_{n,m}$. Writing $ND([0,1])$ for the set of nowhere differentiable functions, this says

$$C([0,1]) \setminus ND([0,1]) \;\subseteq\; \bigcup_{n \geq 1,\, m \geq 1} A_{n,m}.$$

Each $A_{n,m}$ is closed. Let $(f_i)_{i \geq 1} \subseteq A_{n,m}$ converge uniformly to $f$, and let $x_i$ be a witness for $f_i$, so that $\lvert (f_i(x_i + h) - f_i(x_i))/h \rvert \leq n$ for all admissible $h$ with $\lvert h \rvert < 1/m$. The witnesses all live in $[0,1]$, which is closed and bounded, so by Bolzano–Weierstrass some subsequence converges, say $x_{i_k} \to x \in [0,1]$. Fix $h$ with $0 < \lvert h \rvert < 1/m$ and $x + h \in [0,1]$; for large $k$ the point $x_{i_k} + h$ lies in $[0,1]$ too, after shrinking $\lvert h \rvert$ by an arbitrarily small amount if $x+h$ is an endpoint. Then

$$\left\lvert \frac{f(x+h) - f(x)}{h} \right\rvert \;=\; \lim_{k \to \infty} \left\lvert \frac{f_{i_k}(x_{i_k} + h) - f_{i_k}(x_{i_k})}{h} \right\rvert \;\leq\; n,$$

where the equality uses that $f_{i_k} \to f$ uniformly and $x_{i_k} \to x$ with $f$ continuous. So $x$ is a witness for $f$, giving $f \in A_{n,m}$.

Each $A_{n,m}$ contains no open ball. Take any $f \in A_{n,m}$ and any $\varepsilon > 0$; we produce a member of $B(f, \varepsilon)$ that is not in $A_{n,m}$. My first thought here was to reach for Fourier series, on the grounds that a truncated series wiggles around the function it approximates.

That does not quite do the job: a partial sum is smooth with bounded derivative, so nothing stops it from lying in $A_{n,m}$ as well. Sawtooth functions do the job, because their slopes are ours to choose. The piecewise-linear functions are dense in $C([0,1])$ — $f$ is uniformly continuous on a compact interval, so a fine enough polygonal interpolation stays within $\varepsilon$ of it — and a polygonal path can be replaced by a zigzag with the same endpoints, arbitrarily fine teeth, and slopes as steep as we please.

So choose a sawtooth $g$ with $\lVert f - g \rVert_\infty < \varepsilon$, every linear piece of slope greater than $n$ in absolute value, and every piece of width less than $1/m$. Given any $x \in [0,1]$, take $h$ small enough that $x$ and $x + h$ lie on one linear piece. The difference quotient is then exactly that piece's slope, so it exceeds $n$ while $0 < \lvert h \rvert < 1/m$. No $x$ can witness $g \in A_{n,m}$, so $g \notin A_{n,m}$ although $g \in B(f, \varepsilon)$. Hence $A_{n,m}$ contains no open ball, and being closed it is nowhere dense by Lemma 1.2(b). Detailed proofs of the density claims are in Green's notes [1].

Conclusion. The family $\{A_{n,m}\}$ is indexed by $\mathbb{N}_{\geq 1} \times \mathbb{N}_{\geq 1}$, which is countable, and each member is closed and nowhere dense. By Theorem 2.1 their union contains no open ball. Since $C([0,1]) \setminus ND([0,1])$ is a subset of that union, it contains no open ball either, and Lemma 2.3 turns this into the statement that $ND([0,1])$ is dense in $C([0,1])$. In particular it is not empty. □

Proof. Let $(X, \lVert \cdot \rVert_X)$ be a Banach space, $(Y, \lVert \cdot \rVert_Y)$ a normed space, $A$ an index set, and assume $T = (f_\alpha)_{\alpha \in A} \subseteq \mathcal{B}(X,Y)$ is pointwise bounded at every $x \in X$. Partition the domain by how large the operators get:

$$E_k := \Bigl\{ x \in X : \sup_{f \in T} \lVert f(x) \rVert_Y \leq k \Bigr\} \;=\; \bigcap_{f \in T} \bigl\{ x : \lVert f(x) \rVert_Y \leq k \bigr\}, \qquad k \in \mathbb{N}.$$

By pointwise boundedness (Definition 3.1) every $x \in X$ has a bound $C_x$, so $x \in E_k$ for any $k \geq C_x$, and therefore

$$\bigcup_{k \geq 1} E_k = X.$$

Each set $\{x : \lVert f(x) \rVert_Y \leq k\}$ is the preimage of a closed set under a continuous map, hence closed, and an intersection of closed sets is closed, so every $E_k$ is closed. By Corollary 2.2 the $E_k$ cannot all be nowhere dense, since a complete space is not a countable union of closed nowhere dense sets. So there is some $N \in \mathbb{N}$ for which $E_N$ is not nowhere dense, and since $E_N$ is closed, Lemma 1.2(b) says it contains an open ball. That ball's closure stays inside the closed set $E_N$, so we also get a closed ball: there are $b \in E_N$ and $\varepsilon > 0$ with $\overline{B}(b, \varepsilon) \subseteq E_N$.

Now take any $z \in X$ with $z \neq 0$ and rescale it to length exactly $\varepsilon$:

$$z_\varepsilon = \frac{\varepsilon}{\lVert z \rVert_X}\, z, \qquad \lVert (b + z_\varepsilon) - b \rVert_X = \lVert z_\varepsilon \rVert_X = \varepsilon.$$

This is the point of the rescaling: it makes $b + z_\varepsilon$ land in the closed ball, hence in $E_N$, which bounds $\sup_{f \in T} \lVert f(b + z_\varepsilon) \rVert_Y \leq N$. Since $b \in E_N$ as well, linearity and the triangle inequality give

$$\sup_{f \in T} \lVert f(z_\varepsilon) \rVert_Y = \sup_{f \in T} \lVert f(b + z_\varepsilon) - f(b) \rVert_Y \leq \sup_{f \in T} \lVert f(b + z_\varepsilon) \rVert_Y + \sup_{f \in T} \lVert f(b) \rVert_Y \leq 2N.$$

The scaled-down $z_\varepsilon$ now has a concrete bound, so scale back up using $z = \frac{\lVert z \rVert_X}{\varepsilon} z_\varepsilon$:

$$\sup_{f \in T} \lVert f(z) \rVert_Y \;=\; \frac{\lVert z \rVert_X}{\varepsilon} \sup_{f \in T} \lVert f(z_\varepsilon) \rVert_Y \;\leq\; \frac{2N \lVert z \rVert_X}{\varepsilon}.$$

Dividing by $\lVert z \rVert_X$ and taking the supremum over $z \neq 0$, the definition of the operator norm gives

$$\lVert f \rVert_{\mathcal{B}} \;=\; \sup_{z \neq 0} \frac{\lVert f(z) \rVert_Y}{\lVert z \rVert_X} \;\leq\; \frac{2N}{\varepsilon} \qquad \text{for every } f \in T.$$

Since $2N/\varepsilon$ is a constant independent of $f$, the collection $T$ is uniformly bounded by Definition 3.0. □