Equivalent norms, density and compactness in infinite dimensions

The lecture studies the topology of normed spaces of infinite dimension.

Equivalent norms#

Definition 1 (Equivalent norms). Let $X$ be a vector space and let $\|\cdot\|_a$ and $\|\cdot\|_b$ be two norms on ${X.}$ These norms are equivalent if there exist ${C_1, C_2 > 0}$ such that

$$C_1 \|x\|_a \le \|x\|_b \le C_2 \|x\|_a \qquad \text{for every } x \in X.$$

The first inequality means that ${B_r^{\|\cdot\|_b}(0) \subset B_{r/C_1}^{\|\cdot\|_a}(0),}$ and the second means that ${B_r^{\|\cdot\|_a}(0) \subset B_{C_2 r}^{\|\cdot\|_b}(0),}$ for every ${r > 0.}$ Here ${B_r^{\|\cdot\|_a}(0) = \{ x \in X : \|x\|_a < r \}}$ and ${B_{C_2 r}^{\|\cdot\|_b}(0) = \{ x \in X : \|x\|_b < C_2 r \}}$ are balls of the distances induced by the two norms.

Example 2. (i) Let $N$ be a positive integer. One can prove that, for all ${p, q \in [1, +\infty],}$ the norms $\|\cdot\|_p$ and $\|\cdot\|_q$ on $\mathbb{R}^N$ of Example 5 (ii) of the tenth post are equivalent.
(ii) Let ${-\infty < a < b < +\infty}$ and ${p \in [1, +\infty].}$ The norms $\|\cdot\|_{W^{1,p}}$ and $\|\cdot\|_{W_0^{1,p}}$ on ${W_0^{1,p}((a, b))}$ of Examples 19 and 21 of the tenth post are equivalent, as in Remark 6 of the previous post.

Remark 3. Let $X$ be a vector space and let $\|\cdot\|_a$ and $\|\cdot\|_b$ be two norms on ${X.}$ The norm $\|\cdot\|_a$ induces a distance, as in Definition 1 of the tenth post, hence balls and a topology ${\tau_a,}$ as in Definition 19 of the first post. In the same way $\|\cdot\|_b$ induces a topology ${\tau_b.}$ If $\|\cdot\|_a$ is equivalent to ${\|\cdot\|_b,}$ then ${\tau_a = \tau_b.}$ In particular:
(i) ${(X, \|\cdot\|_a)}$ and ${(X, \|\cdot\|_b)}$ also have the same closed sets and the same compact sets;
(ii) a function ${f \colon X \to \mathbb{R}}$ is continuous as a map ${f \colon (X, \|\cdot\|_a) \to \mathbb{R}}$ if and only if it is continuous as a map ${f \colon (X, \|\cdot\|_b) \to \mathbb{R}.}$

Proof (sketch). Only the inclusion ${\tau_b \subset \tau_a}$ is proved, with ${C_1, C_2 > 0}$ as in Definition 1. Let $U$ be open in ${\tau_b.}$ For every ${x \in U}$ there exists ${r > 0}$ such that ${B_r^{\|\cdot\|_b}(x) \subset U.}$ Hence ${B_{r/C_2}^{\|\cdot\|_a}(x) \subset B_r^{\|\cdot\|_b}(x) \subset U,}$ and $U$ is open in ${\tau_a.}$ Indeed, if ${\|y - x\|_a < r/C_2,}$ then ${\|y - x\|_b \le C_2 \|y - x\|_a < r}$ by Definition 1. $\blacksquare$

Two norms on $C([-1, 1])$#

Example 4. Let ${X = C([-1, 1]),}$ with the norm $\|\cdot\|_\infty$ of Example 5 (iii) of the tenth post and the norm ${\|\cdot\|_1.}$

(1) For every ${f \in C([-1, 1]),}$

$$\|f\|_1 = \int_{-1}^{1} |f| \, d\lambda \le 2 \|f\|_\infty.$$

(2) There exists no ${C > 0}$ such that ${\|f\|_\infty \le C \|f\|_1}$ for every ${f \in C([-1, 1]).}$

To prove Example 4 (2), for every integer ${n \ge 1}$ let ${f_n \in C([-1, 1])}$ be equal to 0 on ${[-1, 0],}$ linear on $[0, 1/n]$ and equal to 1 on ${[1/n, 1],}$ as in Figure 1. Let $H$ be equal to 0 on $[-1, 0]$ and to 1 on ${(0, 1].}$

n

3

Figure 1. The drawing of Example 4: on the left the function $f_n$ for the $n$ of the slider and, dashed where it differs from ${f_n,}$ the function ${H,}$ with the region between the two graphs shaded, of area ${\|f_n - H\|_1;}$ on the right the function ${H.}$

Then, as in Example 10 of the previous post, ${\|f_n - H\|_1 \to 0}$ as ${n \to \infty.}$ Hence $(f_n)$ is Cauchy in ${L^1,}$ as in the proof of Remark 18 of the tenth post: for every ${\varepsilon > 0}$ there exists $N$ such that ${\|f_n - f_m\|_1 < \varepsilon}$ for all ${n, m \ge N.}$

If there were ${C > 0}$ such that ${\|f\|_\infty \le C \|f\|_1}$ for every ${f \in C([-1, 1]),}$ then $(f_n)$ would be Cauchy with respect to ${\|\cdot\|_\infty,}$ and $(f_n)$ should converge uniformly to a continuous function.

This is false, so no such $C$ exists. In detail, a uniform limit is also a pointwise limit, and $f_n(x)$ tends to $H(x)$ at every ${x \in [-1, 1],}$ while $H$ is not continuous at 0.

Spaces of finite dimension#

Remark 5. Let $X$ be a vector space of finite dimension, that is, with a basis ${\mathcal{B} = \{ e_1, \dots, e_N \} \subset X.}$ Then one can consider

$$\begin{aligned} \phi \colon X &\to \mathbb{R}^N, \\ e_i &\mapsto (0, \dots, 0, 1, 0, \dots, 0), \\ \phi(a_1 e_1 + \dots + a_N e_N) &= (a_1, \dots, a_N) \in \mathbb{R}^N, \end{aligned}$$

where the 1 is in the $i$-th place. The map $\phi$ is an isomorphism between $X$ and ${\mathbb{R}^N,}$ so any finite-dimensional vector space can be identified with ${\mathbb{R}^N.}$

Theorem 6. Let $X$ be a vector space of finite dimension and let $\|\cdot\|_a$ and $\|\cdot\|_b$ be two norms on ${X.}$ Then $\|\cdot\|_a$ and $\|\cdot\|_b$ are equivalent.

Corollary 7. Let $X$ be a vector space of finite dimension and let $\|\cdot\|$ be a norm on ${X.}$ Then a set ${K \subset X}$ is compact if and only if $K$ is closed and bounded.

Proof (sketch). By Theorem 6, $\|\cdot\|$ is equivalent to the Euclidean norm ${x \mapsto |\phi(x)|,}$ with $\phi$ as in Remark 5. Hence, by Remark 3 (i), $K$ is compact in ${(X, \|\cdot\|)}$ if and only if $K$ is compact with respect to the Euclidean topology. In detail, by the Heine–Borel theorem, Theorem 15 of the first post, via ${\phi,}$ this holds if and only if $K$ is closed and bounded for the Euclidean norm, hence for $\|\cdot\|$ by Remark 3 (i) and Definition 1. $\blacksquare$

Subspaces and density#

Definition 8 (Dense subset). Let ${(X, d)}$ be a metric space. A subset ${D \subset X}$ is dense if, as in Figure 2,

$$\forall x_0 \in X \quad \forall r > 0 \quad \exists x \in D \ \text{ such that } \ d(x, x_0) < r, \ \text{ that is, } \ x \in B_r(x_0).$$

Figure 2. The drawing of Definition 8: the ball $B_r(x_0)$ and a point ${x \in D}$ in it.

Remark 9. Let ${(X, d)}$ be a metric space and let ${D \subset X.}$ The following are equivalent.
(i) $D$ is dense.
(ii) For every ${x_0 \in X}$ there exists ${(x_n) \subset D}$ such that ${x_n \to x_0}$ in ${X.}$
(iii) ${\overline{D} = X,}$ where $\overline{D}$ is the closure of $D$ of Definition 12 (i) of the first post.

Example 10. The set $\mathbb{Q}$ is dense in ${\mathbb{R},}$ with the distance ${d(x, y) = |x - y|.}$

Example 11. Let $N$ be a positive integer and let $V$ be a subspace of $\mathbb{R}^N$ with ${V \neq \mathbb{R}^N.}$
(i) $V$ is closed in ${\mathbb{R}^N.}$
(ii) ${(V, \|\cdot\|_2)}$ is a Banach space.
(iii) $V$ is not dense.
(iv) Let ${x \in \mathbb{R}^N \setminus V.}$ Then ${d(x, V) = \inf_{v \in V} |x - v| > 0,}$ and there exists ${\bar{v} \in V}$ such that ${|x - \bar{v}| = d(x, V),}$ as in Figure 3: $\bar{v}$ is the closest point to $x$ in ${V.}$

x₁

1.40

x₂

1.90

Figure 3. The drawing of Example 11, an instance with ${N = 2}$ and $V$ a line: the point $x$ of the sliders, the point $\bar{v}$ of $V$ closest to ${x,}$ at distance ${|x - \bar{v}| = d(x, V),}$ and a ball around a point $x_0$ that does not meet ${V.}$

The continuous functions in $L^1((0, 1))$#

Example 12. Consider ${C([0, 1]) \subset L^1((0, 1)),}$ as in Remark 16 of the sixth post, that is, ${(C([0, 1]), \|\cdot\|_1) \subset (L^1((0, 1)), \|\cdot\|_1).}$ Here ${V = C([0, 1])}$ is a subspace of ${L^1,}$ and ${V \neq L^1.}$
(i) ${(V, \|\cdot\|_1)}$ is not a Banach space, as in Example 10 of the previous post.
(ii) ${V \subset L^1}$ is not closed.
(iii) $V$ is dense in ${L^1,}$ by Theorem 13.

Theorem 13. For every ${f \in L^1((0, 1))}$ there exists ${(f_n) \subset V = C([0, 1])}$ such that ${f_n \to f}$ in ${L^1((0, 1)).}$

Proof (sketch). The idea is drawn in Figure 4. We know that $f$ can be approximated by simple functions, as in Proposition 17 of the fourth post, and any simple function can be approximated by continuous functions. $\blacksquare$

Figure 4. The drawing of the proof of Theorem 13, an instance: a function ${f \in L^1,}$ below it a simple function constant on three intervals, and a continuous function $f_n$ close to the simple function.

In particular, by Theorem 13, if ${f \in L^1 \setminus V,}$ then ${d(f, V) = \inf_{g \in V} \|f - g\|_1 = 0,}$ and it makes no sense to speak of the closest point to $f$ in ${V.}$

Closed and dense subspaces#

Remark 14. Let ${(X, \|\cdot\|)}$ be a normed space.
(1) If ${V \subset X}$ is a subspace and ${\dim V < +\infty,}$ then $V$ is closed.
(2) If ${(X, \|\cdot\|)}$ is a Banach space and ${V \subset X}$ is a subspace, then ${(V, \|\cdot\|)}$ is a Banach space if and only if ${V \subset X}$ is closed.

Remark 15 (Dense subspaces). (1) (Stone–Weierstrass) Let ${-\infty < a < b < +\infty.}$ The subspace $V$ of the polynomials is dense in ${(C([a, b]), \|\cdot\|_\infty).}$
(2) Let ${1 \le p < +\infty}$ and let ${U \subset \mathbb{R}^N}$ be open, with the Lebesgue measure. Then the set of the simple functions that belong to $L^p(U)$ is dense in ${L^p(U),}$ $C_c(U)$ is dense in $L^p(U)$ and $C_c^\infty(U)$ is dense in ${L^p(U).}$ Here $C_c(U)$ and $C_c^\infty(U)$ are defined as in the notation of the fifth post, with $U$ in place of ${(a, b).}$

The continuous functions in $L^\infty((-1, 1))$#

Example 16. Remark 15 (2) is false for ${p = +\infty{:}}$ ${C([-1, 1]) \subset L^\infty((-1, 1))}$ is not dense.

To prove Example 16, let $H$ be the function of Example 4 and let ${B_{1/3}(H) = \{ f \in L^\infty((-1, 1)) : \|f - H\|_\infty < 1/3 \}.}$ We claim that ${f \notin C([-1, 1])}$ for every ${f \in B_{1/3}(H),}$ as in Figure 5.

Figure 5. The drawing of Example 16: the function ${H,}$ the dashed levels $-1/3$ and $1/3$ over ${(-1, 0)}$ and $2/3$ and $4/3$ over ${(0, 1),}$ and between them a function $f$ of ${B_{1/3}(H),}$ an instance. At 0 a bracket marks the gap between the levels $1/3$ and ${2/3.}$

If ${f \in B_{1/3}(H),}$ then

$$\begin{aligned} &\text{(i)} && {-\tfrac{1}{3}} \le f \le \tfrac{1}{3} && \text{almost everywhere on } (-1, 0), \\ &\text{(ii)} && \tfrac{2}{3} \le f \le \tfrac{4}{3} && \text{almost everywhere on } (0, 1), \end{aligned}$$

hence ${f \notin C([-1, 1]).}$

Suppose that $f$ has a continuous representative, still written ${f.}$ Then

$$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x),$$

where the left-hand side is at most $1/3$ and the right-hand side is at least ${2/3,}$ which is impossible.

Compactness#

Remark 17. Let $\|\cdot\|$ be a norm on ${\mathbb{R}^N,}$ a space of finite dimension. Then a set ${K \subset \mathbb{R}^N}$ is compact if and only if $K$ is closed and bounded. In particular, the closed ball ${\overline{B}_1(0) = \{ x \in \mathbb{R}^N : \|x\| \le 1 \}}$ is compact.

Theorem 18 (Riesz). Let ${(X, \|\cdot\|)}$ be a normed space and let ${\overline{B}_1(0) = \{ x \in X : \|x\| \le 1 \}.}$ Then $\overline{B}_1(0)$ is compact if and only if ${\dim X < +\infty.}$ In particular, $\overline{B}_1(0)$ is never compact in a normed space of infinite dimension.

Proof (complete). If ${\dim X < +\infty,}$ then $\overline{B}_1(0)$ is compact, as in Remark 17. Indeed, $\overline{B}_1(0)$ is closed and bounded, hence compact by Corollary 7.

Conversely, let $\overline{B}_1(0)$ be compact. For ${x \in X}$ and a subspace $M$ of $X$ let ${d(x, M) = \inf_{m \in M} \|x - m\|.}$

Step 1. Let ${M \subset X}$ be a closed subspace with ${M \neq X.}$ Then for every ${\varepsilon > 0}$ there exists ${v \in X}$ such that, as in Figure 6,

$$\|v\| = 1 \qquad \text{and} \qquad d(v, M) \ge 1 - \varepsilon.$$

Figure 6. The drawing of Step 1 in the proof of Theorem 18, an instance in $\mathbb{R}^2$ with the Euclidean norm: a closed subspace ${M,}$ a line, the dashed circle of the vectors of norm ${1,}$ and a vector $v$ on it with ${d(v, M) = 1.}$

Proof of Step 1. Since ${M \neq X,}$ there exists ${x_0 \in X \setminus M.}$ Notice that ${d(x_0, M) > 0.}$ Suppose, by contradiction, that ${d(x_0, M) = 0.}$ Then there exists ${(x_n) \subset M}$ such that ${d(x_0, x_n) \to 0,}$ that is, $(x_n)$ converges to ${x_0 \in X.}$ Since $M$ is closed, ${x_0 \in M}$ by Lemma 20 (i) of the first post, but ${x_0 \in X \setminus M,}$ a contradiction.

Now fix ${\varepsilon \in (0, 1).}$ By the definition of the distance ${d(x_0, M),}$ there exists ${m_0 \in M}$ such that

$$0 < \|x_0 - m_0\| \le \frac{1}{1 - \varepsilon} \, d(x_0, M).$$

Indeed, the right-hand side is larger than the infimum ${d(x_0, M) > 0,}$ because ${1/(1 - \varepsilon) > 1,}$ and ${\|x_0 - m_0\| > 0}$ because $x_0$ is not in ${M.}$ Take, as in Figure 7,

$$v := \frac{x_0 - m_0}{\|x_0 - m_0\|}.$$

ε

0.30

m₀

0.00

Figure 7. The drawing of the proof of Step 1, an instance in $\mathbb{R}^2$ with the Euclidean norm: the point ${x_0 \notin M,}$ the thick part of $M$ where ${\|x_0 - m\| \le d(x_0, M)/(1 - \varepsilon),}$ a point $m_0$ in it, the vector $v$ and the dashed line at distance ${1 - \varepsilon}$ from ${M.}$ The sliders set $\varepsilon$ and the position of ${m_0.}$

Then ${\|v\| = 1,}$ and for every ${m \in M}$

$$\begin{aligned} \|v - m\| &= \Big\| \frac{x_0 - m_0}{\|x_0 - m_0\|} - m \Big\| \\ &= \frac{1}{\|x_0 - m_0\|} \, \Bigl\|\, x_0 - m_0 - m \, \|x_0 - m_0\| \,\Bigr\| \\ &\ge \frac{1}{\|x_0 - m_0\|} \, d(x_0, M) \\ &\ge \frac{1 - \varepsilon}{d(x_0, M)} \, d(x_0, M) = 1 - \varepsilon. \end{aligned}$$

The first inequality holds because ${m_0 + m \, \|x_0 - m_0\|}$ belongs to the subspace ${M,}$ so the norm in the second line is that of $x_0$ minus a vector of ${M,}$ which is at least ${d(x_0, M).}$ The second inequality holds because ${\|x_0 - m_0\| \le d(x_0, M)/(1 - \varepsilon).}$ We got ${\|v - m\| \ge 1 - \varepsilon}$ for every ${m \in M,}$ hence ${d(v, M) \ge 1 - \varepsilon.}$

Step 2. We prove that ${\dim X < +\infty.}$ Suppose, by contradiction, that ${\dim X = +\infty.}$ We construct inductively a sequence of vectors.

(i) Fix ${v_0 \in X}$ with ${\|v_0\| = 1}$ and let ${M_0 = \operatorname{Span}(v_0) = \{ \alpha v_0 : \alpha \in \mathbb{R} \}.}$ Then ${M_0 \neq X,}$ because $M_0$ has dimension 1 and $X$ has infinite dimension, and $M_0$ is closed by Remark 14 (1), because it has finite dimension.

(ii) Apply Step 1 to ${M = M_0}$ with ${\varepsilon = 1/2.}$ It gives ${v_1 \in X}$ with ${\|v_1\| = 1}$ and ${d(v_1, M_0) \ge 1 - \varepsilon = 1/2.}$ Now consider ${M_1 = \operatorname{Span}(v_0, v_1).}$ Then ${M_1 \neq X,}$ because $M_1$ has dimension 2 and $X$ has infinite dimension, and $M_1$ is closed, because it has finite dimension. So the construction can be iterated.

(iii) Let ${M_{n-1} = \operatorname{Span}(v_0, v_1, \dots, v_{n-1})}$ be given, of finite dimension, closed and different from ${X.}$ Then Step 1 gives ${v_n \in X}$ such that ${\|v_n\| = 1}$ and ${d(v_n, M_{n-1}) \ge 1/2.}$

In the end we found vectors $v_n$ of $X$ such that ${\|v_n\| = 1}$ for every $n$ and ${d(v_n, M_{n-1}) \ge 1/2}$ for every ${n \ge 1.}$ However, ${\|v_n\| = 1}$ for every $n$ gives ${(v_n) \subset \overline{B}_1(0),}$ which is compact by assumption. Hence, by Lemma 20 (ii) of the first post, there exist a subsequence $(v_{n_k})$ and ${w \in X}$ such that ${v_{n_k} \to w}$ in $X$ as ${k \to \infty.}$ That is, ${d(w, v_{n_k}) \to 0.}$

Therefore $(v_{n_k})$ is Cauchy in ${X{:}}$ for every ${\varepsilon_0 > 0}$ there exists $N$ such that ${d(v_{n_k}, v_{n_{k-1}}) < \varepsilon_0}$ for every ${k \ge N.}$ But, for every ${k \ge 1,}$

$$d(v_{n_k}, v_{n_{k-1}}) \ge d(v_{n_k}, M_{n_k - 1}) \ge \frac{1}{2},$$

because ${v_{n_{k-1}} \in M_{n_{k-1}} \subset M_{n_k - 1}.}$ This is a contradiction for ${\varepsilon_0 < 1/2.}$ $\blacksquare$

Compactness in $C([a, b])$#

Let ${-\infty < a < b < +\infty.}$ A family $\mathcal{F}$ of functions of $C([a, b])$ is bounded if there exists ${R > 0}$ such that ${\|f\|_\infty < R}$ for every ${f \in \mathcal{F}.}$ It is uniformly equicontinuous if for every ${\varepsilon > 0}$ there exists ${\delta > 0}$ such that

$$x, y \in [a, b], \ \ |x - y| < \delta \quad \Longrightarrow \quad |f(x) - f(y)| < \varepsilon \quad \text{for every } f \in \mathcal{F}.$$

Theorem 19 (Ascoli–Arzelà). Let ${-\infty < a < b < +\infty,}$ consider ${(C([a, b]), \|\cdot\|_\infty)}$ and let ${\mathcal{F} \subset C([a, b]).}$ Then $\mathcal{F}$ is precompact, that is, $\overline{\mathcal{F}}$ is compact, as in Definition 14 of the first post, if and only if $\mathcal{F}$ is bounded and uniformly equicontinuous.

In particular, let ${(f_n) \subset C([a, b])}$ be such that ${\sup_n \|f_n\|_\infty < +\infty}$ and ${\mathcal{F} = \{ f_n : n \in \mathbb{N} \}}$ is uniformly equicontinuous. Then there exist a subsequence $(f_{n_k})$ and ${g \in C([a, b])}$ such that ${f_{n_k} \to g}$ in $\|\cdot\|_\infty$ as ${k \to \infty.}$

Remark 20. Let ${-\infty < a < b < +\infty.}$
(i) Let ${(f_n) \subset C([a, b])}$ be equilipschitz: there exists ${L > 0}$ such that, for every $n$ and all $x$ and $y$ in ${[a, b],}$ ${|f_n(x) - f_n(y)| \le L |x - y|.}$ Then $(f_n)$ is uniformly equicontinuous.
(ii) Let ${p \in (1, +\infty]}$ and let ${(f_n) \subset W^{1,p}((a, b))}$ be bounded in ${W^{1,p},}$ that is, ${\sup_n \|f_n\|_{W^{1,p}} < +\infty,}$ with the norm of Example 19 of the tenth post. Then $(f_n)$ is bounded in $\|\cdot\|_\infty$ and uniformly equicontinuous, by estimates analogous to those of Example 4 and Remark 6 of the previous post, so there exists a subsequence $(f_{n_k})$ that converges uniformly, by Theorem 19. Here each $f_n$ is identified with its absolutely continuous representative, as in Example 21 of the tenth post.

Proof (sketch). Only (i) is proved. Let ${\varepsilon > 0}$ and choose ${\delta > 0}$ with ${\delta < \varepsilon/L.}$ Then, for every $n$ and all ${x, y \in [a, b]}$ with ${|x - y| < \delta,}$

$$|f_n(x) - f_n(y)| \le L |x - y| < L \delta < \varepsilon. \tag*{$\blacksquare$}$$