Banach spaces, the Hölder inequality and the completeness of $L^p$

Functional analysis studies infinite-dimensional vector spaces (Brezis).

Normed and Banach spaces#

Definition 1 (Norm). Let $X$ be a vector space. A norm on $X$ is a function ${\|\cdot\| \colon X \to [0, +\infty)}$ such that, for all ${x, y \in X}$ and ${\alpha \in \mathbb{R}{:}}$
(i) ${\|x\| = 0}$ if and only if ${x = 0;}$
(ii) ${\|\alpha x\| = |\alpha| \, \|x\|;}$
(iii) ${\|x + y\| \le \|x\| + \|y\|,}$ the triangular inequality, drawn in Figure 1.
A norm $\|\cdot\|$ induces on $X$ the distance ${d \colon X \times X \to [0, +\infty)}$ given by ${d(x, y) = \|x - y\|,}$ and $(X, d)$ is a metric space.

Figure 1. The drawing of Definition 1 (iii), an instance in the plane: two vectors $x$ and ${y,}$ their sum ${x + y}$ and the vector $y$ drawn again from the end of ${x.}$ Dashed, the segment from the end of $y$ to the end of ${x + y.}$

Remark 2 (Normed space). If $\|\cdot\|$ is a norm on a vector space ${X,}$ the couple $(X, \|\cdot\|)$ is called a normed space. Balls, open, closed and compact sets in $X$ and continuous functions ${f \colon X \to \mathbb{R}}$ are then well defined: they are those of the metric space $(X, d)$ of Definition 1.

Remark 3. Let $(X, \|\cdot\|)$ be a normed space. Then the norm ${\|\cdot\| \colon X \to [0, +\infty)}$ is continuous.

Proof (complete). The norm satisfies the reverse triangle inequality ${\bigl|\, \|x\| - \|y\| \,\bigr| \le \|x - y\|}$ for all ${x, y \in X.}$ Let ${(x_n) \subset X}$ and ${x_0 \in X}$ be such that ${x_n \to x_0}$ in ${X.}$ This means ${d(x_n, x_0) \to 0,}$ that is, ${\|x_n - x_0\| \to 0.}$ We want to prove that ${\|x_n\| \to \|x_0\|,}$ and the reverse triangle inequality gives it: ${\bigl|\, \|x_n\| - \|x_0\| \,\bigr| \le \|x_n - x_0\| \to 0.}$ $\blacksquare$

Definition 4 (Cauchy sequence, Banach space). Let $(X, \|\cdot\|)$ be a normed space.
(i) A sequence $(x_n)$ of points of $X$ is Cauchy if for every ${\varepsilon > 0}$ there exists $N$ such that ${\|x_n - x_m\| < \varepsilon}$ for all ${n, m \ge N.}$ Here ${\|x_n - x_m\| = d(x_n, x_m).}$
(ii) If every Cauchy sequence converges in ${X,}$ we say that $(X, \|\cdot\|)$ is a Banach space.

Norms on Euclidean space and on continuous functions#

Example 5. (i) Let $N$ be a positive integer. The Euclidean space ${(\mathbb{R}^N, |\cdot|)}$ is a Banach space, where

$$|(v_1, \dots, v_N)| = \sqrt{\sum_{i=1}^{N} (v_i)^2}.$$

(ii) For ${p \in [1, +\infty),}$ consider ${(\mathbb{R}^N, \|\cdot\|_p),}$ where

$$\|(v_1, \dots, v_N)\|_p = \Big( \sum_{i=1}^{N} |v_i|^p \Big)^{\frac{1}{p}}, \qquad \text{and also} \qquad \|(v_1, \dots, v_N)\|_\infty = \max_{i = 1, \dots, N} |v_i|.$$

Notice that ${|\cdot| = \|\cdot\|_2.}$ The spaces ${(\mathbb{R}^N, \|\cdot\|_p)}$ are Banach spaces.

(iii) Let ${-\infty < a < b < +\infty,}$ and let $C([a, b])$ be the set of the continuous functions on ${[a, b],}$ with

$$\|u\|_\infty = \operatorname*{ess\,sup}_{[a, b]} |u| = \max_{[a, b]} |u| \qquad \text{for } u \in C([a, b]).$$

The essential supremum, with respect to the Lebesgue measure, is a maximum because $u$ is continuous. The space ${(C([a, b]), \|\cdot\|_\infty)}$ is a Banach space.

In Example 5 with ${N = 2,}$ the balls ${B_1(0),}$ of radius 1 and centre ${0,}$ are drawn in Figure 2 for the norms ${|\cdot|,}$ $\|\cdot\|_\infty$ and ${\|\cdot\|_1,}$ and in Figure 3 for ${\|\cdot\|_p.}$ If ${v_1, v_2 > 0,}$ then ${\|(v_1, v_2)\|_1 = v_1 + v_2.}$ Hence the points of this quadrant with ${\|(v_1, v_2)\|_1 \le 1}$ are those with ${v_1 + v_2 \le 1.}$

B₁(0) for |·|

B₁(0) for ‖·‖_∞

B₁(0) for ‖·‖₁

Figure 2. The drawing of Example 5 with ${N = 2{:}}$ the balls $B_1(0)$ for the norms ${|\cdot|,}$ $\|\cdot\|_\infty$ and ${\|\cdot\|_1,}$ with the corners of the last two.

p

3.0

Figure 3. The drawing of Example 5 for $\|\cdot\|_p$ with ${N = 2{:}}$ the ball $B_1(0)$ for the $p$ of the slider and, dashed, the balls for $\|\cdot\|_1$ and $\|\cdot\|_\infty$ of Figure 2.

For the idea behind Example 5 (iii), which is not a proof, let ${(u_n) \subset C([a, b])}$ and ${u \colon [a, b] \to \mathbb{R}.}$ If ${u_n \to u}$ in ${C([a, b]),}$ this means ${\|u_n - u\|_\infty \to 0,}$ that is, ${\lim_{n \to \infty} \max_{[a, b]} |u_n - u| = 0.}$ That is, ${u_n \to u}$ uniformly, and therefore $u$ is continuous.

Convergent series#

Definition 6 (Convergent series). Let $(X, \|\cdot\|)$ be a normed space and let $(x_n)$ be a sequence of points of ${X.}$ For every $N$ let

$$S_N = \sum_{n=0}^{N} x_n.$$

If there exists ${S \in X}$ such that ${S_N \to S}$ in ${X,}$ that is, ${\|S_N - S\| \to 0,}$ we say that the series $\sum_n x_n$ converges to $S$ in ${X.}$

Proposition 7 (Criterion to distinguish Banach spaces). Let $(X, \|\cdot\|)$ be a normed space. Then $(X, \|\cdot\|)$ is a Banach space if and only if, for every sequence $(x_n)$ of points of $X$ such that ${\sum_n \|x_n\| < +\infty,}$ the series $\sum_n x_n$ converges in ${X.}$

The $L^p$ spaces (part 2)#

The spaces $L^p(X, \mathcal{M}, \mu)$ of Definitions 14 and 19 of the sixth post are now studied with the notions of Definitions 1 and 4.

The spaces $\ell^p$#

Example 8. Let ${p \in [1, +\infty].}$
(i) Examples of the spaces $L^p(X, \mathcal{M}, \mu)$ are ${L^p(\mathbb{R}^N),}$ $L^p((a, b))$ and so on, with the Lebesgue measure.

(ii) Let ${\ell^p := L^p(\mathbb{N}, \mathcal{P}(\mathbb{N}), \mu_\#),}$ where $\mu_\#$ is the counting measure on the subsets of ${\mathbb{N}.}$ The elements ${x \in \ell^p}$ are sequences, written ${x = (x^{(k)}),}$ where ${x^{(k)} \in \mathbb{R}}$ is the $k$-th value of the sequence ${x.}$ Indeed, the only set of counting measure zero is the empty set, so each class of $\ell^p$ contains exactly one function from $\mathbb{N}$ to ${\mathbb{R}.}$ For ${p < +\infty}$ the $\ell^p$ norm of ${x \in \ell^p}$ is

$$\|x\|_p = \Big( \sum_{k=0}^{\infty} |x^{(k)}|^p \Big)^{\frac{1}{p}}.$$

A sequence in $\ell^p$ is written ${(x_n) \subset \ell^p,}$ with ${x_n \in \ell^p}$ for every ${n,}$ and the $k$-th value of $x_n$ is ${x_n^{(k)} \in \mathbb{R}.}$

Conjugate exponents and the Hölder inequality#

Definition 9 (Conjugate exponent). Let ${p \in [1, +\infty].}$ The conjugate of $p$ is the exponent $p'$ such that:
(i) if ${p \in (1, +\infty),}$ then ${p' \in (1, +\infty)}$ and ${1/p + 1/p' = 1;}$
(ii) if ${p = 1,}$ then ${p' = +\infty;}$
(iii) if ${p = +\infty,}$ then ${p' = 1.}$

Remark 10. (i) The exponent $p'$ is the conjugate of $p$ if and only if $p$ is the conjugate of ${p'.}$
(ii) If ${p = 2,}$ then ${p' = 2.}$

Lemma 11 (Young inequality). Let ${1 < p < +\infty,}$ let $p'$ be its conjugate, and let ${a, b \in \mathbb{R}}$ with ${a, b > 0.}$ Then

$$ab \le \frac{a^p}{p} + \frac{b^{p'}}{p'}.$$

Proof (complete). The logarithm is concave, as in Figure 4, that is, for all ${\alpha, \beta > 0}$ and ${t \in (0, 1),}$

$$\log\big( t\alpha + (1 - t)\beta \big) \ge t \log \alpha + (1 - t) \log \beta.$$

Since ${1/p + 1/p' = 1,}$ concavity applies with ${t = 1/p}$ and ${1 - t = 1/p'.}$ With ${\alpha = a^p}$ and ${\beta = b^{p'}}$ it gives

$$\log\Big( \frac{a^p}{p} + \frac{b^{p'}}{p'} \Big) \ge \frac{1}{p} \log(a^p) + \frac{1}{p'} \log(b^{p'}) = \log a + \log b = \log(ab).$$

Applying the exponential function to both sides gives ${e^{\log(a^p/p + b^{p'}/p')} \ge e^{\log(ab)},}$ which is the claim. $\blacksquare$

a

0.70

b

1.25

p

3.00

Figure 4. The drawing of the proof of Lemma 11: the graph of $\log$ and its chord between the points with abscissas ${\alpha = a^p}$ and ${\beta = b^{p'},}$ for the ${a,}$ $b$ and $p$ of the sliders. Above the abscissa ${t\alpha + (1 - t)\beta}$ with ${t = 1/p,}$ the chord is at height $\log(ab)$ and the graph at height ${\log(a^p/p + b^{p'}/p').}$

Proposition 12 (Hölder inequality). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let ${p \in [1, +\infty]}$ with conjugate ${p',}$ and let ${u, v \colon X \to \mathbb{R}}$ be measurable. Then

$$\int_X |uv| \, d\mu \le \|u\|_p \, \|v\|_{p'},$$

where $\|u\|_p$ and $\|v\|_{p'}$ are given by the formulas of Definitions 17 and 19 of the sixth post and may equal ${+\infty,}$ and the product follows Definition 5 of the first post.

Proof (complete). Without loss of generality ${\|u\|_p > 0}$ and ${\|v\|_{p'} > 0.}$ Otherwise one of the two functions is $0$ almost everywhere, so that ${|uv| = 0}$ almost everywhere, and the inequality reads ${0 \le 0.}$

Without loss of generality ${\|u\|_p < +\infty}$ and ${\|v\|_{p'} < +\infty.}$ Otherwise the right-hand side is ${+\infty,}$ and the inequality is true.

Without loss of generality ${\|u\|_p = \|v\|_{p'} = 1.}$ To reduce to this case, for $u$ and $v$ as above let

$$\tilde{u} = \frac{u}{\|u\|_p}, \qquad \tilde{v} = \frac{v}{\|v\|_{p'}}.$$

Then ${\|\tilde{u}\|_p = \|\tilde{v}\|_{p'} = 1,}$ and the inequality for functions with norms equal to $1$ gives

$$\begin{aligned} \int_X |\tilde{u} \tilde{v}| \, d\mu &\le \|\tilde{u}\|_p \, \|\tilde{v}\|_{p'} = 1, \\ \int_X |\tilde{u} \tilde{v}| \, d\mu &= \frac{1}{\|u\|_p \, \|v\|_{p'}} \int_X |uv| \, d\mu. \end{aligned}$$

From now on ${\|u\|_p = \|v\|_{p'} = 1.}$ If ${1 < p < +\infty,}$ then

$$\begin{aligned} \int_X |uv| \, d\mu &\le \int_X \Big( \frac{|u|^p}{p} + \frac{|v|^{p'}}{p'} \Big) \, d\mu = \frac{1}{p} \|u\|_p^p + \frac{1}{p'} \|v\|_{p'}^{p'} \\ &= \frac{1}{p} + \frac{1}{p'} = 1 = \|u\|_p \, \|v\|_{p'}, \end{aligned}$$

where the inequality is the Young inequality, Lemma 11, at every point of ${X.}$ Indeed, at a point $x$ where ${u(x) = 0}$ or ${v(x) = 0}$ the pointwise inequality holds because its left-hand side is ${0.}$ If ${p = 1,}$ then ${p' = +\infty}$ and

$$\int_X |uv| \, d\mu = \int_X |u| \, |v| \, d\mu \le \|v\|_\infty \int_X |u| \, d\mu = \|v\|_\infty \, \|u\|_1,$$

since ${|v| \le \|v\|_\infty}$ almost everywhere. Indeed, for ${p = +\infty}$ we have ${p' = 1,}$ and the case ${p = 1}$ applies with $u$ and $v$ exchanged. $\blacksquare$

Inclusions and interpolation#

Corollary 13. Let ${(X, \mathcal{M}, \mu)}$ be a measure space such that ${\mu(X) < +\infty,}$ and let ${1 \le p \le q \le +\infty.}$ Then ${L^q \subset L^p.}$

Proof (complete). Let ${f \in L^q.}$ We apply the Hölder inequality, Proposition 12, to ${u = |f|^p}$ and ${v = 1,}$ with the exponent ${q/p \ge 1}$ and its conjugate ${r.}$ Here ${1/r = 1 - 1/(q/p) = 1 - p/q.}$ This gives

$$\begin{aligned} \int_X |f|^p \, d\mu = \int_X |f|^p \cdot 1 \, d\mu &\le \Big( \int_X |f|^{p \frac{q}{p}} \, d\mu \Big)^{\frac{p}{q}} \Big( \int_X 1^r \, d\mu \Big)^{\frac{1}{r}} \\ &= \|f\|_q^p \, \mu(X)^{\frac{1}{r}} < +\infty, \end{aligned}$$

since ${\|f\|_q < +\infty}$ and ${\mu(X) < +\infty.}$ In detail, for ${p = q}$ there is nothing to prove, and for ${q = +\infty}$ Proposition 12 gives in the same way ${\int_X |f|^p \, d\mu \le \|f\|_\infty^p \, \mu(X).}$ $\blacksquare$

Remark 14. For all ${1 \le p \le q \le +\infty,}$ ${\ell^1 \subset \ell^p \subset \ell^q \subset \ell^\infty.}$

Corollary 15 (Interpolation). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let ${1 \le p \le r \le q \le +\infty,}$ and let ${u \colon X \to \mathbb{R}}$ be measurable. Then

$$\|u\|_r \le \|u\|_p^{\alpha} \, \|u\|_q^{1 - \alpha} \qquad \text{with} \qquad \frac{1}{r} = \frac{\alpha}{p} + \frac{1 - \alpha}{q},$$

where ${\alpha \in [0, 1],}$ ${1/(+\infty) := 0}$ and the norms are as in Proposition 12. In particular, ${L^p \cap L^q \subset L^r.}$

The Minkowski inequality and the Riesz–Fischer theorem#

Proposition 16 (Minkowski inequality). Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let ${1 \le p \le +\infty}$ and let ${u, v \in L^p.}$ Then

$$\|u + v\|_p \le \|u\|_p + \|v\|_p.$$

In particular, ${(L^p(X), \|\cdot\|_p)}$ is a normed space.

Proof (sketch). The case ${p = +\infty}$ is not proved. If ${p = 1,}$ then

$$\int_X |u + v| \, d\mu \le \int_X (|u| + |v|) \, d\mu = \|u\|_1 + \|v\|_1.$$

Let ${1 < p < +\infty.}$ Then

$$\begin{aligned} \|u + v\|_p^p &= \int_X |u + v|^p \, d\mu = \int_X |u + v| \, |u + v|^{p-1} \, d\mu \\ &\le \int_X |u| \, |u + v|^{p-1} \, d\mu + \int_X |v| \, |u + v|^{p-1} \, d\mu, \end{aligned}$$

since ${|u + v| \le |u| + |v|.}$ The conjugate of $p$ satisfies ${1/p' = 1 - 1/p = (p - 1)/p,}$ that is, ${p' = p/(p - 1),}$ and the Hölder inequality, Proposition 12, applied to each of the last two integrals gives

$$\begin{aligned} &\int_X |u| \, |u + v|^{p-1} \, d\mu + \int_X |v| \, |u + v|^{p-1} \, d\mu \\ &\quad \le \|u\|_p \Big( \int_X \big( |u + v|^{p-1} \big)^{p'} \, d\mu \Big)^{\frac{1}{p'}} + \|v\|_p \Big( \int_X \big( |u + v|^{p-1} \big)^{p'} \, d\mu \Big)^{\frac{1}{p'}} \\ &\quad = \big( \|u\|_p + \|v\|_p \big) \Big( \int_X |u + v|^p \, d\mu \Big)^{\frac{p-1}{p}} \\ &\quad = \big( \|u\|_p + \|v\|_p \big) \, \|u + v\|_p^{p-1}, \end{aligned}$$

where ${(p - 1) p' = p.}$ All in all,

$$\|u + v\|_p^p \le \big( \|u\|_p + \|v\|_p \big) \, \|u + v\|_p^{p-1},$$

which is equivalent to ${\|u + v\|_p \le \|u\|_p + \|v\|_p.}$ Indeed, ${\|u + v\|_p < +\infty}$ because ${u + v \in L^p}$ by Remark 18 (i) of the sixth post, and for ${\|u + v\|_p > 0}$ one divides by ${\|u + v\|_p^{p-1}.}$ $\blacksquare$

Theorem 17 (Riesz–Fischer). Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let ${1 \le p \le +\infty.}$ Then $L^p(X, \mathcal{M}, \mu)$ is a Banach space.

Proof (sketch). Only the case ${1 \le p < +\infty}$ is treated, with the criterion of Proposition 7. Let ${(f_n) \subset L^p}$ be such that ${\sum_n \|f_n\|_p < +\infty,}$ and let

$$S_N = \sum_{n=0}^{N} f_n.$$

The goal is to show that $(S_N)$ converges in ${L^p.}$ Define

$$g_N(x) = \sum_{n=0}^{N} |f_n(x)|,$$

so that ${g_N \ge 0}$ and ${g_N \in L^p.}$ Since ${g_{N+1} \ge g_N}$ almost everywhere, we can define ${g(x) = \lim_{N \to \infty} g_N(x)}$ almost everywhere. For every ${N,}$ by the Minkowski inequality, Proposition 16,

$$\|g_N\|_p = \Bigl\| \sum_{n=0}^{N} |f_n| \Bigr\|_p \le \sum_{n=0}^{N} \bigl\|\, |f_n| \,\bigr\|_p = \sum_{n=0}^{N} \|f_n\|_p \le \sum_{n=0}^{\infty} \|f_n\|_p < +\infty.$$

By the monotone convergence theorem, Theorem 4 of the sixth post, and by this bound,

$$\begin{aligned} \int_X |g|^p \, d\mu &= \lim_{N \to +\infty} \int_X |g_N|^p \, d\mu = \lim_{N \to +\infty} \|g_N\|_p^p \\ &\le \lim_{N \to +\infty} \Big( \sum_{n=0}^{\infty} \|f_n\|_p \Big)^p < +\infty, \end{aligned}$$

hence ${g \in L^p.}$ Now, at almost every ${x \in X,}$

$$|S_N(x)| = \Big| \sum_{n=0}^{N} f_n(x) \Big| \le \sum_{n=0}^{N} |f_n(x)| = g_N(x) \le g(x),$$

with ${g \in L^p.}$ Moreover, at almost every ${x,}$

$$g(x) = \lim_{N \to \infty} g_N(x) = \sum_{n=0}^{\infty} |f_n(x)| < +\infty,$$

because ${g \in L^p,}$ so that the series of the $f_n(x)$ is an absolutely converging series. Hence at almost every $x$ the series of the $f_n(x)$ converges, that is, the limit ${\lim_{N \to \infty} S_N(x)}$ exists. Thus $(S_N)$ converges almost everywhere, and we call ${S(x) := \lim_{N \to \infty} S_N(x)}$ at almost every ${x.}$

By the dominated convergence theorem in ${L^p,}$ Theorem 8 of the seventh post, ${S_N \to S}$ in ${L^p.}$ $\blacksquare$

Remark 18. Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let ${p \in [1, +\infty],}$ let ${(f_n) \subset L^p}$ and let ${f \colon X \to \mathbb{R}}$ be measurable. If ${f_n \to f}$ in ${L^p,}$ that is, ${\|f_n - f\|_p \to 0,}$ then ${f \in L^p.}$

Proof (complete). The sequence $(f_n)$ is Cauchy in ${L^p.}$ For all $n$ and ${m,}$

$$\|f_n - f_m\|_p \le \|f_n - f\|_p + \|f - f_m\|_p,$$

where the first term on the right tends to $0$ as ${n \to \infty}$ and the second as ${m \to \infty.}$ Also, for every ${\varepsilon > 0}$ there exists $N$ such that ${\|f_n - f\|_p < \varepsilon/2}$ for every ${n \ge N.}$ Hence, if ${n, m \ge N,}$ then ${\|f_n - f\|_p + \|f - f_m\|_p \le \varepsilon.}$

By the Riesz–Fischer theorem, Theorem 17, there exists ${g \in L^p}$ such that ${f_n \to g}$ in ${L^p.}$ By the uniqueness of limits, Remark 12 of the seventh post, ${g = f}$ almost everywhere, hence ${f \in L^p.}$ $\blacksquare$

The spaces $W^{1,p}((a, b))$#

Example 19. Let ${-\infty < a < b < +\infty}$ and consider the space $W^{1,p}((a, b))$ with ${p \in [1, +\infty].}$ For ${f \in W^{1,p}((a, b))}$ let

$$\|f\|_{W^{1,p}} := \|f\|_p + \|w\|_p,$$

where $w$ is the weak derivative of ${f.}$ Then ${(W^{1,p}((a, b)), \|\cdot\|_{W^{1,p}})}$ is a Banach space.

Remark 20. Since ${\lambda((a, b)) < +\infty,}$ Corollary 13 gives ${L^p((a, b)) \subset L^1((a, b))}$ for every ${p \ge 1.}$ Hence

$$W^{1,p}((a, b)) \subset W^{1,1}((a, b)) = \mathrm{AC}([a, b]),$$

where $\mathrm{AC}([a, b])$ is the set of the absolutely continuous functions on ${[a, b].}$ That is, every function of $W^{1,p}((a, b))$ has an absolutely continuous representative.

Example 21. Let ${a,}$ $b$ and $p$ be as in Example 19. The set

$$W_0^{1,p}((a, b)) = \{ f \in W^{1,p}((a, b)) : f(a) = f(b) = 0 \}$$

is a subspace of ${W^{1,p}((a, b)).}$ In detail, $f(a)$ and $f(b)$ are the values at $a$ and $b$ of the absolutely continuous representative of $f$ of Remark 20.
(i) The space ${(W_0^{1,p}((a, b)), \|\cdot\|_{W^{1,p}})}$ is a Banach space.

(ii) The space ${(W_0^{1,p}((a, b)), \|\cdot\|_{W_0^{1,p}})}$ is a Banach space, where for ${f \in W_0^{1,p}((a, b))}$

$$\|f\|_{W_0^{1,p}} := \|w\|_p,$$

with $w$ the weak derivative of ${f.}$ In fact, ${w = f'}$ almost everywhere, since ${f \in \mathrm{AC}([a, b]).}$