Incubator escapee wiki

Hölder's inequality

Article Discussion
(Redirected from Holder inequality)

Template:Short description In mathematical analysis, Hölder's inequality, named after Otto Hölder, is a fundamental inequality between integrals and an indispensable tool for the study of [[Lp space|Template:Math spaces]].

Template:Math theorem

The numbers Template:Mvar and Template:Mvar above are said to be Hölder conjugates of each other. The special case Template:Math gives a form of the Cauchy–Schwarz inequality.<ref>Template:Harvnb</ref> Hölder's inequality holds even if Template:Math is infinite, the right-hand side also being infinite in that case. Conversely, if Template:Mvar is in Template:Math and Template:Mvar is in Template:Math, then the pointwise product Template:Math is in Template:Math.

Hölder's inequality is used to prove the Minkowski inequality, which is the triangle inequality in the space Template:Math, and also to establish that Template:Math is the dual space of Template:Math for Template:Math Template:Closed-open.

Hölder's inequality (in a slightly different form) was first found by Template:Harvs. Inspired by Rogers' work, Template:Harvtxt gave another proof as part of a work developing the concept of convex and concave functions and introducing Jensen's inequality,<ref>Template:Citation</ref> which was in turn named for work of Johan Jensen building on Hölder's work.<ref>Template:Citation</ref>

RemarksEdit

ConventionsEdit

The brief statement of Hölder's inequality uses some conventions.

<math>\begin{align}

&\left(\int_S |f|^p\,\mathrm{d}\mu\right)^{\frac{1}{p}} \\ &\left(\int_S |g|^q\,\mathrm{d}\mu\right)^{\frac{1}{q}} \end{align}</math>

Estimates for integrable productsEdit

As above, let Template:Mvar and Template:Mvar denote measurable real- or complex-valued functions defined on Template:Mvar. If Template:Math is finite, then the pointwise products of Template:Mvar with Template:Mvar and its complex conjugate function are Template:Mvar-integrable, the estimate

<math>\biggl|\int_S f\bar g\,\mathrm{d}\mu\biggr|\le\int_S|fg|\,\mathrm{d}\mu =\|fg\|_1</math>

and the similar one for Template:Math hold, and Hölder's inequality can be applied to the right-hand side. In particular, if Template:Mvar and Template:Mvar are in the Hilbert space Template:Math, then Hölder's inequality for Template:Math implies

<math>|\langle f,g\rangle| \le \|f\|_2 \|g\|_2,</math>

where the angle brackets refer to the inner product of Template:Math. This is also called Cauchy–Schwarz inequality, but requires for its statement that Template:Math and Template:Math are finite to make sure that the inner product of Template:Mvar and Template:Mvar is well defined. We may recover the original inequality (for the case Template:Math) by using the functions Template:Math and Template:Math in place of Template:Mvar and Template:Mvar.

Generalization for probability measuresEdit

If Template:Math is a probability space, then Template:Math Template:Closed-closed just need to satisfy Template:Math, rather than being Hölder conjugates. A combination of Hölder's inequality and Jensen's inequality implies that

<math>\|fg\|_1 \le \|f\|_p \|g\|_q</math>

for all measurable real- or complex-valued functions Template:Mvar and Template:Mvar on Template:Mvar.

Notable special casesEdit

For the following cases assume that Template:Mvar and Template:Mvar are in the open interval Template:Open-open with Template:Math.

Counting measureEdit

For the <math>n</math>-dimensional Euclidean space, when the set <math>S</math> is <math>\{1,\dots,n\}</math> with the counting measure, we have

<math>\sum_{k=1}^n |x_k\,y_k| \le \biggl( \sum_{k=1}^n |x_k|^p \biggr)^{\frac{1}{p}} \biggl( \sum_{k=1}^n |y_k|^q \biggr)^{\frac{1}{q}}

\text{ for all }(x_1,\ldots,x_n),(y_1,\ldots,y_n)\in\mathbb{R}^n\text{ or }\mathbb{C}^n.</math> Often the following practical form of this is used, for any <math>(r,s)\in\mathbb{R}_+</math>:

<math>\biggl(\sum_{k=1}^n |x_k|^r\,|y_k|^s \biggr)^{r+s}\le \biggl( \sum_{k=1}^n |x_k|^{r+s} \biggr)^{r} \biggl( \sum_{k=1}^n |y_k|^{r+s} \biggr)^{s}.</math>

For more than two sums, the following generalisation (Template:Harvtxt, Template:Harvtxt) holds, with real positive exponents <math> \lambda_i </math> and <math> \lambda_a + \lambda_b+ \cdots +\lambda_z =1</math>:

<math>\sum_{k=1}^n |a_k|^{\lambda_a}\,|b_k|^{\lambda_b} \cdots |z_k|^{\lambda_z} \le \biggl(\sum_{k=1}^n |a_k|\biggr)^{\lambda_a} \biggl(\sum_{k=1}^n |b_k|\biggr)^{\lambda_b} \cdots \biggl(\sum_{k=1}^n |z_k|\biggr)^{\lambda_z} .

</math> Equality holds iff <math> |a_1|: |a_2|: \cdots : |a_n| =|b_1|: |b_2|: \cdots : |b_n| = \cdots = |z_1|: |z_2|: \cdots : |z_n| </math>.

If <math>S=\N</math> with the counting measure, then we get Hölder's inequality for sequence spaces:

<math>\sum_{k=1}^{\infty} |x_k\,y_k| \le \biggl( \sum_{k=1}^{\infty} |x_k|^p \biggr)^{\frac{1}{p}} \left( \sum_{k=1}^{\infty} |y_k|^q \right)^{\frac{1}{q}}

\text{ for all }(x_k)_{k\in\mathbb N}, (y_k)_{k\in\mathbb N}\in\mathbb{R}^{\mathbb N}\text{ or }\mathbb{C}^{\mathbb N}.</math>

Lebesgue measureEdit

If <math>S</math> is a measurable subset of <math>\R^n</math> with the Lebesgue measure, and <math>f</math> and <math>g</math> are measurable real- or complex-valued functions on <math>S</math>, then Hölder's inequality is

<math>\int_S \bigl| f(x)g(x)\bigr| \,\mathrm{d}x \le\biggl(\int_S |f(x)|^p\,\mathrm{d}x\biggr)^{\frac{1}{p}} \biggl(\int_S |g(x)|^q\,\mathrm{d}x\biggr)^{\frac{1}{q}}.</math>

Probability measureEdit

For the probability space <math>(\Omega, \mathcal{F}, \mathbb{P}),</math> let <math>\mathbb{E}</math> denote the expectation operator. For real- or complex-valued random variables <math>X</math> and <math>Y</math> on <math>\Omega,</math> Hölder's inequality reads

<math>\mathbb{E}[|XY|] \leqslant \left (\mathbb{E}\bigl[ |X|^p\bigr]\right)^{\frac{1}{p}} \left(\mathbb{E}\bigl[|Y|^q\bigr]\right)^{\frac{1}{q}}.</math>

Let <math>1 < r < s < \infty</math> and define <math>p = \tfrac{s}{r}.</math> Then <math>q = \tfrac{p}{p-1}</math> is the Hölder conjugate of <math>p.</math> Applying Hölder's inequality to the random variables <math>|X|^r</math> and <math>1_{\Omega}</math> we obtain

<math>\mathbb{E}\bigl[|X|^r\bigr]\leqslant \left(\mathbb{E}\bigl[|X|^s\bigr]\right)^{\frac{r}{s}}.</math>

In particular, if the Template:Mvarth absolute moment is finite, then the Template:Mvar th absolute moment is finite, too. (This also follows from Jensen's inequality.)

Product measureEdit

For two σ-finite measure spaces Template:Math and Template:Math define the product measure space by

<math>S=S_1\times S_2,\quad \Sigma=\Sigma_1\otimes\Sigma_2,\quad \mu=\mu_1\otimes\mu_2,</math>

where Template:Mvar is the Cartesian product of Template:Math and Template:Math, the Template:Nowrap arises as product σ-algebra of Template:Math and Template:Math, and Template:Mvar denotes the product measure of Template:Math and Template:Math. Then Tonelli's theorem allows us to rewrite Hölder's inequality using iterated integrals: If Template:Mvar and Template:Mvar are Template:Nowrap real- or complex-valued functions on the Cartesian product Template:Mvar, then

<math>\int_{S_1}\int_{S_2}|f(x,y)\,g(x,y)|\,\mu_2(\mathrm{d}y)\,\mu_1(\mathrm{d}x) \le\left(\int_{S_1}\int_{S_2}|f(x,y)|^p\,\mu_2(\mathrm{d}y)\,\mu_1(\mathrm{d}x)\right)^{\frac{1}{p}}\left(\int_{S_1}\int_{S_2}|g(x,y)|^q\,\mu_2(\mathrm{d}y)\,\mu_1(\mathrm{d}x)\right)^{\frac{1}{q}}.</math>

This can be generalized to more than two Template:Nowrap measure spaces.

Vector-valued functionsEdit

Let Template:Math denote a Template:Nowrap measure space and suppose that Template:Math and Template:Math are Template:Math-measurable functions on Template:Mvar, taking values in the Template:Mvar-dimensional real- or complex Euclidean space. By taking the product with the counting measure on Template:Math, we can rewrite the above product measure version of Hölder's inequality in the form

<math> \int_S \sum_{k=1}^n|f_k(x)\,g_k(x)|\,\mu(\mathrm{d}x) \le \left(\int_S\sum_{k=1}^n|f_k(x)|^p\,\mu(\mathrm{d}x)\right)^{\frac{1}{p}}\left(\int_S\sum_{k=1}^n|g_k(x)|^q\,\mu(\mathrm{d}x)\right)^{\frac{1}{q}}.</math>

If the two integrals on the right-hand side are finite, then equality holds if and only if there exist real numbers Template:Math, not both of them zero, such that

<math>\alpha \left (|f_1(x)|^p,\ldots,|f_n(x)|^p \right )= \beta \left (|g_1(x)|^q,\ldots,|g_n(x)|^q \right ),</math>

for Template:Mvar-almost all Template:Mvar in Template:Mvar.

This finite-dimensional version generalizes to functions Template:Mvar and Template:Mvar taking values in a normed space which could be for example a sequence space or an inner product space.

Proof of Hölder's inequalityEdit

There are several proofs of Hölder's inequality; the main idea in the following is Young's inequality for products.

Template:Math proof

Alternative proof using Jensen's inequality:

Template:Math proof</math>

where Template:Mvar is any probability distribution and Template:Mvar any Template:Mvar-measurable function. Let Template:Mvar be any measure, and Template:Mvar the distribution whose density w.r.t. Template:Mvar is proportional to <math>g^q</math>, i.e.

<math>\mathrm{d}\nu = \frac{g^q}{\int g^q\,\mathrm{d}\mu}\mathrm{d}\mu</math>

Hence we have, using <math>\frac{1}{p}+\frac{1}{q}=1</math>, hence <math>p(1-q)+q=0</math>, and letting <math>h=fg^{1-q}</math>,

<math> \begin{align}\int fg\,\mathrm{d}\mu = & \left (\int g^q\,\mathrm{d}\mu \right )\int \underbrace{fg^{1-q}}_h\underbrace{\frac{g^q}{\int g^q\,\mathrm{d}\mu}\mathrm{d}\mu}_{\mathrm{d}\nu}\\
\leq & \left (\int g^q\mathrm{d}\mu \right ) \left (\int \underbrace{f^pg^{p(1-q)}}_{h^p}\underbrace{\frac{g^q}{\int g^q\,\mathrm{d}\mu}\,\mathrm{d}\mu}_{\mathrm{d}\nu} \right )^{\frac{1}{p}}\\
= & \left (\int g^q\,\mathrm{d}\mu \right ) \left (\int \frac{f^p}{\int g^q\,\mathrm{d}\mu}\,\mathrm{d}\mu \right )^{\frac{1}{p}} .

\end{align}</math>

Finally, we get

<math>\int fg\,\mathrm{d}\mu \leq \left(\int f^p\,\mathrm{d}\mu \right )^{\frac{1}{p}} \left(\int g^q\,\mathrm{d}\mu \right )^{\frac{1}{q}}</math>

This assumes that Template:Math are real and non-negative, but the extension to complex functions is straightforward (use the modulus of Template:Math). It also assumes that <math>\|f\|_p,\|g\|_q</math> are neither null nor infinity, and that <math>p,q > 1</math>: all these assumptions can also be lifted as in the proof above. }}

We could also bypass use of both Young's and Jensen's inequalities. The proof below also explains why and where the Hölder exponent comes in naturally.

Template:Math proof</math> where <math> \nu(X)=1 </math> and <math>h</math> is <math>\nu</math>-measurable (real or complex) function on <math> X </math>. To prove this, we must bound <math>|h| </math> by <math> |h|^p </math>. There is no constant <math> C </math> that will make <math> |h(x)| ~\leq~ C|h(x)|^p </math> for all <math> x > 0 </math>. Hence, we seek an inequality of the form

<math> |h(x)| ~\leq~ a'|h(x)|^p + b', \quad \text{for all} \quad x>0 </math>

for suitable choices of <math> a' </math> and <math> b' </math>.

We wish to obtain <math> A:=\|f\|_p </math> on the right-hand side after integrating this inequality. By trial and error, we see that the inequality we wish should have the form

<math> |h(x)| ~\leq~ aA^{1-p}|h(x)|^p + bA, \quad \text{for all} \quad x>0, </math>

where <math> a, b </math> are non-negative and <math> a+b=1 </math>. Indeed, the integral of the right-hand side is precisely <math> A </math>. So, it remains to prove that such an inequality does hold with the right choice of <math>a,b.</math>

The inequality we seek would follow from:

<math> \tfrac{y}{A} ~\leq~ a(\tfrac{y}{a})^p + b, \quad \text{for all} \quad y>0, </math>

which, in turn, is equivalent to

<math> (*) \quad z ~\leq~ az^p + b, \quad \text{for all} \quad z>0. </math>

It turns out there is one and only one choice of <math> a, b </math>, subject to <math> a+b=1 </math>, that makes this true: <math> a=\tfrac{1}{p}</math> and, necessarily, <math> b=1-\tfrac{1}{p}</math>. (This is where Hölder conjugate exponent is born!) This completes the proof of the inequality at the first paragraph of this proof. Proof of Hölder's inequality follows from this as in the previous proof. Alternatively, we can deduce Young's inequality and then resort to the first proof given above. Young's inequality follows from the inequality (*) above by choosing <math> z=\tfrac{a}{b^{q-1}}</math> and multiplying both sides by <math> b^{q} </math>. }}

Extremal equalityEdit

StatementEdit

Assume that Template:Math and let Template:Mvar denote the Hölder conjugate. Then for every Template:Math,

<math>\|f\|_p = \max \left \{ \left| \int_S f g \, \mathrm{d}\mu \right | : g\in L^q(\mu), \|g\|_q \le 1 \right\},</math>

where max indicates that there actually is a Template:Mvar maximizing the right-hand side. When Template:Math and if each set Template:Mvar in the Template:Nowrap Template:Math with Template:Math contains a subset Template:Math with Template:Math (which is true in particular when Template:Mvar is Template:Nowrap), then

<math>\|f\|_\infty = \sup \left\{ \left| \int_S f g \,\mathrm{d}\mu \right| : g\in L^1(\mu), \|g\|_1 \le 1 \right \}.</math>

Proof of the extremal equality:

Template:Math proof

Remarks and examplesEdit

  • The equality for <math>p = \infty</math> fails whenever there exists a set <math>A</math> of infinite measure in the <math>\sigma</math>-field <math>\Sigma</math> with that has no subset <math>B \in \Sigma</math> that satisfies: <math>0 < \mu(B) < \infty.</math> (the simplest example is the <math>\sigma</math>-field <math>\Sigma</math> containing just the empty set and <math>S,</math> and the measure <math>\mu</math> with <math>\mu(S) = \infty.</math>) Then the indicator function <math>1_A</math> satisfies <math>\|1_A\|_{\infty} = 1,</math> but every <math>g \in L^1 (\mu)</math> has to be <math>\mu</math>-almost everywhere constant on <math>A,</math> because it is <math>\Sigma</math>-measurable, and this constant has to be zero, because <math>g</math> is <math>\mu</math>-integrable. Therefore, the above supremum for the indicator function <math>1_A</math> is zero and the extremal equality fails.
  • For <math>p = \infty,</math> the supremum is in general not attained. As an example, let <math>S = \mathbb{N}, \Sigma = \mathcal{P}(\mathbb{N})</math> and <math>\mu</math> the counting measure. Define:
<math>\begin{cases} f: \mathbb{N} \to \mathbb{R} \\ f(n) = \frac{n-1}{n} \end{cases}</math>
Then <math>\|f\|_{\infty} = 1.</math> For <math>g \in L^1 (\mu, \mathbb{N})</math> with <math>0 < \|g\|_1 \leqslant 1,</math> let <math>m</math> denote the smallest natural number with <math>g(m) \neq 0.</math> Then
<math>\left |\int_S fg\,\mathrm{d}\mu\right| \leqslant \frac{m-1}{m}|g(m)|+\sum_{n=m+1}^\infty|g(n)| = \|g\|_1-\frac{|g(m)|}m<1.</math>

ApplicationsEdit

<math>\kappa_f(g) = \int_S f g \, \mathrm{d}\mu,\qquad g\in L^q(\mu).</math>
The extremal equality (when true) shows that the norm of this functional Template:Math as element of the continuous dual space Template:Math coincides with the norm of Template:Mvar in Template:Math (see also the Template:Nowrap article).

Generalization with more than two functionsEdit

StatementEdit

Assume that Template:Math Template:Open-closed and Template:Math Template:Open-closed such that

<math>\sum_{k=1}^n \frac1{p_k} = \frac1r</math>

where 1/∞ is interpreted as 0 in this equation, and r=∞ implies Template:Math Template:Open-closed are all equal to ∞. Then, for all measurable real or complex-valued functions Template:Math defined on Template:Mvar,

<math>\left\|\prod_{k=1}^n f_k\right\|_r \le \prod_{k=1}^n \left\|f_k\right\|_{p_k}</math>

where we interpret any product with a factor of ∞ as ∞ if all factors are positive, but the product is 0 if any factor is 0.

In particular, if <math>f_k \in L^{p_k}(\mu)</math> for all <math>k \in \{ 1, \ldots, n \}</math> then <math>\prod_{k=1}^n f_k \in L^r(\mu).</math>

Note: For <math>r \in (0, 1),</math> contrary to the notation, Template:Math is in general not a norm because it doesn't satisfy the triangle inequality.

Proof of the generalization: Template:Math proof \left\|f_n\right\|_{\infty}. \end{align}</math>

Case 2: If <math>p_n < \infty</math> then necessarily <math>r < \infty</math> as well, and then

<math>p := \frac{p_n}{p_n-r}, \qquad q := \frac{p_n}r</math>

are Hölder conjugates in Template:Open-open. Application of Hölder's inequality gives

<math>\left \||f_1 \cdots f_{n-1}|^r\,|f_n|^r\right \|_1 \le \left \||f_1 \cdots f_{n-1}|^r\right\|_p\,\left \||f_n|^r\right \|_q.</math>

Raising to the power <math>1/r</math> and rewriting,

<math>\|f_1 \cdots f_n\|_r \le \|f_1 \cdots f_{n-1}\|_{pr} \|f_n\|_{qr}.</math>

Since <math>q r = p_n</math> and

<math>\sum_{k=1}^{n-1} \frac1{p_k} = \frac1r-\frac1{p_n} = \frac{p_n-r}{rp_n} = \frac1{pr},</math>

the claimed inequality now follows by using the induction hypothesis. }}

InterpolationEdit

Let Template:Math Template:Open-closed and let Template:Math denote weights with Template:Math. Define <math>p</math> as the weighted harmonic mean, that is,

<math> \frac1p = \sum_{k=1}^n \frac{\theta_k}{p_k}.</math>

Given measurable real- or complex-valued functions <math>f_k</math> on Template:Mvar, then the above generalization of Hölder's inequality gives

<math>\left\| |f_1|^{\theta_1}\cdots |f_n|^{\theta_n}\right\|_p \le \left\||f_1|^{\theta_1}\right\|_{\frac{p_1}{\theta_1}}\cdots \left\| |f_n|^{\theta_n}\right\|_{\frac{p_n}{\theta_n}} = \|f_1\|_{p_1}^{\theta_1}\cdots \|f_n\|_{p_n}^{\theta_n}.</math>

In particular, taking <math>f_1 = \cdots = f_n=:f</math> gives

<math>\|f\|_p \leqslant \prod_{k=1}^n \|f\|_{p_k}^{\theta_k}.</math>

Specifying further Template:Math and Template:Math, in the case <math>n = 2,</math> we obtain the interpolation result

Template:Math theorem

An application of Hölder gives

Template:Math theorem \cdot |f_1|^{\frac{p_1 \theta}{p}}\right\|_p^p \le \|f_0\|_{p_0}^{p_0(1-\theta)} \|f_1\|_{p_1}^{p_1\theta}</math>

and in particular

<math display="block">\|f\|_p^p \leqslant \|f\|_{p_0}^{p_0(1-\theta)} \cdot \|f\|_{p_1}^{p_1\theta}.</math> }}

Both Littlewood and Lyapunov imply that if <math>f \in L^{p_0}\cap L^{p_1}</math> then <math>f \in L^p</math> for all <math>p_0 < p < p_1.</math><ref>Template:Cite book</ref>

Reverse Hölder inequalitiesEdit

Two functionsEdit

Assume that Template:Math and that the measure space Template:Math satisfies Template:Math. Then for all measurable real- or complex-valued functions Template:Mvar and Template:Mvar on Template:Mvar such that Template:Math for Template:Nowrap all Template:Math,

<math>\|fg\|_1\geqslant \|f\|_{\frac{1}{p}}\,\|g\|_{\frac{-1}{p-1}}.</math>

If

<math>\|fg\|_1 < \infty \quad \text{and} \quad \|g\|_{\frac{-1}{p-1}} > 0, </math>

then the reverse Hölder inequality is an equality if and only if

<math>\exists \alpha \geqslant 0 \quad |f| = \alpha|g|^{\frac{-p}{p-1}} \qquad \mu\text{-almost everywhere}.</math>

Note: The expressions:

<math> \|f\|_{\frac{1}{p}}</math> and <math>\|g\|_{\frac{-1}{p-1}},</math>

are not norms, they are just compact notations for

<math>\left (\int_S|f|^{\frac{1}{p}}\,\mathrm{d}\mu\right)^{p} \quad \text{and} \quad \left (\int_S|g|^{\frac{-1}{p-1}}\,\mathrm{d}\mu\right)^{-(p-1)}.</math>

Template:Hidden\right \|_1 &= \left \||fg|^{\frac{1}{p}}\,|g|^{-\frac{1}{p}}\right \|_1\\ &\leqslant \left \| |fg|^{\frac{1}{p}} \right \|_p \left \| |g|^{-\frac{1}{p}}\right \|_q \\ &=\|fg\|_1^{\frac{1}{p}}\left \||g|^{\frac{-1}{p-1}}\right \|_1^{\frac{p-1}{p}} \end{align}</math>

Raising to the power Template:Mvar gives us:

<math>\left \||f|^{\frac{1}{p}}\right \|_1^p \leqslant \|fg\|_1 \left \||g|^{\frac{-1}{p-1}}\right \|_1^{p-1}.</math>

Therefore:

<math>\left \||f|^{\frac{1}{p}}\right \|_1^p \left \||g|^{\frac{-1}{p-1}}\right \|_1^{-(p-1)} \leqslant \|fg\|_1 .</math>

Now we just need to recall our notation.

Since Template:Mvar is not almost everywhere equal to the zero function, we can have equality if and only if there exists a constant Template:Math such that Template:Math almost everywhere. Solving for the absolute value of Template:Mvar gives the claim. }}

Multiple functionsEdit

The Reverse Hölder inequality (above) can be generalized to the case of multiple functions if all but one conjugate is negative. That is,

Let <math>p_1,..., p_{m-1} < 0</math> and <math>p_m \in \mathbb{R}</math> be such that <math>\sum_{k=1}^{m} \frac{1}{p_k} = 1</math> (hence <math>0 < p_m < 1</math>). Let <math>f_k</math> be measurable functions for <math>k = 1,...,m</math>. Then
<math>\left\|\prod_{k=1}^m f_k\right\|_1 \ge \prod_{k=1}^m \left\|f_k\right\|_{p_k}.</math>

This follows from the symmetric form of the Hölder inequality (see below).

Symmetric forms of Hölder inequalityEdit

It was observed by Aczél and Beckenbach<ref>Template:Cite book</ref> that Hölder's inequality can be put in a more symmetric form, at the price of introducing an extra vector (or function):

Let <math>f = (f(1),\dots,f(m)) , g = (g(1),\dots, g(m)), h = (h(1),\dots,h(m))</math> be vectors with positive entries and such that <math>f(i) g(i) h(i) = 1</math> for all <math>i</math>. If <math>p,q,r</math> are nonzero real numbers such that <math>\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=0</math>, then:

  • <math>\|f\|_p \|g\|_q \|h\|_r \ge 1</math> if all but one of <math>p,q,r</math> are positive;
  • <math>\|f\|_p \|g\|_q \|h\|_r \le 1</math> if all but one of <math>p,q,r</math> are negative.

The standard Hölder inequality follows immediately from this symmetric form (and in fact is easily seen to be equivalent to it). The symmetric statement also implies the reverse Hölder inequality (see above).

The result can be extended to multiple vectors:

Let <math>f_1, \dots, f_n</math> be <math>n</math> vectors in <math>\mathbb{R}^m</math> with positive entries and such that <math>f_1(i) \dots f_n(i) = 1</math> for all <math>i</math>. If <math>p_1,\dots,p_n</math> are nonzero real numbers such that <math>\frac{1}{p_1}+\dots+\frac{1}{p_n}=0</math>, then:

  • <math>\|f_1\|_{p_1} \dots \|f_n\|_{p_n} \ge 1</math> if all but one of the numbers <math>p_i</math> are positive;
  • <math>\|f_1\|_{p_1} \dots \|f_n\|_{p_n} \le 1</math> if all but one of the numbers <math>p_i</math> are negative.

As in the standard Hölder inequalities, there are corresponding statements for infinite sums and integrals.

Conditional Hölder inequalityEdit

Let Template:Math be a probability space, Template:Math a Template:Nowrap, and Template:Math Template:Open-open Hölder conjugates, meaning that Template:Math. Then for all real- or complex-valued random variables Template:Mvar and Template:Mvar on Template:Math,

<math>\mathbb{E}\bigl[|XY|\big|\,\mathcal{G}\bigr] \le \bigl(\mathbb{E}\bigl[|X|^p\big|\,\mathcal{G}\bigr]\bigr)^{\frac{1}{p}} \,\bigl(\mathbb{E}\bigl[|Y|^q\big|\,\mathcal{G}\bigr]\bigr)^{\frac{1}{q}}

\qquad\mathbb{P}\text{-almost surely.}</math>

Remarks:

<math>\mathbb{E}[Z|\mathcal{G}] = \sup_{n\in\mathbb{N}}\,\mathbb{E}[\min\{Z,n\}|\mathcal{G}]\quad\text{a.s.}</math>
  • On the right-hand side of the conditional Hölder inequality, 0 times ∞ as well as ∞ times 0 means 0. Multiplying Template:Math with ∞ gives ∞.

Proof of the conditional Hölder inequality: Template:Math proof,\qquad V=\bigl(\mathbb{E}\bigl[|Y|^q\big|\,\mathcal{G}\bigr]\bigr)^{\frac{1}{q}}</math>

and note that they are measurable with respect to the Template:Nowrap. Since

<math>\mathbb{E}\bigl[|X|^p1_{\{U=0\}}\bigr] = \mathbb{E}\bigl[1_{\{U=0\}}\underbrace{\mathbb{E}\bigl[|X|^p\big|\,\mathcal{G}\bigr]}_{=\,U^p}\bigr]=0,</math>

it follows that Template:Math a.s. on the set Template:Math. Similarly, Template:Math a.s. on the set Template:Math, hence

<math>\mathbb{E}\bigl[|XY|\big|\,\mathcal{G}\bigr]=0\qquad\text{a.s. on }\{U=0\}\cup\{V=0\}</math>

and the conditional Hölder inequality holds on this set. On the set

<math>\{U=\infty, V>0\}\cup\{U>0, V=\infty\}</math>

the right-hand side is infinite and the conditional Hölder inequality holds, too. Dividing by the right-hand side, it therefore remains to show that

<math>\frac{\mathbb{E}\bigl[|XY|\big|\,\mathcal{G}\bigr]}{UV}\le1

\qquad\text{a.s. on the set }H:=\{0<U<\infty,\,0<V<\infty\}.</math>

This is done by verifying that the inequality holds after integration over an arbitrary

<math>G\in\mathcal{G},\quad G\subset H.</math>

Using the measurability of Template:Mvar with respect to the Template:Nowrap, the rules for conditional expectations, Hölder's inequality and Template:Math, we see that

<math>\begin{align}

\mathbb{E}\biggl[\frac{\mathbb{E}\bigl[|XY|\big|\,\mathcal{G}\bigr]}{UV}1_G\biggr] &=\mathbb{E}\biggl[\mathbb{E}\biggl[\frac{|XY|}{UV}1_G\bigg|\,\mathcal{G}\biggr]\biggr]\\ &=\mathbb{E}\biggl[\frac{|X|}{U}1_G\cdot\frac{|Y|}{V}1_G\biggr]\\ &\le\biggl(\mathbb{E}\biggl[\frac{|X|^p}{U^p}1_G\biggr]\biggr)^{\frac{1}{p}} \biggl(\mathbb{E}\biggl[\frac{|Y|^q}{V^q}1_G\biggr]\biggr)^{\frac{1}{q}}\\ &=\biggl(\mathbb{E}\biggl[\underbrace{\frac{\mathbb{E}\bigl[|X|^p\big|\,\mathcal{G}\bigr]}{U^p}}_{=\,1\text{ a.s. on }G}1_G\biggr]\biggr)^{\frac{1}{p}} \biggl(\mathbb{E}\biggl[\underbrace{\frac{\mathbb{E}\bigl[|Y|^q\big|\,\mathcal{G}\bigr]}{V^p}}_{=\,1\text{ a.s. on }G}1_G\biggr]\biggr)^{\frac{1}{q}}\\ &=\mathbb{E}\bigl[1_G\bigr]. \end{align}</math> }}

Hölder's inequality for increasing seminormsEdit

Let Template:Mvar be a set and let <math>F(S, \mathbb{C})</math> be the space of all complex-valued functions on Template:Mvar. Let Template:Mvar be an increasing seminorm on <math>F(S, \mathbb{C}),</math> meaning that, for all real-valued functions <math>f, g \in F(S, \mathbb{C})</math> we have the following implication (the seminorm is also allowed to attain the value ∞):

<math> \forall s \in S \quad f(s) \geqslant g(s) \geqslant 0 \qquad \Rightarrow \qquad N(f) \geqslant N(g).</math>

Then:

<math>\forall f, g \in F(S, \mathbb{C}) \qquad N(|fg|) \leqslant \bigl(N(|f|^p)\bigr)^{\frac{1}{p}} \bigl(N(|g|^q)\bigr)^{\frac{1}{q}},</math>

where the numbers <math>p</math> and <math>q</math> are Hölder conjugates.<ref>For a proof see Template:Harv.</ref>

Remark: If Template:Math is a measure space and <math>N(f)</math> is the upper Lebesgue integral of <math>|f|</math> then the restriction of Template:Mvar to all Template:Nowrap functions gives the usual version of Hölder's inequality.


Distances based on Hölder inequalityEdit

Hölder inequality can be used to define statistical dissimilarity measures<ref>Template:Cite journal</ref> between probability distributions. Those Hölder divergences are projective: They do not depend on the normalization factor of densities.

See alsoEdit

CitationsEdit

Template:Reflist

ReferencesEdit

External linksEdit

Template:Lp spaces Template:Measure theory Template:Functional analysis

Retrieved from "https://zalansite.site/mediawiki/index.php?title=Hölder%27s_inequality&oldid=135222"