View source for Young's inequality for products

{{Short description|Mathematical concept}}
[[File:Young.png|thumb|The area of the rectangle a,b can't be larger than sum of the areas under the functions <math>f</math> (red) and <math>f^{-1}</math> (yellow)]]

In [[mathematics]], '''Young's inequality for products''' is a [[inequality (mathematics)|mathematical inequality]] about the product of two numbers.<ref>{{citation|first1 = W. H.|last1 = Young|author-link = William Henry Young|year = 1912|title = On classes of summable functions and their Fourier series|journal = [[Proceedings of the Royal Society A]]|volume=87|pages=225–229|jfm = 43.1114.12|jstor = 93236|doi = 10.1098/rspa.1912.0076|issue = 594|bibcode = 1912RSPSA..87..225Y|doi-access = free}}</ref> The inequality is named after [[William Henry Young]] and should not be confused with [[Young's convolution inequality]].

Young's inequality for products can be used to prove [[Hölder's inequality]]. It is also widely used to estimate the norm of nonlinear terms in [[partial differential equation|PDE theory]], since it allows one to estimate a product of two terms by a sum of the same terms raised to a power and scaled.

==Standard version for conjugate Hölder exponents==

The standard form of the inequality is the following, which can be used to prove [[Hölder's inequality]].

{{Math theorem|name=Theorem|math_statement=
If <math>a \geq 0</math> and <math>b \geq 0</math> are [[nonnegative]] [[real number]]s and if <math>p > 1</math> and <math>q > 1</math> are real numbers such that <math>\frac{1}{p} + \frac{1}{q} = 1,</math> then
<math display=block>a b ~\leq~ \frac{a^p}{p} + \frac{b^q}{q}.</math>

Equality holds if and only if <math>a^p = b^q.</math>
}}

{{Math proof|drop=hidden|title=Proof<ref>{{cite web |last1=Pearse |first1=Erin |title=Math 209D - Real Analysis Summer Preparatory Seminar Lecture Notes |url=https://pi.math.cornell.edu/~erin/analysis/lectures.pdf |access-date=17 September 2022}}</ref>|proof=
Since <math>\tfrac{1}{p} + \tfrac{1}{q} = 1,</math> <math>p - 1 = \tfrac{1}{q-1}.</math> A graph <math>y = x^{p-1}</math> on the <math>x y</math>-plane is thus also a graph <math>x = y^{q-1}.</math> From sketching a visual representation of the integrals of the area between this curve and the axes, and the area in the rectangle bounded by the lines <math>x=0, x=a, y=0, y=b,</math> and the fact that <math>y</math> is always increasing for increasing <math>x</math> and vice versa, we can see that <math>\int^a_0 x^{p-1} \mathrm{d}x </math> upper bounds the area of the rectangle below the curve (with equality when <math>b\ge a^{p-1}</math>) and <math>\int^b_0 y^{q-1} \mathrm{d}y</math> upper bounds the area of the rectangle above the curve (with equality when <math>b\le a^{p-1}</math>). Thus, <math>\int^a_0 x^{p-1} \mathrm{d}x + \int^b_0 y^{q-1} \mathrm{d}y \geq ab,</math> with equality when <math>b=a^{p-1}</math> (or equivalently, <math>a^p=b^q</math>). Young's inequality follows from evaluating the integrals. (See [[Young's inequality for products#Standard version for increasing functions|below]] for a generalization.)
}}

A second proof is via [[Jensen's inequality]].

{{Math proof|drop=hidden|title=Proof{{sfn|Bahouri|Chemin|Danchin|2011|p=}}|proof=
The claim is certainly true if <math>a = 0</math> or <math>b = 0</math> so henceforth assume that <math>a > 0</math> and <math>b > 0.</math> 
Put <math>t = 1/p </math> and <math>(1 - t) = 1/q.</math> 
Because the [[logarithm]] function is [[concave function|concave]],
<math display=block>\ln\left(t a^p + (1-t) b^q\right) ~\geq~ t \ln\left(a^p\right) + (1-t) \ln\left(b^q\right) = \ln(a) + \ln(b) = \ln(ab)</math>
with the equality holding if and only if <math>a^p = b^q.</math> 
Young's inequality follows by exponentiating.
}}
Yet another proof is to first prove it with <math>b = 1</math> an then apply the resulting inequality to <math>\tfrac{a}{b^q}  </math>. The proof below illustrates also why Hölder conjugate exponent is the only possible parameter that makes Young's inequality hold for all non-negative values. The details follow:

{{Math proof|drop=hidden|title=Proof|proof=
Let <math> 0 <\alpha<1</math> and <math>\alpha + \beta = 1</math>. The inequality 
<math display=block> x ~\leq~ \alpha x^p + \beta, \qquad\, for \quad\ all \quad\ x ~\geq~ 0 </math>
holds if and only if <math> \alpha = \tfrac{1}{p}</math> (and hence <math> \beta= \tfrac{1}{q}</math>). This can be shown by convexity arguments or by simply minimizing the single-variable function.

To prove full Young's inequality, clearly we assume that <math> a > 0 </math> and <math>b > 0</math>.  Now, we apply the inequality above to <math> x=\tfrac{a}{b^s}</math> to obtain:
<math display=block> \tfrac{a}{b^s} ~\leq~ \tfrac{1}{p}\tfrac{a^p}{b^{sp}} + \tfrac{1}{q}.</math>
It is easy to see that choosing <math> s = q-1 </math> and multiplying both sides by <math> b^q </math> yields Young's inequality.
}}

Young's inequality may equivalently be written as
<math display=block>a^\alpha b^\beta \leq \alpha a + \beta b, \qquad\, 0 \leq \alpha, \beta \leq 1, \quad\ \alpha + \beta = 1.</math>

Where this is just the concavity of the [[logarithm]] function.
Equality holds if and only if <math>a = b</math> or <math>\{\alpha, \beta\} = \{0, 1\}.</math>
This also follows from the weighted [[AM-GM Inequality|AM-GM inequality]].

===Generalizations===

{{Math theorem|name=Theorem{{sfn|Jarchow|1981|pp=47-55}}|math_statement=
Suppose <math>a > 0</math> and <math>b > 0.</math>
If <math>1 < p < \infty</math> and <math>q</math> are such that <math>\tfrac{1}{p} + \tfrac{1}{q} = 1</math> then
<math display=block>a b ~=~ \min_{0 < t < \infty} \left(\frac{t^p a^p}{p} + \frac{t^{-q} b^q}{q}\right).</math>

Using <math>t := 1</math> and replacing <math>a</math> with <math>a^{1/p}</math> and <math>b</math> with <math>b^{1/q}</math> results in the inequality:
<math display=block>a^{1/p} \, b^{1/q} ~\leq~ \frac{a}{p} + \frac{b}{q},</math>
which is useful for proving [[Hölder's inequality]]. 
}}

{{Math proof|drop=hidden|title=Proof{{sfn|Jarchow|1981|pp=47-55}}|proof=
Define a real-valued function <math>f</math> on the positive real numbers by
<math display=block>f(t) ~=~ \frac{t^p a^p}{p} + \frac{t^{-q} b^q}{q}</math>
for every <math>t > 0</math> and then calculate its minimum. 
}}

{{Math theorem|name=Theorem|math_statement=
If <math>0 \leq p_i \leq 1</math> with <math>\sum_i p_i = 1</math> then
<math display=block>\prod_i {a_i}^{p_i} ~\leq~ \sum_i p_i a_i.</math>
Equality holds if and only if all the <math>a_i</math>s with non-zero <math>p_i</math>s are equal.
}}

==Elementary case==

An elementary case of Young's inequality is the inequality with [[exponent]] <math>2,</math>
<math display=block>a b \leq \frac{a^2}{2} + \frac{b^2}{2},</math>
which also gives rise to the so-called Young's inequality with <math>\varepsilon</math> (valid for every <math>\varepsilon > 0</math>), sometimes called the Peter–Paul inequality.
<ref>{{citation
|last1       = Tisdell
|first1      = Chris
|title      = The Peter Paul Inequality
|publisher  = YouTube video on Dr Chris Tisdell's YouTube channel
|year       = 2013
|url        = https://www.youtube.com/watch?v=C_bjbjTzHP4
}},</ref>  This name refers to the fact that tighter control of the second term is achieved at the cost of losing some control of the first term – one must "rob Peter to pay Paul"
<math display=block>a b ~\leq~ \frac{a^2}{2 \varepsilon} + \frac{\varepsilon b^2}{2}.</math>

'''Proof''': Young's inequality with exponent <math>2</math> is the special case <math>p = q = 2.</math> However, it has a more elementary proof.

Start by observing that the square of every real number is zero or positive.  Therefore, for every pair of real numbers <math>a</math> and <math>b</math> we can write:
<math display=block>0 \leq (a-b)^2</math> 
Work out the square of the right hand side:
<math display=block>0 \leq a^2 - 2 a b + b^2</math> 
Add <math>2a b </math> to both sides:
<math display=block>2 a b \leq a^2 + b^2</math>
Divide both sides by 2 and we have Young's inequality with exponent <math>2:</math>
<math display=block>a b \leq \frac{a^2}{2} + \frac{b^2}{2}</math>

Young's inequality with <math>\varepsilon</math> follows by substituting <math>a'</math> and <math>b'</math> as below into Young's inequality with exponent <math>2:</math> 
<math display=block>a' = a/\sqrt{\varepsilon}, \; b' = \sqrt{\varepsilon} b.</math>

==Matricial generalization==

T. Ando proved a generalization of Young's inequality for complex matrices ordered 
by [[Loewner order]]ing.<ref>{{cite book|author1=T. Ando|editor1-last=Huijsmans|editor1-first=C. B.|editor2-last=Kaashoek|editor2-first=M. A.|editor3-last=Luxemburg|editor3-first=W. A. J.|editor4-last=de Pagter|editor4-first=B.| display-editors = 3|title=Operator Theory in Function Spaces and Banach Lattices|date=1995|publisher=Springer|isbn=978-3-0348-9076-2|pages=33–38|chapter=Matrix Young Inequalities}}</ref> It states that for any pair <math>A, B</math> of complex matrices of order <math>n</math> there exists a [[unitary matrix]] <math>U</math> such that
<math display=block>U^* |A B^*| U \preceq \tfrac{1}{p} |A|^p + \tfrac{1}{q} |B|^q,</math>
where <math>{}^*</math> denotes the [[conjugate transpose]] of the matrix and <math>|A| = \sqrt{A^* A}.</math>

==Standard version for increasing functions==

For the standard version<ref>{{citation
|last1       = Hardy
|first1      = G. H.
|last2       = Littlewood
|first2      = J. E.
|last3       = Pólya
|first3      = G.
|author1-link = Godfrey Harold Hardy
|author2-link = John Edensor Littlewood
|author3-link = George Pólya
|title      = Inequalities
|edition    = 2nd
|series     = Cambridge Mathematical Library
|publisher  = Cambridge University Press
|location   = Cambridge
|year       = 1952
|origyear   = 1934
|pages      = 
|isbn       = 0-521-05206-8 
|url        = http://www.cambridge.org/catalogue/catalogue.asp?isbn=9780521358804
|mr       = 0046395
|zbl       = 0047.05302
}}, Chapter&nbsp;4.8</ref><ref>{{citation
|last1        = Henstock
|first1       = Ralph
|author1-link = Ralph Henstock
|title        = Lectures on the Theory of Integration
|series       = Series in Real Analysis Volume I
|publisher    = World Scientific
|location     = Singapore, New Jersey
|year         = 1988
|isbn         = 9971-5-0450-2
|mr           = 0963249
|zbl          = 0668.28001
|url-access   = registration
|url          = https://archive.org/details/lecturesontheory0000hens
}}, Theorem&nbsp;2.9</ref> of the inequality,
let <math>f</math> denote a real-valued, continuous and strictly increasing function on <math>[0, c]</math> with <math>c > 0</math> and <math>f(0) = 0.</math> Let <math>f^{-1}</math> denote the [[inverse function]] of <math>f.</math> Then, for all <math>a \in [0, c]</math> and <math>b \in [0, f(c)],</math>
<math display=block>a b ~\leq~ \int_0^a f(x)\,dx + \int_0^b f^{-1}(x)\,dx</math>
with equality if and only if <math>b = f(a).</math>

With <math>f(x) = x^{p-1}</math> and <math>f^{-1}(y) = y^{q-1},</math> this reduces to standard version for conjugate Hölder exponents.

For details and generalizations we refer to the paper of Mitroi & Niculescu.<ref>Mitroi, F. C., & Niculescu, C. P. (2011). An extension of Young's inequality. In Abstract and Applied Analysis (Vol. 2011). Hindawi.</ref>

==Generalization using Fenchel–Legendre transforms==

By denoting the [[convex conjugate]] of a real function <math>f</math> by <math>g,</math> we obtain
<math display=block>a b ~\leq~ f(a) + g(b).</math>
This follows immediately from the definition of the convex conjugate. For a convex function <math>f</math> this also follows from the [[Legendre transformation]].

More generally, if <math>f</math> is defined on a real vector space <math>X</math> and its [[convex conjugate]] is denoted by <math>f^\star</math> (and is defined on the [[dual space]] <math>X^\star</math>), then
<math display=block>\langle u, v \rangle \leq f^\star(u) + f(v).</math>
where <math>\langle \cdot , \cdot \rangle : X^\star \times X \to \Reals</math> is the [[dual pair]]ing.

===Examples===

The convex conjugate of <math>f(a) = a^p / p</math> is <math>g(b) = b^q / q</math> with <math>q</math> such that <math>\tfrac{1}{p} + \tfrac{1}{q} = 1,</math> and thus Young's inequality for conjugate Hölder exponents mentioned above is a special case.

The  Legendre transform of <math>f(a) = e^a - 1</math> is <math>g(b) = 1 - b + b \ln b</math>, hence <math>a b \leq e^a - b + b \ln b</math>  for all non-negative <math>a</math> and <math>b.</math> This estimate is useful in [[large deviations theory]] under exponential moment conditions, because <math>b \ln b</math> appears in the definition of [[relative entropy]], which is the [[rate function]] in [[Sanov's theorem]].

==See also==

* {{annotated link|Convex conjugate}}
* {{annotated link|Integral of inverse functions}}
* {{annotated link|Legendre transformation}}
* {{annotated link|Young's convolution inequality}}

==Notes==

{{reflist}}

==References==
{{sfn whitelist|CITEREFBahouriCheminDanchin2011}}
* {{Jarchow Locally Convex Spaces}} <!--{{sfn|Jarchow|1981|p=}}-->
* {{Bahouri Chemin Danchin Fourier Analysis and Nonlinear Partial Differential Equations 2011}} <!--{{sfn|Bahouri|Chemin|Danchin|2011|p=}}-->

==External links==

* [http://planetmath.org/encyclopedia/YoungsInequality.html ''Young's Inequality''] at [[PlanetMath]]
* {{MathWorld|title=Young's Inequality|urlname=YoungsInequality}}

{{Measure theory}}

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[[Category:Articles containing proofs]]
[[Category:Inequalities]]