Hölder's Inequality
Hölder's inequality is a statement about sequences that generalizes the Cauchy-Schwarz inequality to multiple sequences and different exponents.
Definition
Hölder's inequality states that, for sequences \( {a_i}, {b_i}, \ldots , {z_i} , \) the inequality
\[ (a_1 + a_2 + \cdots + a_n)^{ \lambda_a} \cdots (z_1 + z_2 + \cdots + z_n)^{ \lambda_z } \geq a_1^{ \lambda_a } b_1^{ \lambda_b} \cdots z_1^{ \lambda_z } + \cdots + a_n^{ \lambda_a } b_n^{ \lambda_b } \cdots z_n^{ \lambda_z } \]
holds for all \( \lambda_a + \lambda_b + \cdots + \lambda_z = 1 .\) For instance, in the case of \( \lambda_a = \lambda_b = \frac{1}{2},\) Hölder's inequality reduces to
\[ (a_1 + a_2 + \cdots + a_n)^{ \frac{1}{2}} (b_1 + b_2 + \cdots + b_n)^{ \frac{1}{2} } \geq (a_1 b_1)^{\frac{1}{2}} + (a_2 b_2)^{\frac{1}{2}} + \cdots + (a_n b_n)^{ \frac{1}{2}}, \]
which is the Cauchy-Schwarz inequality.
Proof
Hölder's inequality can be proven using Jensen's inequality. In particular, we may seek to prove the following form of Hölder's inequality:
\[ \prod _{j=1} ^z \left( \int f _j(x) ^{p_j} \, dx \right) ^{1 /p _j} \ge \int \prod _{j=1} ^z f _j(x) \, dx, \]
where \( \frac{1}{p _j} = \lambda _j,\ \sum_{j=1}^z \lambda _j = 1,\ \big\{ f _1(x_i) ^{p _1} dx \big\} = \big\{ a_1, a_2, \dots , a_n,\big\},\ \big\{ f _2(x_i) ^{p _2} dx \big\} = \big\{ b_1, b_2, \dots , b_n \big\}, \dots, \) and \( \big\{ f _z(x_i) ^{p _z} dx \big\} = \{ z_1, z_2, \dots , z_n \}. \)
First, please refer to the proof presented in the Wikipedia article on Hölder's inequality for the simpler case of \( \left( \int f ^p d\mu \right) ^ {1 /p} \left( \int g ^q d\mu \right) ^{1 /q} \ge \int fg\, d\mu \), where \( \frac {1} {p} +\frac {1} {q} = 1 \).
Recall the Jensen's inequality for the convex function \( x^p \) \((\)it is convex because obviously \( p\geq1): \) \( \int h\,d\nu\leq\left(\int h^pd\nu \right )^{\frac{1}{p}} \), where \(\nu\) is any probability distribution and \(h\) is any \(\nu\)-measurable function. Let \(\mu\) be any measure, and \(\nu\) the distribution whose density w.r.t. \(\mu\) is proportional to \( g^q \), i.e. \( d\nu = \frac{g^q}{\int g^q\,d\mu}d\mu \). Then we have, using \( \frac{1}{p}+\frac{1}{q}=1 \), \( p(1-q)+q=0 \), and letting \( h=fg^{1-q} \) gives \[ \int fg\,d\mu = \left (\int g^q\,d\mu \right )\int \underbrace{fg^{1-q}}_h\underbrace{\frac{g^q}{\int g^q\,d\mu}d\mu}_{d\nu} \leq \left (\int g^qd\mu \right ) \left (\int \underbrace{f^pg^{p(1-q)}}_{h^p}\underbrace{\frac{g^q}{\int g^q\,d\mu}\,d\mu}_{d\nu} \right )^{\frac{1}{p}} = \left (\int g^q\,d\mu \right ) \left (\int \frac{f^p}{\int g^q\,d\mu}\,d\mu \right )^{\frac{1}{p}}. \] Finally, we get \( \int fg\,d\mu \leq \left(\int f^p\,d\mu \right )^{\frac{1}{p}} \left(\int g^q\,d\mu \right )^{\frac{1}{q}} \). \(_\square\)
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Next, we prove that \( \left( \int f ^p d\mu \right) ^ {1 /p} \left( \int g ^q d\mu \right) ^{1 /q} \ge \left( \int (fg)^r d\mu \right) ^{1 /r} \) for any \(p\), \( q \gt 0 \), where \( \frac {1} {p} +\frac {1} {q} = \frac {1} {r} \).
Because \( \frac {r} {p} +\frac {r} {q} = \frac {r} {r} = 1 \), therefore \( \textstyle \left( \int F ^{p /r} d\mu \right) ^ {r /p} \left( \int G ^{q /r} d\mu \right) ^{r /q} \ge \int FG\, d\mu \), using the proof above for the simpler case.
Now, let \( f = F ^{1/r} \) and \( g = G ^{1/r} \). With some algebra, we can see that \( \textstyle \left( \left( \int f ^p d\mu \right) ^ {r /p} \left( \int g ^q d\mu \right) ^{r /q} \right) ^{1/r} \ge \left( \int f ^r g ^r d\mu \right) ^{1 /r} \).
Given the above, we can show that
\[\left( \int f _1 ^{p _1} d\mu \right) ^{1 /p _1} \left( \int f _2 ^{p _2} d\mu \right) ^{1 /p _2} \left( \int f _3 ^{p _3} d\mu \right) ^{1 /p _3} \ge \left( \int (f _1 f _2) ^{\frac {1} {(1 /p _1) +(1 /p _2)}} d\mu \right) ^{(1 /p _1) + (1 /p_2)} \left( \int f _3 ^{p _3} d\mu \right) ^{1 /p _3} \ge \left( \int (f _1 f _2 f _3) ^{\frac {1} {(1 /p _1) +(1 /p _2) +(1 /p _3)}} d\mu \right) ^{(1 /p _1) +(1 /p_2) +(1 /p _3)}. \]
In fact, define \( s(m) = \sum_{j=1}^{m} \frac{1}{p _j}. \) Given
\[\prod _{j=1}^{m-1} \left( \int f _j ^{p _j} d\mu \right) ^{1 /p _j} \ge \left( \int \left( \prod_{j=1}^{m-1} f _j \right) ^{\frac {1} {s(m-1)}} d\mu \right) ^{s(m-1)}, \]
we have
\[ \prod _{j=1}^{m-1} \left( \int f _j ^{p _j} d\mu \right) ^{1 /p _j} \left( \int f _m ^{p _m} d\mu \right) ^ {1 /p _m} \ge \left( \int \left( \prod_{j=1}^{m-1} f _j \right) ^{\frac {1} {s(m-1)}} d\mu \right) ^{s(m-1)} \left( \int f _m ^{p _m} d\mu \right) ^ {1 /p _m} \ge \left( \int \left( \prod_{j=1}^{m} f _j \right) ^{\frac {1} {s(m)}} d\mu \right) ^{s(m)}. \]
Therefore, by induction,
\[ \prod _{j=1}^{z} \left( \int f _j ^{p _j} d\mu \right) ^{1 /p _j} \ge \left( \int \left( \prod_{j=1}^{z} f _j \right) ^{\frac {1} {s(z)}} d\mu \right) ^{s(z)}. \]
In the case that \( s(z) = 1 \), we arrive at the original claim that we've sought to prove.
Hölder's inequality may also be proven using Young's inequality. Young's inequality states that if \(p,q\) are positive reals satisfying \(\frac{1}{p}+\frac{1}{q}=1\), then
\[ab \leq \frac{a^p}{p}+\frac{b^q}{q}\]
for all nonnegative reals \(a,b\).
It follows from the concavity of the logarithm function and Jensen's inequality; in particular,
\[\log\left(\frac{1}{p}a^{p}+\frac{1}{q}b^q\right) \geq \frac{1}{p}\log a^p+\frac{1}{q}\log b^q=\log a+\log b=\log ab\]
and exponentiating gives Young's inequality.
Strategies and Applications
Hölder's inequality is often used to deal with square (or higher-power) roots of expressions in inequalities since those can be eliminated through successive multiplication. Here is an example:
Let \(a,b,c\) be positive reals satisfying \(a+b+c=3\). What is the minimum possible value of
\[\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}?\]
By Hölder's inequality,
\[\begin{align} \left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)^{1/3}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)^{1/3}(a+b+c)^{1/3} &\geq 1^{1/3}+1^{1/3}+1^{1/3} \\ &= 3 \\ \Rightarrow \left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)(a+b+c) &\geq 3^3. \end{align}\]
Since \(a+b+c=3\), the minimum possible value of \(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\) is \(3\). \(_\square\)
(2001 IMO, Problem 2)
Let \(a,b,c\) be positive real numbers. Prove that
\[\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1.\]
By Hölder's inequality,
\[\left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum a(a^{2}+8bc)\right)\ge (a+b+c)^{3}.\]
Thus it suffices to show that
\[(a+b+c)^{3}\ge a^{3}+b^{3}+c^{3}+24abc,\]
which follows immediately from the famous inequality (incidentally, easily provable with Hölder's)
\[(a+b)(b+c)(c+a)\ge 8abc.\ _\square\]
A cycling strategy can also be used, as in the demonstration that Hölder's inequality implies AM-GM:
By Hölder's inequality,
\[(a_1+a_2+\ldots+a_n)^{\frac{1}{n}}(a_2+a_3+\cdots+a_1)^{\frac{1}{n}}\cdots(a_n+a_1+\cdots+a_{n-1})^{\frac{1}{n}} \geq n\sqrt[n]{a_1a_2\cdots a_n},\]
which rearranges to AM-GM immediately. \(_\square\)
Another useful strategy is to insert constants (especially 1) as members of a sequence, especially to "reduce" powers. For instance,
Let \(a,b\) be positive real numbers. Show that
\[4\big(a^3+b^3\big) \geq (a+b)^3.\]
By Hölder's inequality,
\[\big(a^3+b^3\big)^{\frac{1}{3}}(1+1)^{\frac{1}{3}}(1+1)^{\frac{1}{3}} \geq \big(a^3\big)^{\frac{1}{3}}1^{\frac{1}{3}}1^{\frac{1}{3}}+\big(b^3\big)^{\frac{1}{3}}1^{\frac{1}{3}}1^{\frac{1}{3}},\]
which rearranges to the desired inequality. \(_\square\)
If \(a\) and \(b\) are positive real numbers such that \(a^{2015}+b^{2015}=2015,\) find the maximum value of \(a+b\).
\[a^2 +b^2+c^2+d^2\]
If \(a,b,c\) and \(d\) are positive reals satisfying \(a^{2016}+b^{2016}+c^{2016}+d^{2016}=4032\), maximize the expression above to 3 decimal places
Minkowski's Inequality
Minkowski's inequality easily follows from Holder's inequality.
Minkowski's inequality states that
\[\left(\sum _{ n=1 }^{ k } ({ x }_{ n }+{ y }_{ n })^p\right)^{ \frac { 1 }{ p } }\le \left(\sum _{ n=1 }^{ k }{ { x }_{ n }^{ p } } \right) ^{ \frac { 1 }{ p } }+\left(\sum _{ n=1 }^{ k }{ { y }_{ n }^{ p } } \right)^{ \frac { 1 }{ p } }\]
for \(p>1\) and \({x}_{n},{y}_{n}\ge0.\)
We know
\[\sum _{ n=1 }^{ k }{ ({ x }_{ n }+{ y }_{ n })^{ p }=\displaystyle\sum _{ n=1 }^{ k }{ { x }_{ n }({ x }_{ n }+{ y }_{ n })^{ p-1 } } } +\displaystyle\sum _{ n=1 }^{ k }{ { y }_{ n }({ x }_{ n }+{ y }_{ n } } )^{ p-1 }.\]
Let's define \(a\) as \(a=\frac { p }{ p-1 } ,\) then by Holder's inequality we have
\[\sum_{n \mathop = 1}^k x_n \left({x_n + y_n}\right)^{p-1} + \sum_{n \mathop = 1}^k y_n \left({x_n + y_n}\right)^{p-1} \le \left[ \left(\sum _{ n=1 }^{ k }{ { x }_{ n }^{ p } } \right)^{ \frac { 1 }{ p } }+\left(\sum _{ n=1 }^{ k }{ { y }_{ n }^{ p } } \right)^{ \frac { 1 }{ p } }\right] \left[ \sum _{ n=1 }^{ k }{ ({ x }_{ n }+{ y }_{ n })^{ p } } \right]^{ \frac { 1 }{ a } }.\]
Dividing by \(\left[\sum _{ n=1 }^{ k }{ ({ x }_{ n }+{ y }_{ n })^{ p } } \right]^{ \frac { 1 }{ a }}, \) we get
\[\left(\sum _{ n=1 }^{ k } ({ x }_{ n }+{ y }_{ n })^p\right)^{ \frac { 1 }{ p } }\le \left(\sum _{ n=1 }^{ k }{ { x }_{ n }^{ p } } \right) ^{ \frac { 1 }{ p } }+\left(\sum _{ n=1 }^{ k }{ { y }_{ n }^{ p } } \right)^{ \frac { 1 }{ p } }.\] Hence proved. \(_\square\)