Hölder's Inequality

Hölder's inequality is a statement about sequences that generalizes the Cauchy-Schwarz inequality to multiple sequences and different exponents.

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.

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 $\int fg\,d\mu \leq \left(\int f^p\,d\mu \right )^{\frac{1}{p}} \left(\int g^q\,d\mu \right )^{\frac{1}{q}}$. $_\square$

$$
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.

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

$$\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}?$$

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By Hölder's inequality,

$$\begin{aligned} \left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)^{1/3}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)^{1/3}(a+b+c)^{1/3} &\geq 1^{1/3}+1^{1/3}+1^{1/3} \\ &= 3 \\ \Rightarrow \left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)(a+b+c) &\geq 3^3. \end{aligned}$$

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

$$\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1.$$

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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

$$4\big(a^3+b^3\big) \geq (a+b)^3.$$

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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. $_\square$

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. $_\square$

• Inequalities - Convexity Arguments

• Classical Inequalities Problem Solving - Basic

• Classical Inequalities Problem Solving - Intermediate

• Classical Inequalities Problem Solving - Advanced