algebra911.com

Holders Inequality

Q: What happens if p=1?

If you try to set $p=1$, the conjugate equation $1/1 + 1/q = 1$ would mean $1/q=0$, which is impossible for a finite number $q$. This is treated as a special case where $q$ is considered to be infinity, leading to a simpler inequality that's easy to prove on its own.

Q: Is this related to the Pythagorean theorem?

In a way, yes! The Cauchy-Schwarz case ($p=q=2$) is deeply connected to the geometry of vectors and the concept of dot products. The length of a vector is calculated using sums of squares, just like in the inequality, which is directly related to the Pythagorean theorem in higher dimensions.

Download as PDF

Ever wondered how to compare the sum of products with the product of sums? Hölder's Inequality provides a powerful and elegant answer. This fundamental tool in mathematics helps us understand the relationship between different sums and is a key stepping stone to more advanced concepts.

Holders Inequality — an original Algebra911 reference diagram defining holders inequality with its key formula and a worked example. — Hölder's Inequality: A Complete Guide

What Is Hölder's Inequality?

Hölder's Inequality is a fundamental mathematical principle that establishes a relationship between the sum of the product of two sequences of numbers and the product of the sums of those numbers raised to certain powers. It provides an upper bound for the sum of products, guaranteeing that one value will not be larger than another.

Let's break down the official formula. For two sequences of non-negative real numbers, $a_{1}, a_{2}, \dots, a_{n}$ and $b_{1}, b_{2}, \dots, b_{n}$ , and for real numbers $p, q > 1$ such that $\frac{1}{p} + \frac{1}{q} = 1$ , the inequality states:

\sum_{k = 1}^{n} a_{k} b_{k} \leq {(\sum_{k = 1}^{n} a_{k}^{p})}^{1 / p} {(\sum_{k = 1}^{n} b_{k}^{q})}^{1 / q}

This looks complicated, but we can unpack it piece by piece.

$\sum_{k = 1}^{n}$ : This is the summation symbol, sigma. It means we sum up a series of terms. For example, $\sum_{k = 1}^{3} a_{k}$ is just $a_{1} + a_{2} + a_{3}$ .
$a_{k}$ and $b_{k}$ : These represent the terms in our two sequences of numbers. For instance, if our first sequence is $(2, 5, 8)$ , then $a_{1} = 2$ , $a_{2} = 5$ , and $a_{3} = 8$ .
The Left-Hand Side (LHS): $\sum_{k = 1}^{n} a_{k} b_{k}$ . This side tells us to multiply the corresponding terms of our two sequences first ( $a_{1} b_{1}$ , $a_{2} b_{2}$ , etc.) and then add all those products together.
The Right-Hand Side (RHS): ${(\sum_{k = 1}^{n} a_{k}^{p})}^{1 / p} {(\sum_{k = 1}^{n} b_{k}^{q})}^{1 / q}$ . This side is more complex. It tells us to:
1. Take every term in the first sequence ( $a_{k}$ ), raise it to the power of $p$ , and then add them all up.
2. Take the $p$ -th root of that sum (which is the same as raising it to the power of $1 / p$ ).
3. Do the same for the second sequence ( $b_{k}$ ), but using the power $q$ .
4. Finally, multiply these two results together.
$p$ and $q$ : These are the special exponents, which we'll explore next. They must be greater than $1$ and satisfy a unique relationship.

In simple words, Hölder's Inequality guarantees that the number you get on the left side will never be larger than the number you get on the right side.

What Are Conjugate Exponents?

The engine of Hölder's Inequality is the special relationship between the exponents $p$ and $q$ . They are known as conjugate exponents or Hölder conjugates.

Two real numbers $p$ and $q$ are conjugate exponents if they are both greater than $1$ and their reciprocals add up to exactly $1$ .

\frac{1}{p} + \frac{1}{q} = 1

This simple equation creates a partnership between $p$ and $q$ . If you know one, you can always find the other. Let's say you know $p$ . To find its conjugate $q$ , you can rearrange the formula:

\frac{1}{q} = 1 - \frac{1}{p}

\frac{1}{q} = \frac{p - 1}{p}

q = \frac{p}{p - 1}

Let's look at some common pairs of conjugate exponents:

If we choose $p = 2$ , then $\frac{1}{q} = 1 - \frac{1}{2} = \frac{1}{2}$ , which means $q = 2$ . This is a very special case that leads to another famous inequality.
If we choose $p = 3$ , then $\frac{1}{q} = 1 - \frac{1}{3} = \frac{2}{3}$ , which means $q = \frac{3}{2}$ .
If we choose $p = 4$ , then $\frac{1}{q} = 1 - \frac{1}{4} = \frac{3}{4}$ , which means $q = \frac{4}{3}$ .

Notice a pattern? As $p$ gets larger, $q$ gets closer to $1$ . Here is a table showing some more pairs:

Value of $p$	Calculation for $q$	Value of $q$
$2$	$q = \frac{2}{2 - 1}$	$2$
$3$	$q = \frac{3}{3 - 1}$	$\frac{3}{2}$ or $1.5$
$4$	$q = \frac{4}{4 - 1}$	$\frac{4}{3}$ or $\approx 1.33$
$5 / 2$ or $2.5$	$q = \frac{5 / 2}{(5 / 2) - 1} = \frac{5 / 2}{3 / 2}$	$\frac{5}{3}$ or $\approx 1.67$
$10$	$q = \frac{10}{10 - 1}$	$\frac{10}{9}$ or $\approx 1.11$

Understanding this relationship is crucial because you can't just pick any two exponents for Hölder's Inequality; they must be a conjugate pair.

A Familiar Friend: The Cauchy-Schwarz Inequality

Before diving into a full Hölder's example, let's look at its most famous special case. What happens when we pick the conjugate exponents $p = 2$ and $q = 2$ ?

Let's substitute these values into the Hölder's Inequality formula:

\sum_{k = 1}^{n} a_{k} b_{k} \leq {(\sum_{k = 1}^{n} a_{k}^{2})}^{1 / 2} {(\sum_{k = 1}^{n} b_{k}^{2})}^{1 / 2}

This is the celebrated Cauchy-Schwarz Inequality! You may have seen it before. Raising something to the power of $1 / 2$ is the same as taking the square root. So, we can write it like this:

\sum_{k = 1}^{n} a_{k} b_{k} \leq \sqrt{\sum_{k = 1}^{n} a_{k}^{2}} \cdot \sqrt{\sum_{k = 1}^{n} b_{k}^{2}}

Often, people square both sides to get rid of the square roots (since both sides are non-negative, this is allowed):

{(\sum_{k = 1}^{n} a_{k} b_{k})}^{2} \leq (\sum_{k = 1}^{n} a_{k}^{2}) (\sum_{k = 1}^{n} b_{k}^{2})

Thinking of Hölder's Inequality as a "supercharged" version of the Cauchy-Schwarz Inequality can make it feel less intimidating. It generalizes the same core idea to work with any pair of conjugate exponents, not just $(2, 2)$ .

How Do You Apply Hölder's Inequality? (Example 1)

Let's put the inequality to the test with a concrete example. We'll follow a clear, step-by-step process.

Example 1

Let's verify Hölder's Inequality for the sequences $A = (1, 3)$ and $B = (4, 2)$ using the conjugate exponents $p = 4$ and $q = 4 / 3$ .

Step 1: Identify your sequences and exponents.

Our sequences are $a_{k} = (1, 3)$ and $b_{k} = (4, 2)$ . So, $a_{1} = 1$ , $a_{2} = 3$ , $b_{1} = 4$ , and $b_{2} = 2$ .

Our exponents are $p = 4$ and $q = 4 / 3$ . Let's check if they are conjugates: $\frac{1}{4} + \frac{1}{4 / 3} = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1$ . Yes, they are!

Step 2: Calculate the Left-Hand Side (LHS).

The LHS is $\sum_{k = 1}^{2} a_{k} b_{k}$ .

LHS = a_{1} b_{1} + a_{2} b_{2}

LHS = (1) (4) + (3) (2)

LHS = 4 + 6 = 10

So, the left side of our inequality is $10$ .

Step 3: Calculate the first part of the Right-Hand Side (RHS).

The first part is ${(\sum_{k = 1}^{2} a_{k}^{p})}^{1 / p}$ . We use $p = 4$ .

First, the sum: $\sum a_{k}^{4} = a_{1}^{4} + a_{2}^{4} = 1^{4} + 3^{4} = 1 + 81 = 82$ .

Now, we take the $1 / p$ -th power: $(82)^{1 / 4}$ . This is the fourth root of $82$ . Using a calculator, $\sqrt[4]{82} \approx 3.0092$ .

Step 4: Calculate the second part of the Right-Hand Side (RHS).

The second part is ${(\sum_{k = 1}^{2} b_{k}^{q})}^{1 / q}$ . We use $q = 4 / 3$ .

First, the sum: $\sum b_{k}^{4 / 3} = b_{1}^{4 / 3} + b_{2}^{4 / 3} = 4^{4 / 3} + 2^{4 / 3}$ .

This can be tricky to calculate by hand. $4^{4 / 3} = (4^{4})^{1 / 3} = 256^{1 / 3} \approx 6.3496$ . And $2^{4 / 3} = (2^{4})^{1 / 3} = 16^{1 / 3} \approx 2.5198$ .

The sum is approximately $6.3496 + 2.5198 = 8.8694$ .

Now, we take the $1 / q$ -th power, which is the $3 / 4$ power: $(8.8694)^{3 / 4}$ . Using a calculator, this is approximately $5.093$ .

Step 5: Compare the LHS and RHS.

The final step is to multiply the two parts of the RHS together.

RHS \approx (3.0092) \times (5.093) \approx 15.3258

Now we check the inequality: Is $LHS \leq RHS$ ?

10 \leq 15.3258

Yes, it is! The inequality holds true. This process demonstrates how the inequality provides an upper bound for the sum of the products.

A Second Worked Example with Three Terms

Let's try another example, this time with sequences of length three, to solidify our understanding.

Example 2

Verify Hölder's Inequality for the sequences $A = (2, 1, 4)$ and $B = (3, 5, 2)$ using the special case of the Cauchy-Schwarz Inequality, where $p = 2$ and $q = 2$ .

Step 1: Identify sequences and exponents.

Sequences: $a_{k} = (2, 1, 4)$ and $b_{k} = (3, 5, 2)$ .

Exponents: $p = 2$ and $q = 2$ . We already know they are conjugates.

Step 2: Calculate the LHS.

LHS = \sum_{k = 1}^{3} a_{k} b_{k} = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}

LHS = (2) (3) + (1) (5) + (4) (2)

LHS = 6 + 5 + 8 = 19

Step 3: Calculate the first part of the RHS.

This is ${(\sum_{k = 1}^{3} a_{k}^{p})}^{1 / p}$ with $p = 2$ .

The sum: $\sum a_{k}^{2} = a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = 2^{2} + 1^{2} + 4^{2} = 4 + 1 + 16 = 21$ .

The $1 / p$ -th power: $(21)^{1 / 2} = \sqrt{21}$ .

Step 4: Calculate the second part of the RHS.

This is ${(\sum_{k = 1}^{3} b_{k}^{q})}^{1 / q}$ with $q = 2$ .

The sum: $\sum b_{k}^{2} = b_{1}^{2} + b_{2}^{2} + b_{3}^{2} = 3^{2} + 5^{2} + 2^{2} = 9 + 25 + 4 = 38$ .

The $1 / q$ -th power: $(38)^{1 / 2} = \sqrt{38}$ .

Step 5: Compare the LHS and RHS.

The full RHS is the product of the two parts: $\sqrt{21} \cdot \sqrt{38} = \sqrt{21 \times 38} = \sqrt{798}$ .

Now we check the inequality: Is $19 \leq \sqrt{798}$ ? To check this without a calculator, we can square both sides.

19^{2} = 361

(\sqrt{798})^{2} = 798

Is $361 \leq 798$ ? Yes, it is. The inequality holds.

Key formulas for holders inequality by Algebra911.

When Does Equality Hold?

In mathematics, it's always important to ask when an inequality becomes an equality. For Hölder's Inequality, the less-than-or-equal-to sign $\leq$ becomes a simple equals sign $=$ only under a very specific condition.

Equality holds if and only if the sequence of $a_{k}^{p}$ is proportional to the sequence of $b_{k}^{q}$ . This means that there is a constant, non-negative number $c$ such that for every term $k$ :

a_{k}^{p} = c \cdot b_{k}^{q}

This is a strict condition. If even one pair of terms ( $a_{k}, b_{k}$ ) doesn't fit this rule, the inequality will be strict (the left side will be strictly less than the right side). Let's construct an example where equality holds.

Example 3

Show that equality holds for the sequences $A = (1, 2)$ and $B = (1, 4)$ with exponents $p = 3$ and $q = 3 / 2$ .

Step 1: Check the condition for equality.

The condition is $a_{k}^{p} = c \cdot b_{k}^{q}$ . Here, $p = 3$ and $q = 3 / 2$ . We need to see if there is a constant $c$ such that $a_{k}^{3} = c \cdot b_{k}^{3 / 2}$ .

For $k = 1$ : $a_{1}^{3} = 1^{3} = 1$ . And $b_{1}^{3 / 2} = 1^{3 / 2} = 1$ . So $1 = c \cdot 1$ , which implies $c = 1$ .

For $k = 2$ : Let's check if $c = 1$ also works here. We need to check if $a_{2}^{3} = 1 \cdot b_{2}^{3 / 2}$ .

$a_{2}^{3} = 2^{3} = 8$ .

$b_{2}^{3 / 2} = 4^{3 / 2} = (\sqrt{4})^{3} = 2^{3} = 8$ .

Since $8 = 8$ , the condition holds with $c = 1$ . This means we should find that the LHS equals the RHS.

Step 2: Calculate the LHS and RHS to confirm.

LHS: $\sum a_{k} b_{k} = (1) (1) + (2) (4) = 1 + 8 = 9$ .

RHS Part 1: ${(\sum a_{k}^{p})}^{1 / p} = (1^{3} + 2^{3})^{1 / 3} = (1 + 8)^{1 / 3} = 9^{1 / 3}$ .

RHS Part 2: ${(\sum b_{k}^{q})}^{1 / q} = (1^{3 / 2} + 4^{3 / 2})^{1 / (3 / 2)} = (1 + 8)^{2 / 3} = 9^{2 / 3}$ .

Full RHS: $9^{1 / 3} \cdot 9^{2 / 3} = 9^{(1 / 3 + 2 / 3)} = 9^{1} = 9$ .

Since LHS = $9$ and RHS = $9$ , we have confirmed that equality holds. This happens precisely because the sequences satisfied the proportionality condition.

What Are Some Common Mistakes?

Hölder's Inequality is powerful, but it's easy to make small mistakes when you're first learning. Here are some common pitfalls to watch out for:

Forgetting the Conjugate Condition: The most common error is picking exponents $p$ and $q$ that don't satisfy $\frac{1}{p} + \frac{1}{q} = 1$ . The inequality is only guaranteed to be true if they are a conjugate pair. Always check this first!
Mixing Up Exponents: It's easy to get the exponents confused. Remember, on the right-hand side, the terms inside the summation are raised to the power of $p$ or $q$ , but the entire sum is then raised to the power of $1 / p$ or $1 / q$ . Don't swap them.
Errors with Fractional Exponents: Calculating terms like $8^{2 / 3}$ can be tricky. Remember the rule: $x^{a / b} = (\sqrt[b]{x})^{a}$ . Break it down into two steps: take the root first (the denominator), then apply the power (the numerator). Taking the root first keeps the numbers smaller and easier to manage.
Assuming Equality: Don't assume the two sides are equal. The vast majority of the time, the left side will be strictly smaller than the right side. Equality is a very special case.
Negative Numbers: The standard form of Hölder's Inequality that we've discussed applies to sequences of non-negative real numbers. Applying it to negative numbers requires using absolute values: $\sum | a_{k} b_{k} | \leq (\sum | a_{k} |^{p})^{1 / p} (\sum | b_{k} |^{q})^{1 / q}$ . For now, just stick to positive numbers until you're comfortable.

Quick Summary

Feeling a bit overwhelmed? Let's boil it all down to the essentials. Here's what you need to remember about Hölder's Inequality.

The Main Idea: It provides an upper limit on the sum of products.

The Formula:

\sum_{k = 1}^{n} a_{k} b_{k} \leq {(\sum_{k = 1}^{n} a_{k}^{p})}^{1 / p} {(\sum_{k = 1}^{n} b_{k}^{q})}^{1 / q}

The Key Conditions:

The terms in the sequences ( $a_{k}$ and $b_{k}$ ) must be non-negative.
The exponents $p$ and $q$ must be conjugate exponents, meaning $p > 1$ , $q > 1$ , and they must satisfy the crucial equation: $\frac{1}{p} + \frac{1}{q} = 1$ .

Special Cases:

When $p = 2$ and $q = 2$ , Hölder's Inequality becomes the more famous Cauchy-Schwarz Inequality.

When Equality Occurs:

The two sides are exactly equal if and only if the sequence $a_{k}^{p}$ is a constant multiple of the sequence $b_{k}^{q}$ .

Frequently Asked Questions

Why is Hölder's Inequality important?

It's a foundational tool in many areas of higher mathematics, like analysis and probability theory. It helps prove other important results and provides a way to measure how 'related' two sequences are, which is a concept that appears in geometry and data science.

What happens if p=1?

If you try to set $p = 1$ , the conjugate equation $1 / 1 + 1 / q = 1$ would mean $1 / q = 0$ , which is impossible for a finite number $q$ . This is treated as a special case where $q$ is considered to be infinity, leading to a simpler inequality that's easy to prove on its own.

Is this related to the Pythagorean theorem?

In a way, yes! The Cauchy-Schwarz case ( $p = q = 2$ ) is deeply connected to the geometry of vectors and the concept of dot products. The length of a vector is calculated using sums of squares, just like in the inequality, which is directly related to the Pythagorean theorem in higher dimensions.

Do the sequences have to be the same length?

Yes, for the standard formula to work, both sequences must have the same number of terms, $n$ . This is because you are pairing up terms ( $a_{1}$ with $b_{1}$ , $a_{2}$ with $b_{2}$ , etc.), so you can't have any terms left over.

Can I use this inequality for more than two sequences?

Absolutely! There is a generalized version of Hölder's Inequality that works for three or more sequences. It involves three or more exponents whose reciprocals must add up to 1, like $\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 1$ .

Where does the name Hölder come from?

The inequality is named after Otto Hölder, a German mathematician who published it in 1889. However, a similar result was discovered a year earlier by Leonard Rogers, so you might sometimes see it called the Rogers-Hölder inequality.

Do I need a calculator to solve problems with this?

For many textbook problems, the numbers are chosen carefully so that roots and powers work out nicely, like $\sqrt[3]{8} = 2$ . However, as our first example showed, applying the inequality to random sequences will often result in numbers that require a calculator to evaluate.