# Hölder’s Inequality

Hölder’s inequality, a generalized form of Cauchy Schwarz inequality, is an inequality of sequences that generalizes multiple sequences and different exponents.

It states that if {an}, {bn}, …, {zn} are the sequences and λa + λb + … + λz = 1, then the inequality

${\left( a_{1}+a_{2}+\ldots +a_{n}\right) ^{\lambda _{a}}\ldots \left( z_{1}+z_{2}+\ldots +z_{n}\right) ^{\lambda _{z}}\geq a_{1}^{\lambda _{a}}b_{1}^{\lambda _{b}}\ldots z_{1}^{\lambda _{z}}+\ldots +a_{n}^{\lambda _{a}}b_{n}^{\lambda _{b}}\ldots z_{n}^{\lambda _{z}}}$

Here, if λa = λb = ${\dfrac{1}{2}}$, Hölder’s inequality reduces to the elementary form of Cauchy Schwarz inequality.

Thus, Hölder’s inequality can be stated as:

If {an}, {bn} are two sequences of positive real numbers and ${\left\{ \lambda _{i}\right\} _{i=1}^{n}}$ is the sequence for p, q > 1 such that ${\dfrac{1}{p}=\lambda _{a}}$, ${\dfrac{1}{q}=\lambda _{b}}$ and ${\Sigma \lambda =1}$, then

${\sum ^{n}_{i=1}a_{i}b_{i}\leq \left( \sum ^{n}_{i=1}a_{i}^{p}\right) ^{\dfrac{1}{p}}\left( \sum ^{n}_{i=1}b_{i}^{q}\right) ^{\dfrac{1}{q}}}$, which is the elementary form of Hölder’s inequality for sums.

## Proof

Let us define ${a=\dfrac{a_{i}}{\left( \sum ^{n}_{i=1}a_{i}^{p}\right) ^{\dfrac{1}{p}}}}$,

${b=\dfrac{b_{i}}{\left( \sum ^{n}_{i=1}b_{i}^{q}\right) ^{\dfrac{1}{q}}}}$, here i = 1, 2, …, n

By Young’s inequality, we have

${\dfrac{a_{i}b_{i}}{\left( \sum ^{n}_{i=1}a_{i}^{p}\right) ^{\dfrac{1}{p}}\left( \sum ^{n}_{i=1}b_{i}^{q}\right) ^{\dfrac{1}{q}}}\leq \dfrac{1}{p}\dfrac{a_{i}^{p}}{\sum ^{n}_{i=1}a_{i}^{p}}+\dfrac{1}{q}\dfrac{b_{i}^{q}}{\sum ^{n}_{i=1}b_{i}^{q}}}$ …..(i)

On adding the inequalities for i = 1, 2, …, n, we get

${\dfrac{\sum ^{n}_{i=1}a_{i}b_{i}}{\left( \sum ^{n}_{i=1}a_{i}^{p}\right) ^{\dfrac{1}{p}}\left( \sum ^{n}_{i=1}b_{i}^{q}\right) ^{\dfrac{1}{1}}}\leq \dfrac{1}{p}\dfrac{\sum ^{n}_{i=1}a_{i}^{p}}{\sum ^{n}_{i=1}a_{i}^{p}}+\dfrac{1}{q}\dfrac{\sum ^{n}_{i=1}b_{i}^{q}}{\sum ^{n}_{i=1}b_{i}^{q}}}$

Since ${\dfrac{1}{p}+\dfrac{1}{q}=1}$

Now, ${\dfrac{\sum ^{n}_{i=1}a_{i}b_{i}}{\left( \sum ^{n}_{i=1}a_{i}^{p}\right) ^{\dfrac{1}{p}}\left( \sum ^{n}_{i=1}b_{i}^{q}\right) ^{\dfrac{1}{1}}}\leq 1}$

⇒ ${\sum ^{n}_{i=1}a_{i}b_{i}\leq \left( \sum ^{n}_{i=1}a_{i}^{p}\right) ^{\dfrac{1}{p}}\left( \sum ^{n}_{i=1}b_{i}^{q}\right) ^{\dfrac{1}{q}}}$, here the equality occurs if and only if ${\dfrac{a_{1}^{p}}{b_{1}^{q}}=\dfrac{a_{2}^{p}}{b_{2}^{q}}=\ldots =\dfrac{a_{n}^{p}}{b_{n}^{q}}}$

Thus, Hölder’s inequality is proved.

## Alternative Form

Moreover, if ${\dfrac{1}{p}+\dfrac{1}{q}=1}$ and p, q > 1, then Hölder’s inequality for integrals states that

${\int ^{b}_{a}\left| f\left( x\right) g\left( x\right) \right| dx\leq \left[ \int ^{b}_{a}\left| f\left( x\right) \right| ^{p}dx\right] ^{\dfrac{1}{p}}\left[ \int ^{b}_{a}\left| g\left( x\right) \right| ^{q}dx\right] ^{\dfrac{1}{q}}}$

Here, f Є Lp, g Є Lq, and fg Є L1

⇒ ${\left\| fg\right\| _{1}\leq \left\| f\right\| _{p}\left\| g\right\| _{q}}$

### Proof

If ||fp|| = 0, then f = 0 almost everywhere.

Similarly, if ||gq|| = 0, then g = 0 almost everywhere.

Let us consider for all x the functions f(x), g(x) Є ℝ

Also, ${\dfrac{1}{p}=\lambda _{a}}$, ${\dfrac{1}{q}=\lambda _{b}}$,

${a=\dfrac{\left| f\left( x\right) \right| ^{p}}{\left\| f\right\| _{p}^{p}}}$

${b=\dfrac{\left| g\left( x\right) \right| ^{q}}{\left\| g\right\| _{q}^{q}}}$

By Young’s inequality, we have

${\dfrac{\left| f\left( x\right) \right| }{\left\| f\right\| _{p}}\dfrac{\left| g\left( x\right) \right| }{\left\| g\right\| _{q}}\leq \dfrac{1}{p}\dfrac{\left| f\left( x\right) \right| ^{p}}{\left\| f\right\| _{p}^{p}}+\dfrac{1}{q}\dfrac{\left| g\left( x\right) \right| ^{q}}{\left\| g\right\| _{q}^{q}}}$

By integrating both sides, we get

${\dfrac{\left\| fg\right\| _{1}}{\left\| f\right\| _{p}\left\| g\right\| _{q}}\leq \dfrac{1}{p}\dfrac{\left\| f\left( x\right) \right\| ^{p}}{\left\| f\right\| _{p}^{p}}+\dfrac{1}{q}\dfrac{\left\| g\left( x\right) \right\| ^{q}}{\left\| g\right\| _{q}^{q}}}$

⇒ ${\dfrac{\left\| fg\right\| _{1}}{\left\| f\right\| _{p}\left\| g\right\| _{q}}\leq \dfrac{1}{p}+\dfrac{1}{q}}$

⇒ ${\dfrac{\left\| fg\right\| _{1}}{\left\| f\right\| _{p}\left\| g\right\| _{q}}\leq 1}$ (since ${\dfrac{1}{p}+\dfrac{1}{q}=1}$)

⇒ ${\left\| fg\right\| _{1}\leq \left\| f\right\| _{p}\left\| g\right\| _{q}}$

Thus, Hölder’s inequality is proved.