# Total Internal Reflection

Technology and Art

# Important Inequalities in Functional Analysis

Avishek Sen Gupta on 27 September 2021

Continuing my self-study of Functional Analysis, this post describes proofs for the following important inequalities in the subject:

• Young’s Inequality
• Hölder’s Inequality
• Minkowski’s Inequality

The paths of the proofs closely follow Erwin Kreyszig’s Introductory Functional Analysis with Applications.

## Young’s Inequality

We begin with the idea of conjugate exponents, which we call $$p$$ and $$q$$, related like so:

$\frac{1}{p} + \frac{1}{q} = 1 \\ \Rightarrow q + p = pq \\ \Rightarrow q + p = pq \\ \Rightarrow p(q - 1)= q \\ \Rightarrow p=\frac{q}{q-1} \\ \Rightarrow 1-p=1-\frac{q}{q-1} \\ \Rightarrow 1-p=\frac{q-1-q}{q-1}$ $$$\Rightarrow 1-p=\frac{1}{1-q} \label{eq:conjugate-exponents-property-1}$$$

where $$p>1$$. We now use these exponents in the following identity:

$u=t^{p-1}$

We also record the following implication.

$t=u^{\frac{1}{p-1}} \\ \Rightarrow t=u^{q-1}$

For $$p=2$$, we get $$u=t$$, which is the equation of the straight line $$x=y$$. The above follows from $$\eqref{eq:conjugate-exponents-property-1}$$.

Let $$\alpha, \beta >0: \alpha, \beta \in \mathbb{R}$$, then in $$\mathbb{R}^2$$, $$\alpha\beta$$ describes the area of a rectangle. Let us plot the graph of $$u=t^{p-1}$$ for $$p=2$$, so that $$u=t$$.

If we integrate $$\displaystyle\int\limits_0^\alpha u.dt=\int\limits_0^\alpha t^{p-1}.dt$$, and $$\displaystyle\int\limits_0^\alpha t.du=\int\limits_0^\alpha u^{q-1}.du$$, we can compute an area like this:

\begin{align*} S &= \int\limits_0^\alpha t^{p-1}.dt + \int\limits_0^\alpha u^{q-1}.du \\ &= \frac{\alpha^2}{2} + \frac{\beta^2}{2} \end{align*}

You will notice that regardless of the choice of $$\alpha, \beta$$, there will always be a small portion of $$S$$ which is bigger than $$\alpha\beta$$. The only situation in which there is no extra area is when $$\alpha=\beta$$. Therefore, we can say that:

$\alpha \beta \leq \frac{\alpha^2}{2} + \frac{\beta^2}{2}$

This less-than-or-equal relation carries over to other values of $$p$$ where $$u=t^{p-1}$$ is an exponential graph. The following two graphs illustrate how this is always true.

Thus, we can conclude that:

$$$\alpha \beta \leq \frac{\alpha^p}{p} + \frac{\beta^q}{q} \label{eq:youngs-inequality}$$$

$$\eqref{eq:youngs-inequality}$$ is called Young’s Equality, and we will use it to prove Hölder’s Inequality next.

A more interesting way to think about (and remember) Young’s Equality is to note that $$\text{log }(x)$$ is a concave function, therefore by the definition of concavity, we have:

$f(\alpha x^p + (1 - \alpha) y^q) \geq \alpha f(x^p) + (1-\alpha) f(y^q) \\ \text{log }(\alpha x^p + (1 - \alpha) y^q) \geq \alpha\text{log }(x^p) + (1-\alpha) \text{log }(y^q) \\ \text{log }\left(\frac {x^p}{p} + \frac {y^q}{q}\right) \geq \frac{\text{log }(x^p)}{p} + \frac{\text{log }(y^q)}{q} = \text{log }(x^{(p/p)}) + \text{log }(y^{(q/q)}) \\ \text{log }\left(\frac {x^p}{p} + \frac {y^q}{q}\right) \geq \text{log }(xy) \\ \Rightarrow xy \leq \left(\frac {x^p}{p} + \frac {y^q}{q}\right)$

## Hölder’s Inequality

We look at $$\ell^p$$ spaces now. Briefly recapping, $$\ell^p$$ spaces are spaces of sequences. A sequence $$\xi=\{\xi_1, \xi_2, \cdots\} \in\ell^p$$ must satisfy the following condition:

$\sum\limits_{i=1}^\infty{|\xi_i|}^p<\infty$

The norm $$\|\bullet\|$$ in $$\ell^p$$ spaces is usually defined as:

$\|\bullet\|={\left(\sum\limits_{i=1}^\infty{|\xi_i|}^p\right)}^{\frac{1}{p}}$

which induces the distance metric between two sequences $$\xi, \eta \in \ell_p$$:

$d(\xi,\eta)={\left(\sum\limits_{i=1}^\infty{|\xi_i-\eta_i|}^p\right)}^{\frac{1}{p}}$

Pick any $$\xi,\eta\in\ell^p$$.

Let us pick any two corresponding terms in $$\xi$$ and $$\eta$$, and let $$\alpha=\vert\xi_i\vert$$ and $$\beta_i=\vert\eta\vert$$, since we cannot guarantee these terms will be positive.

Then, from Young’s Inequality $$\eqref{eq:youngs-inequality}$$, we get:

$|\xi_i||\eta_i| \leq \frac{ {|\xi_i|}^p}{p} + \frac{ {|\eta_i|}^q}{q} \\ \Rightarrow |\xi_i\eta_i| \leq \frac{ {|\xi_i|}^p}{p} + \frac{ {|\eta_i|}^q}{q}$

Summing over all $$i$$, we get:

$$$\displaystyle\sum\limits_{i=1}^\infty|\xi_i\eta_i| \leq \frac{\sum\limits_{i=1}^\infty{|\xi_i|}^p}{p} + \frac{\sum\limits_{i=1}^\infty{|\eta_i|}^q}{q} \label{eq:holders-youngs-inequality-application}$$$

What we’d like to do is prove that: $$\displaystyle\sum\limits_{i=1}^\infty\vert\xi_i\eta_i\vert \leq 1$$. The only identity immediately available is the conjugate exponent identity, namely:

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

In order to be able to set the RHS of $$\eqref{eq:holders-youngs-inequality-application}$$, we need the following condition:

$$$\displaystyle\sum\limits_{i=1}^\infty{ {\vert\xi_i\vert}^p}=\displaystyle\sum_{i=1}^\infty{ {\vert\eta_i\vert}^q}=1 \label{eq:holders-unity-condition}$$$

Let us then assume $$\eqref{eq:holders-unity-condition}$$. Then, $$\eqref{eq:holders-youngs-inequality-application}$$ becomes:

$$$\displaystyle\sum\limits_{i=1}^\infty|\xi_i\eta_i| \leq 1 \label{eq:holders-less-than-unity-condition}$$$

We need to determine what sort of $$\xi$$ and $$\eta$$ can satisfy this condition. Let’s take $$\xi$$ as an example. We have:

$\xi=(\xi_1, \xi_2, \cdots)$

Remember the norm for $$\ell^p$$ spaces? Here it is again:

$\|\bullet\|={\left(\sum\limits_{i=1}^\infty{|\xi_i|}^p\right)}^{\frac{1}{p}}$

If we divide each term in $$\xi$$ by its norm:

$$$\xi_i = \frac{\bar\xi_i}{ {\left(\sum\limits_{i=1}^\infty{|\bar{\xi_i}|}^p\right)}^{\frac{1}{p}}} \label{eq:holders-sequence-scale-factor}$$$

Then, from $$\eqref{eq:holders-unity-condition}$$, we get:

$\displaystyle\sum\limits_{i=1}^\infty{ {|\xi_i|}^p}=\frac{1}{ \sum\limits_{i=1}^\infty{|\bar{\xi_i}|}^p} \left({|\bar{\xi_1}|}^p + {|\bar{\xi_2}|}^p + \cdots\right) \\ =\frac{\sum\limits_{i=1}^\infty{|\bar{\xi_i}|}^p}{\sum\limits_{i=1}^\infty{|\bar{\xi_i}|}^p} = 1$

which satisfies condition $$\eqref{eq:holders-unity-condition}$$, regardless of which sequence $$\bar{\xi}$$ we choose from $$\ell^p$$.

Applying the same logic to $$\eta_i$$, and substituting $$\eqref{eq:holders-sequence-scale-factor}$$ into $$\eqref{eq:holders-less-than-unity-condition}$$, we get:

$\displaystyle\sum\limits_{i=1}^\infty |\bar\xi_i \bar\eta_i| \leq {\left(\displaystyle\sum\limits_{i=1}^\infty{ {|\bar\xi_i|}^p}\right)}^{\frac{1}{p}} \bullet {\left(\displaystyle\sum\limits_{i=1}^\infty{ {|\bar\eta_i|}^q}\right)}^{\frac{1}{q}}$

Removing the overbars from $$\bar\xi_i$$ $$\bar\eta_i$$ to indicate any two sequences in an $$\ell^p$$ space, we get:

$$$\displaystyle\sum\limits_{i=1}^\infty |\xi_i \eta_i| \leq {\left(\displaystyle\sum\limits_{i=1}^\infty{ {|\xi_i|}^p}\right)}^{\frac{1}{p}} \bullet {\left(\displaystyle\sum\limits_{i=1}^\infty{ {|\eta_i|}^q}\right)}^{\frac{1}{q}} \label{eq:holders-inequality}$$$

The result $$\eqref{eq:holders-inequality}$$ is referred to as Hölder’s Inequality.

In the case of $$p=2$$, $$q=2$$; this special case is called the Cauchy-Schwarz Inequality, which is:

$$$\displaystyle\sum\limits_{i=1}^\infty |\xi_i \eta_i| \leq \sqrt{\left(\displaystyle\sum\limits_{i=1}^\infty{ {|\xi_i|}^2}\right) \bullet \left(\displaystyle\sum\limits_{i=1}^\infty{ {|\eta_i|}^2}\right)} \label{eq:cauchy-schwarz-inequality}$$$

Hölder’s Inequality will be used to prove Minkowski’s Inequality next.

## Minkowski’s Inequality

Minkowski’s Inequality is a generalisation of the Triangle Inequality. As usual, we assume $$\xi, \eta \in \ell^p$$. We start with writing (for economy of notation):

$\omega_i=|\xi_i+\eta_i| \\ \Rightarrow {\omega_i}^p={|\xi_i+\eta_i|}^p \\ \Rightarrow {\omega_i}^p={\omega_i}^{p-1}|\xi_i+\eta_i|$

Summing up over $$i$$, we get:

$$$\displaystyle\sum\limits_{i=1}^\infty{\omega_i}^p = \displaystyle\sum\limits_{i=1}^\infty {\omega_i}^{p-1}|\xi_i+\eta_i| \label{eq:minkowski-separated-summed}$$$

Now applying the Triangle Inequality to the second term on the RHS of $$\eqref{eq:minkowski-separated-summed}$$, we get:

$$$\displaystyle\sum\limits_{i=1}^\infty{\omega_i}^p = \displaystyle\sum\limits_{i=1}^\infty\omega^{p-1} \bullet \underbrace{|\xi_i+\eta_i|}_\text{Apply Triangle Inequality} \\ \Rightarrow \displaystyle\sum\limits_{i=1}^\infty{\omega_i}^p \leq \displaystyle\sum\limits_{i=1}^\infty\omega^{p-1}(|\xi_i|+|\eta_i|) \\ \Rightarrow \displaystyle\sum\limits_{i=1}^\infty{\omega_i}^p \leq \displaystyle\sum\limits_{i=1}^\infty\omega^{p-1} |\xi_i| + \displaystyle\sum\limits_{i=1}^\infty\omega^{p-1} |\eta_i| \\ \label{eq:minkowski-separated-summed-inequality}$$$

Apply Hölder’s Inequality $$\eqref{eq:holders-inequality}$$ to each term on the RHS individually in $$\eqref{eq:minkowski-separated-summed-inequality}$$, we have:

$\displaystyle\sum\limits_{i=1}^\infty\omega^{p-1} |\xi_i| \leq \displaystyle\sum\limits_{i=1}^\infty|\omega^{p-1} \xi_i| \leq {\left[\displaystyle\sum\limits_{i=1}^\infty{\left({|\omega_i|}^{p-1}\right)}^q\right]}^\frac{1}{q} {\left[\displaystyle\sum\limits_{i=1}^\infty{|\xi_i|}^p\right]}^\frac{1}{p} \\ \displaystyle\sum\limits_{i=1}^\infty\omega^{p-1} |\xi_i| \leq \displaystyle\sum\limits_{i=1}^\infty|\omega^{p-1} \xi_i| \leq {\left[\displaystyle\sum\limits_{i=1}^\infty{\left({|\omega_i|}^{p-1}\right)}^q\right]}^\frac{1}{q} {\left[\displaystyle\sum\limits_{i=1}^\infty{|\eta_i|}^p\right]}^\frac{1}{p}$

Note that since $$p$$ and $$q$$ are conjugate exponents, we can write:

$p=pq-q$

Then the above inequalities simplify to:

$$$\displaystyle\sum\limits_{i=1}^\infty\omega^{p-1} |\xi_i| \leq \displaystyle\sum\limits_{i=1}^\infty|\omega^{p-1} \xi_i| \leq {\left[\displaystyle\sum\limits_{i=1}^\infty{|\omega_i|}^p\right]}^\frac{1}{q} {\left[\displaystyle\sum\limits_{i=1}^\infty{|\xi_i|}^p\right]}^\frac{1}{p} \label{eq:minkowski-holder-inequality-1}$$$ $$$\displaystyle\sum\limits_{i=1}^\infty\omega^{p-1} |\xi_i| \leq \displaystyle\sum\limits_{i=1}^\infty|\omega^{p-1} \xi_i| \leq {\left[\displaystyle\sum\limits_{i=1}^\infty|{\omega_i|}^p\right]}^\frac{1}{q} {\left[\displaystyle\sum\limits_{i=1}^\infty{|\eta_i|}^p\right]}^\frac{1}{p} \label{eq:minkowski-holder-inequality-2}$$$

Applying $$\eqref{eq:minkowski-holder-inequality-1}$$ and $$\eqref{eq:minkowski-holder-inequality-2}$$ to $$\eqref{eq:minkowski-separated-summed-inequality}$$, we get:

$\displaystyle\sum\limits_{i=1}^\infty{\omega_i}^p \leq {\left[\displaystyle\sum\limits_{i=1}^\infty{|\omega_i|}^p\right]}^\frac{1}{q} {\left[\displaystyle\sum\limits_{i=1}^\infty{|\xi_i|}^p\right]}^\frac{1}{p} + {\left[\displaystyle\sum\limits_{i=1}^\infty|{\omega_i|}^p\right]}^\frac{1}{q} {\left[\displaystyle\sum\limits_{i=1}^\infty{|\eta_i|}^p\right]}^\frac{1}{p} \\ \displaystyle\sum\limits_{i=1}^\infty{\omega_i}^p \leq {\left[\displaystyle\sum\limits_{i=1}^\infty{|\omega_i|}^p\right]}^\frac{1}{q} \left({\left[\displaystyle\sum\limits_{i=1}^\infty{|\xi_i|}^p\right]}^\frac{1}{p} + {\left[\displaystyle\sum\limits_{i=1}^\infty{|\eta_i|}^p\right]}^\frac{1}{p}\right) \\ {\left[\displaystyle\sum\limits_{i=1}^\infty{|\omega_i|}^p\right]}^{1-\frac{1}{q}} \leq {\left[\displaystyle\sum\limits_{i=1}^\infty{|\xi_i|}^p\right]}^\frac{1}{p} + {\left[\displaystyle\sum\limits_{i=1}^\infty{|\eta_i|}^p\right]}^\frac{1}{p}$

Again, noting from the conjugate exponent identity that $$1-\frac{1}{q}=\frac{1}{p}$$, we get:

$$${\left[\displaystyle\sum\limits_{i=1}^\infty{|\omega_i|}^p\right]}^\frac{1}{p} \leq {\left[\displaystyle\sum\limits_{i=1}^\infty{|\xi_i|}^p\right]}^\frac{1}{p} + {\left[\displaystyle\sum\limits_{i=1}^\infty{|\eta_i|}^p\right]}^\frac{1}{p} \label{eq:minkowski-inequality}$$$

$$\eqref{eq:minkowski-inequality}$$ is referred to as Minkowski’s Inequality.

tags: Mathematics - Proof - Functional Analysis - Pure Mathematics