# 1.3 Inner product spaces and hilbert spaces

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Review of inner products and inner product spaces.

## Inner products

We have defined distances and norms to measure whether two signals are different from each other and to measure the “size” of a signal. However, it is possible for two pairs of signals with the same norms and distance to exhibit different behavior - an example of this contrast is to pick a pair of orthogonal signals and a pair of non-orthogonal signals, as shown in [link] .

To obtain a new metric that distinguishes between orthogonal and non-orthogonal we use the inner product , which provides us with a new metric of “similarity”.

Definition 1 An inner product for a vector space $\left(X,R,+,·\right)$ is a function $⟨·,·⟩:X×X\to R$ , sometimes denoted $\left(·|·\right)$ , with the following properties: for all $x,y,z\in X$ and $a\in R$ ,

1. $⟨x,y⟩$ = $\overline{⟨y,x⟩}$ (complex conjugate property),
2. $⟨x+y,z⟩$ = $⟨x,z⟩+⟨y,z⟩$ (distributive property),
3. $⟨\alpha x,y⟩$ = $\alpha ⟨x,y⟩$ (scaling property),
4. $⟨x,x⟩\ge 0$ and $⟨x,x⟩=0$ if and only if $x=0$ .

A vector space with an inner product is called an inner product space or a pre-Hilbert space.

It is worth pointing out that properties (2-3) say that the inner product is linear, albeit only on the first input. However, if $R=\mathbb{R}$ , then the properties (2-3) hold for both inputs and the inner product is linear on both inputs.

Just as every norm induces a distance, every inner product induces a norm: ${||x||}_{i}=\sqrt{⟨x,x⟩}$ .

## Hilbert spaces

Definition 2 An inner product space that is complete under the metric induced by the induced norm is called a Hilbert space .

Example 1 The following are examples of inner product spaces:

1. $X={\mathbb{R}}^{n}$ with the inner product $〈x,y〉={\sum }_{i=1}^{n}{x}_{i}{y}_{i}={y}^{T}x$ . The corresponding induced norm is given by ${||x||}_{i}=\sqrt{〈x,x〉}=\sqrt{{\sum }_{i=1}^{n}{x}_{i}^{2}}={||x||}_{2}$ , i.e., the ${\ell }_{2}$ norm. Since $\left({\mathbb{R}}^{n},\parallel ·{\parallel }_{2}\right)$ is complete, then it is a Hilbert space.
2. $X=C\left[T\right]$ with inner product $〈x,y〉={\int }_{T}x\left(t\right)y\left(t\right)dt$ . The corresponding induced norm is ${||x||}_{i}=\sqrt{{\int }_{T}x{\left(t\right)}^{2}dt}={||x||}_{2}$ , i.e., the ${L}_{2}$ norm.
3. If we allow for $X=C\left[T\right]$ to be complex-valued, then the inner product is defined by $〈x,y〉={\int }_{T}x\left(t\right)\overline{y\left(t\right)}dt$ , and the corresponding induced norm is ${||x||}_{i}=\sqrt{{\int }_{T}x\left(t\right)\phantom{\rule{3.33333pt}{0ex}}\overline{x\left(t\right)}dt}=\sqrt{{\int }_{T}{|x\left(t\right)|}^{2}dt}={||x||}_{2}$ .
4. $X={\mathbb{C}}^{n}$ with inner product $〈x,y〉={\sum }_{i=1}^{n}{x}_{i}\overline{{y}_{i}}={y}^{H}x$ ; here, ${x}^{H}$ denotes the Hermitian of $x$ . The corresponding induced norm is ${||x||}_{i}=\sqrt{{\sum }_{i=1}^{n}{|{x}_{i}|}^{2}}={||x||}_{2}$ .

Theorem 1 (Cauchy-Schwarz Inequality) Assume $X$ is an inner product space. For each $x,y\in X$ , we have that $|〈x,y〉|\le {||x||}_{i}{||y||}_{i}$ , with equality if ( $i$ ) $y=ax$ for some $a\in R$ ; ( $ii$ ) $x=0$ ; or ( $iii$ ) $y=0$ .

Proof: We consider two separate cases.

• if $y=0$ then $⟨x,y⟩=\overline{⟨y,x⟩}=\overline{⟨0·y,x⟩}=\overline{0}\overline{⟨y,x⟩}=0⟨x,y⟩=0={\parallel x\parallel }_{i}{\parallel y\parallel }_{i}$ . The proof is similar if $x=0$ .
• If $x,y\ne 0$ then $0\le ⟨x-ay,x-ay⟩=⟨x,x⟩-a⟨y,x⟩-\overline{a}⟨x,y⟩+a\overline{a}⟨y,y⟩$ , with equality if $x-ay=0$ , i.e., $x=ay$ for some $a\in R$ . Now set $a=\frac{⟨x,y⟩}{⟨y,y⟩}$ , and so $\overline{a}=\frac{⟨y,x⟩}{⟨y,y⟩}$ . We then have
$\begin{array}{cc}\hfill 0& \le ⟨x,x⟩-\frac{⟨x,y⟩}{⟨y,y⟩}⟨y,x⟩-\frac{⟨y,x⟩}{⟨y,y⟩}⟨x,y⟩+\frac{⟨x,y⟩}{⟨y,y⟩}\frac{⟨y,x⟩}{⟨y,y⟩}⟨y,y⟩\hfill \\ & \le ⟨x,x⟩-\frac{\overline{⟨x,y⟩}⟨x,y⟩}{{||y||}^{2}}={||x||}^{2}-\frac{{|⟨x,y⟩|}^{2}}{{||y||}^{2}}.\hfill \end{array}$
This implies $\frac{{|⟨x,y⟩|}^{2}}{{||y||}^{2}}\le {||x||}^{2}$ , and so since all quantities involved are positive we have $|⟨x,y⟩|\le ||x||·||y||$ .

## Properties of inner products spaces

In the previous lecture we discussed norms induced by inner products but failed to prove that they are valid norms. Most properties are easy to check; below, we check the triangle inequality for the induced norm.

Lemma 1 If ${\parallel x\parallel }_{i}=\sqrt{⟨x,x⟩}$ , then ${\parallel x+y\parallel }_{i}\le {\parallel x\parallel }_{i}+{\parallel y\parallel }_{i}$ .

From the definition of the induced norm,

$\begin{array}{cc}\hfill {∥x,+,y∥}_{i}^{2}& =〈x,+,y,,,x,+,y〉,\hfill \\ & =〈x,,,x〉+〈x,,,y〉+〈y,,,x〉+〈y,,,y〉,\hfill \\ & ={∥x∥}_{i}^{2}+〈x,,,y〉+\overline{〈x,,,y〉}+{∥y∥}_{i}^{2}\hfill \\ & ={∥x∥}_{i}^{2}+2\mathrm{real}\left(〈x,,,y〉\right)+{∥y∥}_{i}^{2}.\hfill \end{array}$

At this point, we can upper bound the real part of the inner product by its magnitude: $\mathrm{real}\left(〈x,,,y〉\right)\le |〈x,,,y〉|$ . Thus, we obtain

$\begin{array}{cc}\hfill {∥x,+,y∥}_{i}^{2}& \le {∥x∥}_{i}^{2}+2|〈x,,,y〉|+{∥y∥}_{i}^{2},\hfill \\ & \le {∥x∥}_{i}^{2}+2{∥x∥}_{i}{∥y∥}_{i}+{∥y∥}_{i}^{2},\hfill \\ & \le {\left({∥x∥}_{i}+∥{y}_{i}∥\right)}^{2},\hfill \end{array}$

where the second inequality is due to the Cauchy-Schwarz inequality. Thus we have shown that ${∥x,+,y∥}_{i}\le {∥x∥}_{i}+{∥y∥}_{i}$ . Here's an interesting (and easy to prove) fact about inner products:

Lemma 2 If $〈x,,,y〉=0$ for all $x\in X$ then $y=0$ .

Proof: Pick $x=y$ , and so $〈y,,,y〉=0$ . Due to the properties of an inner product, this implies that $y=0$ .

Earlier, we considered whether all distances are induced by norms (and found a counterexample). We can ask the same question here: are all norms induced by inner products? The following theorem helps us check for this property.

Theorem 2 (Parallelogram Law) If a norm $∥·∥$ is induced by an inner product, then ${∥x,+,y∥}^{2}+{∥x,-,y∥}^{2}=2\left({∥x∥}^{2}+{∥y∥}^{2}\right)$ for all $x,y\in X$ .

This theorem allows us to rule out norms that cannot be induced.

Proof: For an induced norm we have ${∥x∥}^{2}=〈x,,,x〉$ . Therefore,

$\begin{array}{cc}\hfill {∥x,+,y∥}^{2}+{∥x,-,y∥}^{2}& =〈x,+,y,,,x,+,y〉+〈x,-,y,,,x,-,y〉,\hfill \\ & =〈x,,,x〉+〈x,,,y〉+〈y,,,x〉+〈y,,,y〉+〈x,,,x〉-〈x,,,y〉-〈y,,,x〉+〈y,,,y〉,\hfill \\ & =2〈x,,,x〉+2〈y,,,y〉,\hfill \\ & =2\left({∥x∥}^{2}+{∥y∥}^{2}\right).\hfill \end{array}$

Example 2 Consider the normed space $\left(C\left[T\right],{L}_{\infty }\right)$ , and recall that ${∥x∥}_{\infty }={sup}_{t\in T}|x\left(t\right)|$ . If this norm is induced, then the Parallelogram law would hold. If not, then we can find a counterexample. In particular, let $T=\left[0,2\pi \right]$ , $x\left(t\right)=1$ , and $y\left(t\right)=cos\left(t\right)$ . Then, we want to check if ${∥x,+,y∥}^{2}+{∥x,-,y∥}^{2}=2\left({∥x∥}^{2}+{∥y∥}^{2}\right)$ . We compute:

$\begin{array}{cc}\hfill {∥x∥}_{\infty }& =1,\hfill \\ \hfill {∥y∥}_{\infty }& =1,\hfill \\ \hfill {∥x,+,y∥}_{\infty }& =∥1,+,cos,\left(,t,\right)∥=\underset{t\in T}{sup}|1+cos\left(t\right)|=1+1=2,\hfill \\ \hfill {∥x,-,y∥}_{\infty }& =∥1,-,cos,\left(,t,\right)∥=\underset{t\in T}{sup}|1-cos\left(t\right)|=1-\left(-1\right)=2.\hfill \end{array}$

Plugging into the two sides of the Parallelogram law,

$\begin{array}{cc}\hfill {2}^{2}+{2}^{2}& =2\left({1}^{2}+{1}^{2}\right),\hfill \\ \hfill 8& =4,\hfill \end{array}$

and the Parallelogram law does not hold. Thus, the ${L}_{\infty }$ norm is not an induced norm.

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