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\[\langle\ \,\ \ \rangle:V\times V\to\mathbb{F}\]
(the field \(\mathbb{F}\) could be \(\mathbb{R}\) or \(\mathbb{C}\)) and this function has the following properties:
1. \(\langle v,v\rangle\geq 0\) for all \(v\in V\) (positivity).
2. \(\langle v,v\rangle=0\Leftrightarrow v=0\) (definiteness).
3. \(\langle u+v,w\rangle=\langle u,w\rangle+\langle v,w\rangle\) for all \(u,v,w\in V\) (additivity in first slot).
4. \(\langle\alpha v,w\rangle=\alpha\langle v,w\rangle\) for all \(\alpha\in\mathbb{F}\) and all \(v,w\in V\) (homogeneity in first slot).
5. \(\langle v,w\rangle=\overline{\langle w,v\rangle}\) for all \(v,w\in V\) (conjugate symmetry).
Note that if we are in a _real_ inner product space, then \(\overline{\langle w,v\rangle}=\langle w,v\rangle\) because every real number is equal to its complex conjugate. Then for condition \(e\)), we simply write
\[\langle v,w\rangle=\langle w,v\rangle,\]
for all \(v,w\in V\).
**Example 66**.: The following are examples of inner products on various spaces. It is routine to check these functions actually define an inner product.
1. For \(x,y\in\mathbb{R}^{2}\), \[\langle x,y\rangle=2x_{1}y_{1}+3x_{2}y_{2}\] defines an inner product.
2. For \(x,y\in\mathbb{R}^{n}\), \[\langle x,y\rangle=x_{1}y_{1}+x_{2}y_{2}+\cdots+x_{n}y_{n}\] defines an inner product on \(\mathbb{R}^{n}\).
**Example 67**.: Let \(C[a,b]\) be the space of all continuous functions on \([a,b]\), namely,
\[C[a,b]:=\left\{f:\ f:[a,b]\to\mathbb{R},\ f\,\text{is continuous}\right\}.\]
Consider \(f,g\in C[a,b]\). We can define an inner product on \(C[a,b]\) as:
\[\langle f,g\rangle=\int_{a}^{b}f(x)g(x)dx.\]
Check the conditions given in Definition 59.
**Example 68**.: Let \(P_{2}(\mathbb{R})\) be the space of all polynomials of real coefficients with degree at most 2. Given \(p(x),g(x)\in P_{2}(\mathbb{R})\) where \(p(x)=a_{0}+a_{1}x+a_{2}x^{2}\) and \(g(x)=b_{0}+b_{1}x+b_{2}x^{2}\) then
\[\langle p,g\rangle=a_{0}b_{0}+a_{1}b_{1}+a_{2}b_{2},\]
defines an inner product on \(P_{2}(\mathbb{R})\).
### 2.1 Metric, Normed, and Inner Product Spaces
#### Orthogonal Decomposition
**Definition 60** (Orthogonal).: Two vectors \(u,v\in V\) are called _orthogonal_ if
\[\langle u,v\rangle=0.\]
Note that in the above definition order does not matter, since \(\langle u,v\rangle=0\) if and only if \(\langle v,u\rangle=0.\) Furthermore, the zero vector, \(0,\) is orthogonal to every vector and \(0\) is the only vector that is orthogonal to itself. Some mathematicians use the term _perpendicular_ instead of _orthogonal_.
**Theorem 67** (Pythagorean Theorem).: _If \(u\) and \(v\) are orthogonal vectors in \(V\), then_
\[||u+v||^{2}=||u||^{2}+||v||^{2}.\]
Proof
: Suppose \(u\) and \(v\) are orthogonal vectors in \(V\); then using the definition of the norm and the properties of the inner product we have:
\[||u+v||^{2} = \langle u+v,u+v\rangle=\langle u,u\rangle+\langle u,v\rangle+ \langle v,u\rangle+\langle v,v\rangle\] \[= ||u||^{2}+\langle u,v\rangle+\overline{\langle u,v\rangle}+||v|| ^{2}\] \[= ||u||^{2}+2\mathrm{Re}\langle u,v\rangle+||v||^{2}\] \[= ||u||^{2}+||v||^{2}.\qed\]
**Remark 37**.: The converse of the Pythagorean theorem holds in real inner product spaces. Note that in the above proof of Pythagorean theorem, the equality in Theorem 67 holds if and only if
\[\langle u,v\rangle+\langle v,u\rangle=2\mathrm{Re}\langle u,v\rangle=0.\]
However, if we are in a **real** inner product space, then
\[\langle u,v\rangle=\langle v,u\rangle=0.\]
Thus the converse of the Pythagorean theorem holds in real inner product spaces.
We are now ready to focus on orthogonal decomposition. Let \(u,v\in V\). Suppose we want to write \(u=\alpha v+w\) where \(\alpha\) is a scalar and \(w\) is a vector orthogonal to \(v\) (Figure 2.7). Expressing this in another way, we can write
\[u=\alpha v+(u-\alpha v).\]
Figure 2.7: Orthogonal decompositionTo find this scalar \(\alpha\), notice that we need to choose \(\alpha\) so that \(v\bot(u-\alpha v)\) which means that
\[\langle v,u-\alpha v\rangle = 0\] \[\Rightarrow\langle v,u\rangle-\alpha||v||^{2} = 0\] \[\Rightarrow\alpha = \frac{\langle v,u\rangle}{||v||^{2}}.\]
By assuming that \(v\neq 0\) we can write
\[u=\frac{\langle v,u\rangle}{||v||^{2}}v+\left(u-\frac{\langle v,u\rangle}{||v ||^{2}}v\right).\]
Thus, we were able to write \(u\) as a scalar multiple of \(v\) plus a vector \(w=u-\frac{\langle v,u\rangle}{||v||^{2}}v\) which is perpendicular to \(v\). The vector \(\alpha v\) is often named the _projection_ of \(u\) onto \(v\) and defined as:
\[\mathrm{Proj}_{v}(u)=\frac{\langle v,u\rangle}{||v||^{2}}v.\]
We will use this decomposition to prove the Cauchy-Schwarz-Bunyakovsky (CSB) inequality.
### The Cauchy-Schwarz-Bunyakovsky Inequality
The Cauchy-Schwarz-Bunyakovsky inequality is one of the most useful inequalities one can encounter in different settings. It has been used in linear algebra, analysis, and probability theory as well as other disciplines such as physics. In physics, the general formulation of the Heisenberg Uncertainty Principle, a fundamental part of quantum mechanics, is derived by using the CSB inequality in Hilbert spaces. The CSB inequality also has many generalizations: Holder’s Inequality for \(L^{p}\)-spaces is one such example. The inequality for sums was published by Augustin-Louis Cauchy in 1821. The integral analog of Cauchy’s inequality is known as the Schwarz inequality and it was presented by Herman Amandus Schwarz in 1885. It appears that the same result had been independently formulated in 1859 by a Russian mathematician, Victor Yacoulevich Bunyakovsky, and thus also bears his name. We refer the reader to J. M. Steele’s text [48]_The Cauchy-Schwartz Master Class: An Introduction to the Art of Mathematical Inequalities_ for further information. Now, we proceed with the result.
**Theorem 68** (Cauchy-Schwarz-Bunyakovsky Inequality).: _Let \(u\) and \(v\) be vectors in an inner product space \(V\). Then_
\[|\langle u,v\rangle|\leq||u||\cdot||v||,\]
Proof
: If \(v=0\), then the inequality is actually an equality since
\[|\langle u,0\rangle|=0=||u||\cdot||0||.\]
So, let \(v\neq 0\) and consider the orthogonal decomposition
\[u=\frac{\langle v,u\rangle}{||v||^{2}}v+w.\]
By applying the Pythagorean Theorem,
\[||u||^{2} = \left|\left|\frac{\langle v,u\rangle}{||v||^{2}}v\right|\right|^{ 2}+||w||^{2}\] \[= \frac{|\langle v,u\rangle|^{2}}{||v||^{2}}+||w||^{2}\] \[\geq \frac{|\langle v,u\rangle|^{2}}{||v||^{2}}.\]
By multiplying both sides by \(||v||^{2}\) and taking the square roots of both sides, we are left with the CSB inequality.
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