L2 Space
Definition 1
A function space $L^{2}$ is defined as follows.
$$ L^{2} (E) := \left\{ f : \left( \int_{E} | f |^2 dm \right)^{{1} \over {2}} < \infty \right\} $$
Properties
- $L^{2}$ is a metric space. The metric is defined as following. $$ d(f,g) := \left( \int \left| f(x) - g(x) \right|^{2}dx \right)^{\frac{1}{2}} = \left\| f-g \right\|_{2} = \sqrt{\braket{f-g, f-g}} $$
- $L^{2}$ is a vector space.
- $L^{2}$ is a normed space. The norm is defined as following. $$ \left\| f \right\|_{2} := \left( \int \left| f(x) \right|^{2}dx \right)^{\frac{1}{2}} = \sqrt{\braket{f,f}} $$
- $L^{2}$ is a complete space.
- $L^{2}$ is an inner product space. The inner product is defined as following. $$ \braket{f, g} := \int \overline{f(x)}g(x)dx $$
Description
The space $L^{2}$ is a special case when it is $p=2$ of the $L^{p}$ space, and it is the only space among $L^{p}$ spaces where an inner product is defined. A complete inner product space is specifically called a Hilbert space. Therefore, $L^{2}$ is a Hilbert space. Hilbert spaces are important spaces that appear in various fields including partial differential equations and quantum mechanics.
For a generalized proof about $L^{p}$ space, refer here.
Proof
3.
Let’s call $V$ a vector space on $\mathbb{F}$. If a function $\left\| \cdot \right\| : V \to \mathbb{F}$ satisfies the following three conditions for $\mathbf{u}, \mathbf{v} \in V$ and $k \in \mathbb{F}$, then it is defined as the norm on $V$.
- Positivity: $\left\| \mathbf{u} \right\| \ge 0$ and $\mathbf{u} = \mathbb{0} \iff \left\| \mathbf{u} \right\| = 0$
- Homogeneity: $\left\|k \mathbf{u} \right\| = | k | \left\| \mathbf{u} \right\| $
- Triangle inequality: $\left\| \mathbf{u} + \mathbf{v}\right\| \le \left\|\mathbf{v} \right\| + \left\| \mathbf{u} \right\|$
Let’s define the norm of $L ^{2}$ as follows $\displaystyle \left\| f \right\|_{2} := \left( \int_{E} | f |^2 dm \right)^{{1} \over {2}}$.
Part 1. Positivity
Since $| f | \ge 0$, if $\left\| f \right\|_{2} \ge 0$ almost everywhere, then $\left\| f \right\|_{2} = 0$. Conversely, if $\left\| f \right\|_{2} = 0$, then almost everywhere, it must be $f = 0$.
Part 2. Homogeneity
$$ \left\| c f \right\|_{2} = \left( \int_{E} | c f |^2 dm \right)^{{1}\over {2}} =\left( |c|^2 \int_{E} | f |^2 dm \right)^{{1}\over {2}} = |c| \left( \int_{E} | f |^2 dm \right)^{{1}\over {2}} = |c| \left\| f\right\|_{2} $$
Part 3. Triangle inequality
$$ \begin{align*} \left\| f + g \right\|_{2}^{2} =& \int_{E} | f + g |^2 dm \\ =& \int_{E} ( f + g ) \overline{( f + g )} dm \\ =& \int_{E} | f |^2 dm + \int_{E} ( f \overline{g} + \overline{f} g ) dm +\int_{E} | g |^2 dm \end{align*} $$
By the Cauchy-Schwarz inequality, we get the following.
$$ \begin{align*} \int_{E} ( f \overline{g} + \overline{f} g ) dm \le & 2 \int_{E} | fg | dm \le 2 | f |_{2} | g |_{2} \\ =& | f + g | _{2}^{2} \le | f | _{2} + 2 | f | _{2} | g | _{2} + | g | _{2} \\ =& \left( | f |_{2} + | g |_{2} \right)^{2} \end{align*} $$
To summarize
$$ \left\| f + g \right\|_{2} \le \left\| f \right\|_{2} + | g |_{2} $$
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5.
Let us call $H$ a vector space. A function that satisfies the following conditions for $x,y,z \in H$ and $\alpha, \beta \in \mathbb{C}$
$$ \langle \cdot , \cdot \rangle : H \times H \to \mathbb{C} $$
is defined as inner product, and $\left( H, \langle \cdot ,\cdot \rangle \right)$ is called an inner product space.
- Linearity: $\langle \alpha x + \beta y ,z \rangle =\alpha \langle x,z\rangle + \beta \langle y,z\rangle$
- Conjugate Symmetry: $\langle x,y \rangle = \overline{ \langle y,x \rangle}$
- Positive-definiteness: $\langle x,x \rangle \ge 0 \quad \text{and} \quad \langle x,x \rangle = 0\iff x=0$
Let’s define the inner product of $L ^{2}$ as follows $\displaystyle \langle f , g \rangle := \int_{E} f \overline{g} dm$.
Part 1. Linearity
$$ \langle f + g , h \rangle = \int_{E} ( f + g ) \overline{g} dm = \int_{E} f \overline{g} dm + \int_{E} g \overline{g} dm = \langle f , h \rangle + \langle g , h \rangle $$
And
$$ \langle c f , g \rangle = \int_{E} c f \overline{g} dm = c \int_{E} f \overline{g} dm = c \langle f , g \rangle $$
Part 2. Conjugate Symmetry
$$ \langle f , g \rangle = \int_{E} f \overline{g} dm = \overline{ \int_{E} \overline{f} g dm} = \overline{ \int_{E} g \overline{f} dm} = \overline{ \langle f , g \rangle } $$
Part 3. Positive-definiteness
$$ \langle f, f \rangle = \int_{E} f \overline{f} dm = \int_{E} | f |^2 dm = \sqrt{ | f |_{2} } $$
The proof concludes with Part 1 of Property 3..
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See Also
Capinski. (1999). Measure, Integral and Probability: p131. ↩︎