(a) Show that the set of all square-integrable functions is a vector space (refer to Section A.1 for the definition). Hint: The main problem is to show that the sum of two square-integrable functions is itself square-integrable. Use Equation 3.7. Is the set of all normalized functions a vector space?

(b) Show that the integral in Equation 3.6satisfies the conditions for an inner product (Section A.2).

Equation 3.7 follows Schwarz inequality:

$\left|{\int }_a^{b}f\left(x\right)*g\left(x\right)dx\right|\le \sqrt{{\int }_a^b{\left|f\left(x\right)\right|}^{2}dx{\int }_a^b{\left|g\left(x\right)\right|}^{2}dx}$

Equation 3.6 define the inner product of two functions

a)

Let $f\left(x\right)$and $g\left(x\right)$be square-integrable, then we need to prove that$h\left(x\right)=f\left(x\right)+g\left(x\right)$is also square-integrable,

$$\begin{gathered} {\left|h\right|}^2=\left(f+g\right)*\left(f+g\right) \\ ={\left|f\right|}^2+{\left|g\right|}^2+f*g+g*f \end{gathered}$$

The integration is,

$\int {\left|h\right|}^{2}dx=\int {\left|f\right|}^{2}dx+\int {\left|g\right|}^{2}dx+\int f*gdx+\left(\int f*gdx\right)*$

since both $f\left(x\right)$and$g\left(x\right)$are square-integrable, then the first two terms are finite, using Schwarz inequality, we can write the third and the fourth integrals in terms of the first and the second integral:

$\left|{\int }_a^{b}f\left(x\right)*g\left(x\right)dx\right|\le \sqrt{{\int }_a^b{\left|f\left(x\right)\right|}^{2}dx{\int }_a^b{\left|g\left(x\right)\right|}^{2}dx}$

Therefore, the last two integrals are finite too. Hence $\int {\left|h\right|}^{2}dx$is finite, therefore Two square-integrable functions add up to a square-integrable function. Now let $\psi \left(x\right)$be a vector with a value or component for every value of x . If a set of vectors meets two criteria, it can be called a vector space:

- If a vector ${\psi }_1\left(x\right)$is in the set, then so is $A\psi \left(x\right)$for any complex scalar $A$

- If two vectors ${\psi }_1\left(x\right)$and ${\psi }_2\left(x\right)$are in the set, then so is their sum ${\psi }_1\left(x\right)+{\psi }_2\left(x\right)$

these two conditions can be combined by saying that if two vectors ${\psi }_1\left(x\right)$and ${\psi }_2\left(x\right)$are in the set, then so is their linear combination $A{\psi }_1\left(x\right)+B{\psi }_2\left(x\right)$ , for any complex scalars and $A$We can see from this definition that the above-mentioned collection of all normalizable functions is not a vector space. For example, if 1 is true for a vector$\psi \left(x\right)$ , then it is not true if we multiply $\psi \left(x\right)$by any scalar $A$where $\left|A\right|\ne 1$.

b)

The definition of an inner product of two vectors requires that it satisfies three conditions:

the first condition is trivial to prove, as:

For the second condition, we can use:

To prove that and if and only if $\backslash f>=0$. For the third condition, we have:

therefore, Equation 3.6's integral satisfies the requirements for an inner product.