**If the elements of a given set are such that they can be submitted to the
same operations we can do on n-dimensional real vectors, they are vectors and the set is
a vector space**.

This statement can be verified in particular for some set of functions of a given form.

Let's consider, for example, the set Σ_{2} of the functions
with *a* and *b* real numbers and defined for *x* ∈ [-π ; π].
For simplicity sake, these functions will be shortly written
*σ _{2}(a,b)* and sometime only

Whichever two functions of Σ_{2} can be summed in the usual algebraic way
giving a function of Σ_{2}, so Σ_{2} is closed under the sum.
The sum is associative and commutative. The neutral element of the sum is obviously
*σ _{2}(0,0)* and for every

*σ _{2}(a,b)* can be multiplied by a real number

Moreover, given whichever functions *σ _{2}(a,b)* and

The Wolfram Integrator speeds the calculation of this and other integrals in these pages.

The value of this integral coincides, apart from the constant factor π, with the scalar
product between the real plane vectors *(a,b)* and *(c,d)*.
It seems therefore logical to take this integral as the expression of the scalar product
between *σ _{2}(a,b)* and

This definition allows to define a norm ||*σ _{2}*||
for the vector |σ

So the set Σ_{2}, like the set of the real plane vectors, is a
normed vector space.

A linear operator L on Σ_{2} must transform univocally a vector
into another vector in the same space and must have the same properties of the
real square matrices with respect to real plane vectors hightlighted in
section 2, that is

The second derivative of the functions of Σ_{2} with respect to the variable
*x* has such properties; in fact the derivative of a sum of functions is the sum of
the derivatives of each function and the derivative of the product of a function by a real
number *k* is the product of the derivative of the function by *k*. Moreover

Therefore the double differentiation operator is a linear operator on Σ_{2}.
In order to calculate its eigenvalues and its eigenvectors (here its
eigenfunctions)
one must solve the equation

that is

This equation can be satisfied only if

So the eigenvalues of the double differentiation operator are
*λ _{1}*=-1 and

Like in real plane vector space, these eigenfunctions can be normalized if we assume
that the scalar factor *a* and *b* are positive and divide each vector
for its norm. We obtain

The normalized set {*σ _{2,1}* ,

If we consider the set Γ_{2} of the functions
defined for *x* ∈ [-π ; π], we reach analogous results.

The analogy persists also if we consider the set Φ_{2} of the functions
which are such that
.

This integral gives a value that, apart from the constant factor *π*,
coincides with the scalar product of the two 4-dimensional vectors
*(a,b,c,d)* e *(h,k,l,m)*.

By induction, we can infer that Φ_{n}, that is the set of the functions
,
is a normed vector space in which the norm is given by
.

In particular, for whichever pair of functions we have

Therefore the union of the sets of the functions
is an orthonormal basis of the vector space Φ_{n} and each function of
Φ_{n} can be expressed in the following way

In conclusion, if we let

we have