Multiplication of Functions
Function is a correspondence $f$ between elements of a space $X$ and those of a space $Y$ such that any element $x$ of $X$ has a unique corresponding element $y$ of $Y$ which is denoted $y = f(x).$ The following is a widely used shorthand for "a function $f$ from $X$ to $Y$"
$f:\,X\rightarrow Y.$
In most cases $f(x)$ is defined for all $x\in X,$ but sometimes its domain of definition is smaller. The reason to allow such aberration is that most of the elementary functions are real $(X = R)$ and realvalued $(Y = \mathbb{R})$ and are defined by some kind of formulas. E.g., $f(x) = x^{2}  10.$ However, some formulas are not defined for all real values as, for example, $f(x) = 1/(x1)$ which is not defined for $x = 1.$ The right way to described this latter function is $(\infty,1)\cup (1,\infty)\rightarrow\mathbb{R}.$ Which is a little awkward. Thus we often write $1/(x1): \mathbb{R}\rightarrow \mathbb{R}$ and try not to forget that $x = 1$ is an exceptional point that does not belong to the domain of definition of that function.
Returning to the question of function multiplication, I can think of four quite different operations meaningfully and fruitfully defined for functional spaces (i.e., spaces that contain functions, or whose points are functions.)
Componentwise multiplication. If $Y$ is a multiplicative group we may look at functions $f: X\rightarrow Y$ as vectors with components in $Y$ indexed by elements from $X.$ In this case, we can sensibly define a componentwise multiplication:
$(fg)(x) = f(x)g(x)$
where, as is customary, I have dropped the symbol of multiplication. For example, if $f(x) = x+1$ and $g(x) = x  1$ then $(fg)(x) = x^{2}  1.$
Associativity, commutativity and the distributive law are inherited from $Y.$ However, if, for whatever reason, $Y$ is not a multiplicative group, the function multiplication won't be a group operation either. If $Y$ has a unit element $1,$ then the function $f(x)$ which is identically equal to $1,$ will serve as the unit element for the componentwise multiplication. If $Y = \mathbb{R}$ where all elements but $0$ have inverses, then a function $f: X\rightarrow \mathbb{R}$ among whose values there is a $0,$ won't have an inverse.
Whatever the case, the important thing is that we only can multiply functions from the same space $Y^{X}.$ Componentwise multiplication is not defined for functions with different domains of definition.

Assume we are given two functional spaces $Y^{X}$ and $Z^{Y}.$ We may construct elements from $Z^{X}$ in the following manner: if $y = g(x)$ and $z = f(y)$ then $(fg)(x) = z.$ In other words, $(fg)(x) = f(g(x)).$ The most interesting case is when $X = Y = Z$ so that we deal with a single space $X^{X}$ throughout. If, for a given $f: X\rightarrow X,$ the inverse function exists, it naturally belongs to the same space and constitutes the inverse element $f^{1}.$ A function $f$ has an inverse iff it's 11 and on. The latter means that the inverse function is defined on the whole of $X.$ The identity function $f(x) = x$ serves as the unit element for the composition. A common notation for the composition $f(g(x))$ is $(g\circ f)(x).$
Associativity is straightforward, but the commutative law does not hold. For $f(x) = x+5$ and $g(x) = x^{2},$ $f(g(x)) = x^{2}+5,$ while $g(f(x)) = (x+5)^{2}.$

Scalar product for functions is pretty much the same as the scalar product for vectors. If $X$ is an infinite set then a finite sum must be replaced with a sum of an infinite series (if $X$ is countable) or an integral (if $X$ is not.) In both cases it may not be defined for every $f\in Y^{X}.$ Subspaces of $Y^{X}$ for which the scalar product is defined are analogous to finitedimensional vector spaces. We may consider orthogonal functions, and posit a problem of finding a basis in such a space. The quest for bases in function spaces with scalar product led to the theory of Fourier series and Fourier transform, to the theory of generalized functions and most recently to the discovery of wavelets.
As in the finitedimensional case, scalar product is not a group operation because its result does not belong to the space itself.

Assume $X = Y = R$. Then along with every function $f: \mathbb{R}\rightarrow\mathbb{R}$ we may consider a whole family of functions $f_{a}(x) = f(a  x).$ These are reflected shifts of the function $f.$ If we now restrict ourselves to the subspaces of $\mathbb{R}^{\mathbb{R}}$ on which a scalar product is defined, we may form a product $g*f = g\cdot f_{a},$ where by $h\cdot k$ I denote the scalar product of the functions $h$ and $k.$ Strictly speaking, $g*f$ depends on $a\in\mathbb{R}$ and is itself a real number. Therefore, it may be regarded as a function $g*f: \mathbb{R}\rightarrow\mathbb{R}.$ This function is known as the convolution of $f$ and $g.$ Convolutions arise naturally when differential equations are being replaced with integral ones. Fourier transform of a convolution coincides with the (componentwise) product of Fourier transforms. Thus integral equations with convolutions may be further reduced to algebraic equations. The unit element for the convolution is the wellknown Dirac $\delta$ function. This is a "function" that was invented by P. Dirac in 1920s. It's zero everywhere except for the origin where its value is such that the total integral of the function equals $1!$ Later S. L. Sobolyev and L. Schwartz developed a framework of generalized functions in which such a definition indeed made sense.
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