Homogeneity. For any constant $c$ and function $u$ with domain $[0, 1]$,
$$
\begin{equation}
\eqalign{
f(cu) &= \int_0^1 c u(x) dx \\
&= c \int_0^1 u(x) dx \\
&= c f(u)
}
\end{equation}
$$
keenan
Terrific.
By the way, it seems some of the align/array features of TeX are not supported...
Misaka-10032
@keenan. Aha, just figured that out! Use eqalign and four backslashes (\\\\) rather than two for line break.
keenan
@Misaka-10032 Ah, great! We may also be able to fix the PHP here...
alsdkjlaksdjl
I think we should add some constraints on the function space (for example, smoothness, or integrable function). Since there are all kinds of weird functions that make the statement untrue.
Define u(x) = 1 for x in rational, u(x) = 0 for x in irrational.
Define v(x) = 0 for x in rational, v(x) = 1 for x in irrational.
If we use Riemann sum as the definition for integral, the f(x) and g(x) are both unintegrable.
However u(x) + v(x) = 1, the constant 1 function.
So, f(u, v) = 1;
But f(u) + f(v) is the sum of two integrations of unintegrable function.
keenan
@ alsdkjlaksdjl Yes, definitely—one always has to be more careful when talking about function spaces. Really nice example. (Also, really nice username! :-))
Yes, it is.
$$ \begin{equation} \eqalign{ f(u) + f(v) &= \int_0^1 u(x) dx + \int_0^1 v(x) dx \\ &= \int_0^1 (u(x)+v(x)) dx \\ &= f(u+v) } \end{equation} $$
$$ \begin{equation} \eqalign{ f(cu) &= \int_0^1 c u(x) dx \\ &= c \int_0^1 u(x) dx \\ &= c f(u) } \end{equation} $$
Terrific.
By the way, it seems some of the align/array features of TeX are not supported...
@keenan. Aha, just figured that out! Use eqalign and four backslashes (
\\\\) rather than two for line break.@Misaka-10032 Ah, great! We may also be able to fix the PHP here...
I think we should add some constraints on the function space (for example, smoothness, or integrable function). Since there are all kinds of weird functions that make the statement untrue.
Define u(x) = 1 for x in rational, u(x) = 0 for x in irrational.
Define v(x) = 0 for x in rational, v(x) = 1 for x in irrational.
If we use Riemann sum as the definition for integral, the f(x) and g(x) are both unintegrable.
However u(x) + v(x) = 1, the constant 1 function.
So, f(u, v) = 1;
But f(u) + f(v) is the sum of two integrations of unintegrable function.
@ alsdkjlaksdjl Yes, definitely—one always has to be more careful when talking about function spaces. Really nice example. (Also, really nice username! :-))