Recall for continuous (e.g., smooth) functions , their dot product is
defined to be
(This is not denoted as , to avoid confusion
with usual multiplication of functions.)
When we say and are orthogonal. We write rather than if we want to emphasize . (One can similarly define for any and .)
One thing to keep in mind is that the dot product of two functions is not easily visualized in terms of their graphs, even for the orthogonality condition.
What actually matters about dot products of functions that they does satisfy the usual properties of the dot product that
we are familiar with in , and so allow us to apply to functions
some of the visual intuition we have for geometry with -vectors. For continuous functions and from to , the following hold:
,
,
,
, and precisely when is the zero function.
Moreover, if are continuous functions on that are not identically zero and are pairwise
orthogonal (i.e., when ) then they are linearly independent.
Example 1.
For example, consider the functions and on the interval . We find
,
,
.
So and are orthogonal. What's more, if we define and , then using the linearity of the dot product we find
So and are orthogonal as well.
Below is a visualization of and . Choose by dragging on the unit circle, and see how and change.
Example 2.
Here is another exmaple. Consider the functions
and on the interval . The ugly coefficients
make the 's have length 1: we find
,
,
.
So and are orthogonal. What's more, if we define and , then using the linearity of the dot product, again we find
so and are orthogonal as well.
Below is a visualization of and . Choose by dragging on the unit circle, and see how and change. Notice that it is hard to see orthogonality just by looking at the graphs.