1. Cauchy-Schwarz inequality All the functions here are assumed to be real-valued.
Let f (x), g(x) be two continuous functions on a bounded closed interval [a, b]. Then f (x)2, f (x)g(x), g(x)2 are continuous on [a, b]; they are all Riemnan integrable on [a, b].
Define a function F (t) on R by F (t) =
Z b a
(tf (x) − g(x))2dx.
Assume that f (x) is not the zero function. Expanding (tf (x) − g(x))2, we see that F is a polynomial function:
F (t) =
Z b a
f (x)2dx
t2− 2
Z b a
f (x)g(x)dx
t +
Z b a
g(x)2dx
.
Denote A = Z b
a
f (x)2dx and B = Z b
a
f (x)g(x)dx and C = Z b
a
g(x)2dx. Since f is nonzero, A > 0.
Since (tf (x) − g(x))2 is nonnegative for all t, F (t) ≥ 0 for all t. Then the minimum of F is also nonnegative. The critical point of F obeys F0(t) = 0. In this case, F0(t) = 2At − 2B implies that t∗ = B/A. We can check that F00(t∗) = 2A > 0, and hence t∗ is a local minimum of F (t). Since F is a degree two polynomial function, t∗ is the minimum. The minimum of F is given by
F (t∗) = A · B A
2
− 2B ·B
A + C = AC − B2
A .
Since F (t∗) ≥ 0 and A > 0, we obtain AC − B2 ≥ 0. In other words,
Z b a
f (x)g(x)dx
2
≤
Z b a
f (x)2dx
Z b a
g(x)2dx
. The equality holds if and only F (t∗) = 0. In this case,
Z b
a
(t∗f (x) − g(x))2dx = 0.
Since (t∗f (x) − g(x))2 is a nonnegative continuous function on [a, b], the above equality implies that (t∗f (x) − g(x))2 = 0. Then t∗f (x) = g(x) for all x ∈ [a, b]. We conclude that Theorem 1.1. (Cauchy-Schwarz inequlaity) Let f (x), g(x) be continuous functions on [a, b].
Then
Z b a
f (x)g(x)dx
2
≤
Z b a
f (x)2dx
Z b a
g(x)2dx
The equality holds if and only if there exists c ∈ R so that cf (x) = g(x).
1
