Page 68 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 5: Spaces, Hölder

Theorem 4: SCHWARZ or CAUCHY-SCHWARZ INEQUALITY statement: Let f and g be square Notes
integrable, i.e.
f, g L [a, b]; then fg L [a, b] and fg f g .
2
2 2
Proof: Let x [a, b] be arbitrary, then
[|f(x)| – |g(x)|] 0
2
or 2|f(x)|.|g(x)| |f(x) + |g(x)| . 2
2
On integrating, we get

b b b
2
2
2 |f(x)g(x)|dx |f(x) dx |g(x)| dx … (i)
a a a
b b
2
2
2
Now f, g L [a, b] f and g are measurable over [a, b] and |f(x)| dx , |g(x)| dx .
a a
b
Using in (i), we get |f(x) g(x)|dx
a
Thus fg L [a, b].
Let a R be arbitrary. Then
( |f| + |g|) 2 0

b
( |f| |g|) 2 0
a

b b b
2
2
or 2 |f| dx 2 |fg|dx |g| dx 0
a a a
b b b
2
2
Write A = |f| dx, B 2 |fg|dx, C |g| dx
a a a
2
Then we have A + B + C 0 … (ii)
Now, if A = 0, then f(x) = 0 a.e. in [a, b] and hence B = 0 and both sides of the inequality to be
proved are zero. Thus when A = 0, the inequality is trivial.

B
Again, let A 0. Writing = – in (ii), we get
2A

B 2 B
A B C 0 .
2A 2A
2
which gives B 4AC.

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