First of all, for f∈C([a;b]) we see that
∣∣f∣∣2=(∫ab∣f(x)∣2dx)1/2
is well defined and is in [0;+∞) .
Now let's verify the norm properties:
- Separation, ∣∣f∣∣2=0↔f=0 . If f=0,∣∣f∣∣2=0 is obvious. Now suppose f=0,∣f∣2=0 is a continuous positive function and thus ∫ab∣f(x)∣2dx=0, therefore ∣∣f∣∣2=0 .
- Linearity, ∣∣αf∣∣2=∣α∣⋅∣∣f∣∣2 . It follows from linearity of an integral : (∫ab∣αf∣2(x)dx)1/2=(∣α∣2∫ab∣f(x)∣2dx)1/2=∣α∣⋅∣∣f∣∣2
- Triangle inequality, ∣∣f+g∣∣2≤∣∣f∣∣2+∣∣g∣∣2 .
∫ab∣f+g∣2(x)dx≤∫ab∣f∣2(x)dx+∫ab∣g∣2(x)dx+2∫ab∣fg∣(x)dx
We have (∫ab∣fg∣(x)dx)2≤(∫ab∣f(x)∣2dx)⋅(∫ab∣g(x)∣2dx), as
(∫ab∣f(x)∣2dx)1/2⋅(∫ab∣g(x)∣2dx)1/2∣fg∣≤2∫ab∣f(x)∣2dx∣f∣2+∫ab∣g(x)∣2dx∣g∣2 (by applying 2∣ab∣≤a2+b2 to (∫ab∣f(x)∣2dx)1/2∣f∣ and (∫ab∣g(x)∣2dx)1/2∣g∣ ) and thus by integrating on [a;b] we find :
(∫ab∣f(x)∣2dx)1/2⋅(∫ab∣g(x)∣2dx)1/2∫ab∣fg∣(x)dx≤1
Therefore (∫ab∣fg∣(x)dx)2≤(∫ab∣f∣2dx)⋅(∫ab∣g∣2dx) .
And we conclude :
∣∣f+g∣∣22=∫ab∣f+g∣2≤∫ab∣f∣2+∫ab∣g∣2+2∫ab∣fg∣
∣∣f+g∣∣22≤∫ab∣f∣2+∫ab∣g∣2+2(∫ab∣f∣2)1/2(∫ab∣g∣2)1/2
∣∣f+g∣∣22≤(∣∣f∣∣2+∣∣g∣∣2)2
And thus ∣∣f+g∣∣2≤∣∣f∣∣2+∣∣g∣∣2 .
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