Page 210 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 17: The Hahn-Banach Theorem

Hence f (ky) = k [f (x) + r ] = k f (y) … (5) Notes
o o o
From (4) and (5) it follows that f is linear on M .
o o
If y M, then = 0 in the representation for y so that
y = x.

Hence f (x) = f (x) x M.
o
f extends f from M to M .
o o
Next we show that

f = f .
o
If = 0 this is obvious. So we consider when 0. Since M is a subspace of M we then have
o
f = sup {|f (x)| : x M , x 1}
o o o
sup {|f (x)| : x M, x 1}
o
= sup {|f(x)| : x M, x 1} ( f = f on M)
o
= f .
Thus, f f … (A)
o
So our problem now is to choose r such that f f .
o o
Let x , x M. Then we have
1 2
f (x ) – f (x ) = f (x – x )
2 1 2 1
|f (x – x )|
2 1
f (x + x ) – (x + x )
2 o 1 o
f ( x + x + – (x + x ) )
2 o 1 o
= f x + x + f x + x
2 o 1 o
Thus – f (x ) – f x + x – f (x ) + f x + x … (6)
1 1 o 2 2 o
Since this inequality holds for arbitrary x , x M, we see that
1 2
sup f(y) f y x o int f(y) f y x o
y M y M
Choose r to be any real number such that
o
sup f(y) f y x r
o o
y M
inf f(y) f y x o
y M
From this, we get for all y M
sup {– f (y) – f ( y + x )} r
o o
inf {– f (y) + f ( y + x )}
o
x
Let us take y = in the above inequality, we have

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