Page 195 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes So that
sup f(x) f … (7)
x 1
Now consider

f(x)
f = sup
x 0 x

By supremum property, given > 0, x 0
Such that |f (x )| > (||f|| – ) ||x ||

x
Define x .
x
Since f is continuous in ||x|| 1 and reaches its maximum on the boundary ||x|| = 1,
We get

1
sup f(x) f (x) f (x ) f .
x 1 x

sup f(x) f .
x 1

The arbitrary character of yields that
sup f(x) f
x 1 … (8)
Hence from (7) and (8), we get

f = sup f(x) .
x 1

Note If N is a finite dimensional normed linear space, all linear functions are bounded
and hence continuous. For, let N be of dimension n so that any x N is of the form
n
x , where x , x , …, x is a basis of N and , , …, are scalars uniquely determined
i i 1 2 n 1 2 n
i 1
by the basis.
Since f is linear, we have

n
f (x) = i f (x ) so that
i
i 1
n
| f (x)| i |f (x )| … (1)
i
i 1

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