Basic Analysis – Gently Done Topological Vector Spaces

and
−f(x)−µ ≤ p(−x−x 1 ) for x ∈ M, (taking t = −1).
5: Vector Spaces 49
Replacing x by −y and y ∈ M in this last inequality we see that µ must satisfy
and
µ ≤ p(x+x 1 )−f(x) x ∈ M
f(y)−p(y −x 1 ) ≤ µ y ∈ M.
The idea of the proof is to show that such a µ exists, and then to work backwards
to show that f 1 , as defined above, does satisfy the boundedness requirement. Let
x,y ∈ M. Then
Hence
f(x)+f(y) = f(x+y) ≤ p(x+y) by hypothesis
= p(x−x 1 +y +x 1 )
≤ p(x−x 1 )+p(y +x 1 ).
f(x)−p(x−x 1 ) ≤ p(y +x 1 )−f(y), for x,y ∈ M.
Thus the set {f(x)−p(x−x 1 ) : x ∈ M} is bounded from above in R. Let µ denote
its least upper bound. Then we have
Hence
(1)
and
(2)
f(x)−p(x−x 1 ) ≤ µ ≤ p(y +x 1 )−f(y), for x,y ∈ M.
f(x)−µ ≤ p(x−x 1 ) for x ∈ M
f(y)+µ ≤ p(y +x 1 ) for y ∈ M.
Define f 1 on M 1 by f 1 (x + tx 1 ) = f(x) + tµ, x ∈ M, t ∈ R. Then f 1 is linear
on M 1 and f 1 = f on M. To verify that f 1 satisfies the required bound, we use
the inequalities (1) and (2) and reverse the argument of the preamble. For t > 0,
replacing x by t −1 x in (1) gives
f(t −1 x)−µ ≤ p(t −1 x−x 1 )