0.3. The geometry of Hilbert spaces 23

As another consequence we infer that the map . is indeed a norm. In

fact,

(0.46) f + g

2

= f

2

+ f, g + g, f + g

2

≤ ( f + g

)2.

But let us return to C(I). Can we find a scalar product which has the

maximum norm as associated norm? Unfortunately the answer is no! The

reason is that the maximum norm does not satisfy the parallelogram law

(Problem 0.20).

Theorem 0.23 (Jordan–von Neumann). A norm is associated with a scalar

product if and only if the parallelogram law

(0.47) f + g

2

+ f − g

2

= 2 f

2

+ 2 g

2

holds.

In this case the scalar product can be recovered from its norm by virtue

of the polarization identity

(0.48) f, g =

1

4

(

f + g

2

− f − g

2

+ i f − ig

2

− i f + ig

2

)

.

Proof. If an inner product space is given, verification of the parallelogram

law and the polarization identity is straightforward (Problem 0.22).

To show the converse, we define

s(f, g) =

1

4

(

f + g

2

− f − g

2

+ i f − ig

2

− i f + ig

2

)

.

Then s(f, f) = f

2

and s(f, g) = s(g,

f)∗

are straightforward to check.

Moreover, another straightforward computation using the parallelogram law

shows

s(f, g) + s(f, h) = 2s(f,

g + h

2

).

Now choosing h = 0 (and using s(f, 0) = 0) shows s(f, g) = 2s(f,

g

2

) and

thus s(f, g) + s(f, h) = s(f, g + h). Furthermore, by induction we infer

m

2n

s(f, g) = s(f,

m

2n

g); that is, α s(f, g) = s(f, αg) for every positive rational

α. By continuity (which follows from the triangle inequality for . ) this

holds for all α 0 and s(f, −g) = −s(f, g), respectively, s(f, ig) = i s(f, g),

finishes the proof.

Note that the parallelogram law and the polarization identity even hold

for sesquilinear forms (Problem 0.22).

But how do we define a scalar product on C(I)? One possibility is

(0.49) f, g =

b

a

f

∗(x)g(x)dx.