Page 28 - Vector Analysis

24 CHAPTER 1. Linear Algebra

Remark 1.78. When the domain V and the target W under consideration are clear, we
use } ¨ } instead of } ¨ }B(V,W) to simplify the notation of operator norm.

Remark 1.79. If V is finite dimensional, then L (V, W) = B(V, W).

Proposition 1.80. Let (V, } ¨ }V ) and (W, } ¨ }W ) be two normed vector spaces over a scalar
(
field F. Then B(V, W), } ¨ }) with } ¨ } defined by (1.4) is a normed vector space. (Therefore,

} ¨ } is also called an operator norm).

Definition 1.81 (Dual space). Let (V, } ¨ }) be a normed vector space over field F. An
(
element in B(V, F) is called a bounded linear functional on V, and the space B(V, F), } ¨
)
}B(V,F) is called the dual space of (V, } ¨ }), and is usually denoted by V 1.

Definition 1.82. Let (V, } ¨ }V) and (W, } ¨ }W) be two normed vector spaces over a scalar
field F, and L P B(V, W). The collection of all elements v P V such that Lv = 0 is called
the kernel (or the null space) of L and is denoted by ker(L) or Null(L). In other words,

ker(L) = ␣v P V ˇ Lv = 0( .
ˇ

Theorem 1.83 (Riesz Representation Theorem). Let (V, (¨, ¨)V) be an inner product space,
and f : V Ñ R be a bounded linear map. Then there exists a unique w P V such that
f (v) = (v, w)V for all v P V.

Proof. The uniqueness for such a vector w is simply due to the fact that there is no non-

trivial vector which is orthogonal to itself.

Now we show the existence of w. If f (v) = 0 for all v P V, then w = 0 does the job.
Now suppose that ker(f ) Ĺ V. Then there exists u P ker(f )K such that }u}V = 1.

For v P V, consider the vector y = f (v)u ´ f (u)v. Then y P ker(f ); thus y ¨ u = 0.
Therefore,

0 = f (v)}u}2V ´ f (u)(v, u)V = f (v) ´ (v, fĘ(u)u)V

which implies that f (v) = (v, w)V with w = fĘ(u)u. ˝

By the Riesz representation theorem, we conclude the following

Theorem 1.84. Let (V, (¨, ¨)V) and (W, (¨, ¨)W) be two inner product spaces. Then for all
L P B(V, W), there exists a unique L˚ P B(W, V) such that

(Lv, w)W = (v, L˚w)V @ v P V, w P W .