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Measure, Integration & Real Analysis, 2021a

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286 Chapter 10 Linear Maps on Hilbert Spaces<br />

10.15 Example Volterra operator<br />

The Volterra operator is the linear map V : L 2 ([0, 1]) → L 2 ([0, 1]) defined by<br />

(V f )(x) =<br />

∫ x<br />

0<br />

f (y) dy<br />

for f ∈ L 2 ([0, 1]) and x ∈ [0, 1]; here dy means dλ(y), where λ is the usual<br />

Lebesgue measure on the interval [0, 1].<br />

To show that V is a bounded linear map from L 2 ([0, 1]) to L 2 ([0, 1]), let K be the<br />

function on [0, 1] × [0, 1] defined by<br />

{<br />

1 if x > y,<br />

K(x, y) =<br />

0 if x ≤ y.<br />

In other words, K is the characteristic<br />

function of the triangle below the<br />

Vito Volterra (1860–1940) was a<br />

pioneer in developing functional<br />

diagonal of the unit square. Clearly<br />

analytic techniques to study integral<br />

K ∈L 2 (λ × λ) and V = I K as defined<br />

equations.<br />

in 10.6. Thus V is a bounded linear map<br />

from L 2 ([0, 1]) to L 2 ([0, 1]) and ‖V‖ ≤ √ 1 (by 10.8).<br />

2<br />

Because V ∗ = I K ∗ (by 10.9) and K ∗ is the characteristic function of the closed<br />

triangle above the diagonal of the unit square, we see that<br />

10.16 (V ∗ f )(x) =<br />

∫ 1<br />

x<br />

f (y) dy =<br />

∫ 1<br />

0<br />

f (y) dy −<br />

∫ x<br />

0<br />

f (y) dy<br />

for f ∈ L 2 ([0, 1]) and x ∈ [0, 1].<br />

Now we can show that V ∗ is injective. To do this, suppose f ∈ L 2 ([0, 1]) and<br />

V ∗ f = 0. Differentiating both sides of 10.16 with respect to x and using the Lebesgue<br />

Differentiation Theorem (4.19), we conclude that f = 0. Hence V ∗ is injective. Thus<br />

the Volterra operator V has dense range (by 10.14).<br />

Although range V is dense in L 2 ([0, 1]), it does not equal L 2 ([0, 1]) (because<br />

every element of range V is a continuous function on [0, 1] that vanishes at 0). Thus<br />

the Volterra operator V has dense but not closed range in L 2 ([0, 1]).<br />

Invertibility of Operators<br />

Linear maps from a vector space to itself are so important that they get a special name<br />

and special notation.<br />

10.17 Definition operator; B(V)<br />

• An operator is a linear map from a vector space to itself.<br />

• If V is a normed vector space, then B(V) denotes the normed vector space<br />

of bounded operators on V. In other words, B(V) =B(V, V).<br />

<strong>Measure</strong>, <strong>Integration</strong> & <strong>Real</strong> <strong>Analysis</strong>, by Sheldon Axler

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