MA3421 (Functional Analysis 1) Tutorial sheet 8
[November 27, 2014]
Name: Solutions
Fix g ∈ L∞ [0, 1] and define an operator Mg on (a.e equivalence classes of) functions via
Mg (f ) = gf . (We call Mg a multiplication operator. Since we deal with a.e. equivalance
classes we really want to consider M[g] ([f ]) = [gf ]. The first question is to show that this is
well-defined.)
1. Show first that if g1 = g2 almost everywhere and f1 = f2 almost everywhere on [0, 1] then
g1 f1 = g2 f2 almost everywhere. [Hint: The union of two sets of Lebesgue measure zero is
again of measure zero. So are subsets of sets of measure zero.]
Solution: g1 = g2 almost everywhere means thet Eg = {x ∈ [0, 1] : g1 (x) 6= g2 (x)} has
measure zero. Similarly Ef = {x ∈ [0, 1] : f1 (x) 6= f2 (x)} has measure zero.
Now certainly g1 (x)f1 (x) = g2 (x)f2 (x) for x ∈ [0, 1] \ (E1 ∪ E2 ), or {x ∈ [0, 1] :
g1 (x)f1 (x) 6= g2 (x)f2 (x)} ⊆ E1 ∪ E2 . Sincce E1 ∪ E2 has measure zero and so does each
subset, we have g1 f1 = g2 f2 almost everywhere (on [0, 1]).
2. Show that Mg defines a linear operator from L∞ [0, 1] to itself with operator norm kMg kop ≤
kgk∞ [Hint: If [g] ∈ L∞ [0, 1] then (you can take it as known that) there is g1 with
g(x) = g1 (x) almost everywhere and |g1 (x)| ≤ kgk∞ everywhere on [0, 1]. Same for
f .]
Solution: Assuming g1 is in the same almost everywhere equivalence class as g and has
|g1 (x)| ≤ kgk∞ for all x ∈ [0, 1], and f1 is equal to f almost everywhere and has |f1 (x)| ≤
kf k∞ almost everywhere, then certainly |g1 (x)f1 (x)| ≤ kgk∞ kf k∞ for each x ∈ [0, 1].
Since gf equals g1 f1 almost everywhere, we have that the essential supremum kgf k∞ ≤
kgk∞ kf k∞ .
Now it is clear enough that Mg is a linear transformation from L∞ [0, 1] to itself. We can
check that Mg (f + h) = g(f + h) = gf + gh = Mg f + Mg h and that Mg (λf ) = gλf =
λgf = λMg f (for f, h ∈ L∞ [0, 1] and λ ∈ K.
The inequality kMg f k∞ = kgf k∞ ≤ kgk∞ kf k∞ says that kMg kop ≤ kgk∞ .
As indicated in the note at the start, we typically don’t use formal equivalence
class notation for elements of L∞ [0, 1]. So, maybe the point of what we just
argued would be more clear if we wrote M[g] [f ], k[g]k∞ and k[f ]k∞ .
3. Show that in fact Mg : L∞ [0, 1] → L∞ [0, 1] has operator norm kMg kop = kgk∞ . [Hint: If
g is not zero (almost everywhere) take f (x) = g(x)/|g(x)| on the set of x where g(x) 6= 0
and f (x) = 0 elsewhere. You can take it that f will then be measurable.]
Solution: If g(x) = 0 almost everywhere, then kgk∞ = 0 and kMg kop ≤ kgk∞ implies
kMg kop = 0 = kgk∞ . So we can suppose that g is not zero (almost everywhere).
Taking f as suggested we have |f (x)| = |g(x)|/|g(x)| = 1 on the set of x where g(x) 6= 0
and |f (x)| = 0 otherwise. So |f (x)| ≤ 1 everywhere and for any number M < 1 we
have a set of positive measure where |f (x)| > M . So M = 1 is the smallest M with
|f (x)| ≤ M almost everywhere. So kf k∞ = 1.
Also where g(x) 6= 0 we have
(Mg f )(x) = g(x)
g(x)
|g(x)|2
=
= |g(x)|
|g(x)|
|g(x)|
So kMg f k∞ = kgk∞ . Since
kMg f k∞
kgk∞
=
= kgk∞
kf k∞
1
we have kMg kop ≥ kgk∞ .
Hence kMg kop = kgk∞ .
Solution: (better solution).
Ignoring the hint, take f to be the constant function f (x) = 1 (for x ∈ [0, 1]). Then
kf k∞ = 1 (because |f (x)| ≤ 1 always and no smaller number M < 1 has |f (x)| ≤ M
almost everywhere).
Since Mg f = gf = g we have
kgk∞
kMg f k∞
=
= kgk∞
kf k∞
1
we have kMg kop ≥ kgk∞ .
Hence kMg kop = kgk∞ .
4. Let p be in the range 1 ≤ p < ∞. Show that Mg defines a linear operator from Lp [0, 1] to
itself and that the operator norm of Mg : Lp [0, 1] → Lp [0, 1] is also at most kgk∞ .
Solution: We can suppose (by replacing g by a function almost everywhere the same) that
|g(x)| ≤ kgk∞ for all x ∈ [0, 1]. Then for f ∈ Lp [0, 1] we have
Z 1
Z 1
Z 1
p
p
p
p
p
|f (x)|p dxkgkp∞ kf kpp
kMg f kp =
|g(x)f (x)| dx ≤
kgk∞ |f (x)| dx = kgk∞
0
0
0
So Mg f ∈ Lp [0, 1] and (taking pth roots) kMg f kp ≤ kgk∞ kf kp .
Actually, we need also that Mg is a linear transformation from Lp [0, 1] to itself, but that is
true for the same reason that it is linear on L∞ [0, 1].
Richard M. Timoney
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