MA442A (Harmonic Analysis 1) Tutorial sheet 6
[November 26, 2015]
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Define ft (x) = cos(t + x) for 0 ≤ t, x ≤ 1.
1. Show that ft ∈ C[0, 1] for each t
Solution: This is rather obvious as cos is a conntinuous function and so is x 7→ t + x
(where t is a constant). So ft is a composition of continuous functions.
2. Show that t 7→ ft is a continuous map from [0, 1] to C[0, 1]
Solution: For 0 ≤ t, x ≤ 1, 0 ≤ t + x ≤ 2. Since cos is continuous on the (finite) closed
interval [0, 2], it is uniformly continuous on [0, 2] and so given ε > 0 there is δ > 0 such
that
θ, φ ∈ [0, 2], |θ − φ| < δ ⇒ | cos θ − cos φ| < ε
Then if t, s ∈ [0, 1] have |t − s| < δ, we have θ = t + x ∈ [0, 2], φ = s + x ∈ [0, 2] (any
x ∈ [0, 1]) and
|θ − φ| = |t − s| < δ ⇒ |ft (x) − fs (x)| = | cos(t + x) − cos(s + x)| < ε
So we have
|t − s| < δ ⇒ sup |ft (x) − fs (x)| = max |ft (x) − fs (x)| < ε
x∈[0,1]
x∈[0,1]
or
|t − s| < δ ⇒ kft − fs k∞ = sup |ft (x) − fs (x)| < ε
x∈[0,1]
That is more that we need. We really need that if s ∈ [0, 1] and ε > 0 are fixed in advance
then we can find δ > 0 so that
t ∈ [0, 1], |t − s| < δ ⇒ kft − fs k∞ < ε
but we actually have found that δ can be independent of s, which is uniform continuity.
(In fact uniform continuity follows from continuity on a finite closed interval [0, 1] and so
even if we had not proved it, uniform continuity would have to follow from continuity.
Also, if we used the Mean Value Theorem (for derivatives) we know
cos θ − cos φ = cos0 (α)(θ − φ) = − sin(α)(θ − φ)
for some α between θ and φ. So actually
| cos θ − cos φ| = | sin α||θ − φ| ≤ |θ − φ|
and δ = ε works. But that is more than we need.)
Z
3. Show that the value at x ∈ [0, 1] of
1
Z
ft dt is
0
1
ft (x) dt
0
Solution: The map Tx : C[0, 1] → C given by Tx (f ) = f (x) is a continuous linear transformation (because |Tx (f ) − Tx (g)| = |f (x) − g(x)| ≤ kf − gk∞ ).
Then we had a Lemma 2.1.23 in the notes that says
Z 1
Z 1
Z
Tx (ft ) dt =
Tx
ft dt =
0
0
1
ft (x) dt
0
That
Z 1 is what is required here as the left side is the value at x of the C[0, 1]-valued integral
ft dt.
0
Z
1
ft dt
4. Compute (the C[0, 1]-valued integral)
0
Solution: We can calculate
Z 1
Z 1
ft (x) dt =
cos(t + x) dt = [sin(t + x)]1t=0 = sin(1 + x) − sin x
0
0
for each x ∈ [0, 1].
So the C[0, 1]-valued integral is the function whose value at x is sin(1 + x) − sin x.
Richard M. Timoney
2