26.3. WEAK DERIVATIVES 945

Therefore,

φ = Dψφ +

(∫ b

aφ (y)dy

)φ 0

and so

T (φ) = T (Dψφ )+

(∫ b

aφ (y)dy

)T (φ 0) =

∫ b

aT (φ 0)φ (y)dy.

Let C = T φ 0. This proves the lemma.Proof of Theorem 35.2.2 Since f and D f are both in L1 (a,b),

D f (φ)−∫ b

aD f (x)φ (x)dx = 0.

Consider

f (·)−∫ (·)

aD f (t)dt

and let φ ∈C∞c (a,b).

D(

f (·)−∫ (·)

aD f (t)dt

)(φ)

≡−∫ b

af (x)φ

′ (x)dx+∫ b

a

(∫ x

aD f (t)dt

)φ′ (x)dx

= D f (φ)+∫ b

a

∫ b

tD f (t)φ

′ (x)dxdt

= D f (φ)−∫ b

aD f (t)φ (t)dt = 0.

By Lemma 35.2.3, there exists a constant, C, such that(f (·)−

∫ (·)

aD f (t)dt

)(φ) =

∫ b

aCφ (x)dx

for all φ ∈C∞c (a,b). Thus∫ b

a{(

f (x)−∫ x

aD f (t)dt

)−C}φ (x)dx = 0

for all φ ∈C∞c (a,b). It follows from Lemma 26.3.1 in the next section that

f (x)−∫ x

aD f (t)dt−C = 0 a.e. x.

Thus we let f (a) =C and write

f (x) = f (a)+∫ x

aD f (t)dt.

This proves Theorem 35.2.2.