1438 CHAPTER 43. INTERPOLATION IN BANACH SPACE

⟨BX (tm) ,X (tm)⟩−⟨BX0,X0⟩−2m−1

∑j=0

∫ t j+1

t j

⟨Y (u) ,X rk (u)⟩du

Thus, ⟨BX (tm) ,X (tm)⟩ is constant for all k large enough and the integral term converges to∫ tm

0⟨Y (u) ,X (u)⟩du

It follows that the term on the left does converge to something as k→ ∞. It just remains toconsider what it does converge to. However, from the equation solved by X ,

BX(t j+1

)−BX (t j) =

∫ t j+1

t j

Y (u)du

Therefore, this term is dominated by an expression of the form

mk−1

∑j=0

(∫ t j+1

t j

Y (u)du,X(t j+1

)−X (t j)

)

=mk−1

∑j=0

⟨∫ t j+1

t j

Y (u)du,X(t j+1

)−X (t j)

⟩

=mk−1

∑j=0

∫ t j+1

t j

⟨Y (u) ,X

(t j+1

)−X (t j)

⟩du

=mk−1

∑j=0

∫ t j+1

t j

⟨Y (u) ,X

(t j+1

)⟩−

mk−1

∑j=0

∫ t j+1

t j

⟨Y (u) ,X (t j)

⟩=

∫ T

0⟨Y (u) ,X r (u)⟩du−

∫ T

0

⟨Y (u) ,X l (u)

⟩du

However, both X r and X l converge to X in K = Lp (0,T,V ). Therefore, this term mustconverge to 0. Passing to a limit, it follows that for all t ∈ D, the desired formula holds.Thus, for such t ∈ D,

⟨BX (t) ,X (t)⟩= ⟨BX0,X0⟩+2∫ t

0⟨Y (u) ,X (u)⟩du

It remains to verify that this holds for all t. Let t /∈ D and let t (k) ∈Pk be the largestpoint of Pk which is less than t. Suppose t (m)≤ t (k) so that m≤ k. Then

BX (t (m)) = BX0 +∫ t(m)

0Y (s)ds,

a similar formula for X (t (k)) . Thus for t > t (m) ,

BX (t)−BX (t (m)) =∫ t

t(m)Y (s)ds