25.8. PERTURBATION THEOREMS 931

Now from the inclusion satisfied,

0 = ⟨z∗n− z∗m,xn− xm⟩+ ⟨Bnxn−Bmxm,xn− xm⟩ (25.8.98)

Consider that last term. Bnxn ∈ BJnxn similar for Bmxm. Hence this term is of the form

⟨Bnxn−Bmxm,xn− xm⟩=≥0︷ ︸︸ ︷

⟨Bnxn−Bmxm,Jnxn− Jmxm⟩

+⟨Bnxn−Bmxm,(xn− Jnxn)− (xm− Jmxm)⟩

From the estimate 25.8.97,

⟨Bnxn−Bmxm,xn− xm⟩ ≥ ⟨Bnxn−Bmxm,(xn− Jnxn)− (xm− Jmxm)⟩

and

|⟨Bnxn−Bmxm,(xn− Jnxn)− (xm− Jmxm)⟩| ≤ 2C

(√1n+

√1m

)Then from 25.8.98,

0≥ ⟨z∗n− z∗m,xn− xm⟩+ en,m

where en,m→ 0 as n,m→ ∞. Hence

lim supm,n→∞

⟨z∗n− z∗m,xn− xm⟩ ≤ 0

From Lemma 25.8.11,lim sup

n→∞

⟨z∗n,xn− x⟩ ≤ 0

Hence, since A is pseudomonotone, for every y, there exists z∗ (y) ∈ Ax such that

lim infn→∞⟨z∗n,xn− y⟩ ≥ ⟨z∗ (y) ,x− y⟩

In particular, if x = y, this shows that

lim infn→∞⟨z∗n,xn− x⟩ ≥ 0≥ lim sup

n→∞

⟨z∗n,xn− x⟩

showing thatlimn→∞⟨z∗n,xn⟩= ⟨z∗,x⟩

Next, returning to the inclusion solved,

0 = Lxn + z∗n +Bnxn

Act on (xn− x) . Then from monotonicity of L,

0≥ ⟨Lx,xn− x⟩+ ⟨z∗n,xn− x⟩+ ⟨Bnxn,xn− x⟩