# Weak convergence to strong convergence in a Hilbert space (Fredholm Alternative)

This question is related to the post here. It is regarding the closed range argument as in part (ii) of the theorem (Fredholm Alternative) below.

I have a question regarding, how the author arrives at "then (5) implies $$u_{k_j}\to u$$" in the proof, which is bolded below. It is not clear to me, how one can go from weak convergence to strong convergence like this.

I hope someone could clarify the argument here for me. Thank you in advance.

(PDE Evans, Appendix D, Theorem 5)

THEOREM 5 (Fredholm Alternative). Let $$K : H \to H$$ be a compact linear operator. Then

(i) $$N(I-K)$$ is finite dimensional,

(ii) $$R(I-K)$$ is closed,

(iii) $$R(I-K)=N(I-K^*)^\perp$$,

(iv) $$N(I-K)=\{0\}$$ if and only if $$R(I-k)=H$$,

and

(v) $$\dim N(I-K)=\dim N(I-K^*)$$.

proof of (ii): We next claim there exists a constant $$\gamma > 0$$ such that $$\|u-Ku\|\ge \gamma \|u\| \quad \text{for all }u\in N(I-K)^\perp. \tag{4}$$ Indeed, if not, there would exist for $$k=1,\ldots$$ elements $$u_k \in N(I-K)^\perp$$ with $$\|u_k\|=1$$ and $$\|u_k-Ku_k\|<\frac 1k$$. Consequently, $$u_k-Ku_k \to 0. \tag{5}$$ But since $$\{u_k\}_{k=1}^\infty$$ is bounded, there exists a weakly convergent subsequence $$u_{k_j} \rightharpoonup u$$. By compactness $$Ku_{k_j} \to Ku$$, and then (5) implies $$u_{k_j} \to u$$. We therefore have $$u \in N(I-K)$$ and so $$(u_{k_j},u)=0 \quad (j=1,\ldots).$$ Let $$k_j \to \infty$$ to derive a contradiction to (4).

• (5) implies $u_{k_j}$ is norm (and thus weakly) convergent to $Ku$. But, it also converges weakly to $u$. This implies $Ku=u$. Dec 3, 2020 at 16:31
• Hey David, thank you for your reply! I see it now. Dec 3, 2020 at 18:19

Observe that $$\|u_{k_j}-u_{k_\ell}\|\leq \|u_{k_j}-Ku_{k_j}\|+\|Ku_{k_j}-Ku\|+\|Ku-Ku_{k_\ell}\|+\|Ku_{k_\ell}-u_{k_\ell}\|\rightarrow 0,$$ which shows that $$(u_{k_j})$$ is Cauchy. Since $$H$$ is complete, this must converge. Additionally, the strong limit must match the (unique) weak limit, which is $$u$$.
EDIT: Alternatively, as pointed out by @DavidMitra in the comments, you can avoid showing that it's Cauchy by explicitly showing strong convergence to $$Ku$$, via the same sort of argument as before.
Calculate $$\|u_{k_j}-Ku\|\leq \|u_{k_j}-K u_{k_j}\|+\|Ku_{k_j}-Ku\|\rightarrow 0,$$ then argue as above.