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 $R(I-k)=H$,


(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).

  • 2
    $\begingroup$ (5) implies $u_{k_j}$ is norm (and thus weakly) convergent to $Ku$. But, it also converges weakly to $u$. This implies $Ku=u$. $\endgroup$ Dec 3, 2020 at 16:31
  • $\begingroup$ Hey David, thank you for your reply! I see it now. $\endgroup$ Dec 3, 2020 at 18:19

1 Answer 1


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.

This is easier, but my mind went to Cauchy first for some reason.

  • $\begingroup$ Wonderful. Thank you so much! $\endgroup$ Dec 3, 2020 at 18:19

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