Machine-translated from Chinese.
定理 1. $H$为Hilbert空间. 若$K:H\rightarrow H$为紧算子, 则$L=I-K$的值域是闭的.
证: $H=N(L)\oplus N(L)^\perp.$
$L:N(L)^\perp\rightarrow R(L)$是一一的. 断言$\,\exists\,c>0,$ 满足
$$\Vert Lf\Vert\ge c\Vert f\Vert,\quad \,\forall\,f\in N(L)^\perp.$$
不然, $\,\exists\,\{f_n\}\subset N(L)^\perp,$
满足$\Vert f_n\Vert\equiv 1,$ $\Vert Lf_n\Vert\rightarrow 0.$
由于$K$是紧算子, $\{Kf_n\}$有子列收敛, 不妨设为自身收敛. 这样,
$$\widetilde{f}:=\lim_{n\rightarrow \infty}Kf_n+Lf_n=\lim_{n\rightarrow \infty}f_n.$$
结合闭性, 这说明了$\{f_n\}$自身也收敛到$\widetilde{f}\in N(L)^\perp.$
而$L\widetilde{f}=\lim\limits_{n\rightarrow \infty}Lf_n=0,$
$\widetilde{f}\in N(L).$ 这说明$\widetilde{f}=0.$
然而$\Vert f_n\Vert\equiv 1,$ 不可能收敛到$0,$ 矛盾.
接下来, $\,\forall\,\{g_n\}\subset R(L)$满足$g_n\rightarrow g,$
我们需要证明$g\in R(L).$
我们取$\{f_n\}\subset N(L)^\perp$满足$Lf_n=g_n,$
那么$\Vert g_n\Vert\ge c\Vert f_n\Vert.$ 由于$\{g_n\}$构成Cauchy列,
$\Vert f_n\Vert$也是, 记其收敛到$f.$
这样就有$Lf=\lim\limits_{n\rightarrow \infty} Lf_n=g.$
参考: https://www.math.tamu.edu/\~fnarc/m641/m641_notes/closed_range.pdf
文章最后更新于 2021-12-07 16:09:09
Theorem 1. $H$ is a Hilbert space. If $K:H\rightarrow H$ is a compact operator, then the range of $L=I-K$ is closed.
Certificate: $H=N(L)\oplus N(L)^\perp.$
$L:N(L)^\perp\rightarrow R(L)$ is one-to-one. Assert $\,\exists\,c>0,$ satisfies
$$\Vert Lf\Vert\ge c\Vert f\Vert,\quad \,\forall\,f\in N(L)^\perp.$$
Otherwise, $\,\exists\,\{f_n\}\subset N(L)^\perp,$
Meet $\Vert f_n\Vert\equiv 1,$ $\Vert Lf_n\Vert\rightarrow 0.$
Since $K$ is a compact operator and $\{Kf_n\}$ has subsequence convergence, it might as well be set to self-convergence. In this way,
$$\widetilde{f}:=\lim_{n\rightarrow \infty}Kf_n+Lf_n=\lim_{n\rightarrow \infty}f_n.$$
Combined with the closed property, this shows that $\{f_n\}$ itself also converges to $\widetilde{f}\in N(L)^\perp.$
And $L\widetilde{f}=\lim\limits_{n\rightarrow \infty}Lf_n=0,$
$\widetilde{f}\in N(L).$ This means $\widetilde{f}=0.$
However, $\Vert f_n\Vert\equiv 1,$ cannot converge to $0,$, which is a contradiction.
Next, $\,\forall\,\{g_n\}\subset R(L)$ satisfies $g_n\rightarrow g,$
We need to prove $g\in R(L).$
We take $\{f_n\}\subset N(L)^\perp$ to satisfy $Lf_n=g_n,$
Then $\Vert g_n\Vert\ge c\Vert f_n\Vert.$ since $\{g_n\}$ constitutes the Cauchy column,
$\Vert f_n\Vert$ is also the same, remember that it converges to $f.$
This way we have $Lf=\lim\limits_{n\rightarrow \infty} Lf_n=g.$
Reference: https://www.math.tamu.edu/\~fnarc/m641/m641_notes/closed_range.pdf
The article was last updated on 2021-12-07 16:09:09
