$X$ is a Banach space, $X^*$ is dual space of $X$, $T:X\to X$ is a linear compact operator,$ \lambda \neq 0$, Let $S = \lambda I-T$, Then $$ran(S^*) = (ker(S))^\perp$$
From the definition $$ran(S^*) = \{x^* \in X^* : \exists y^* \in X^* \Rightarrow S^*(y^*) = x^*\}$$
\begin{align} (ker(S))^\perp&= \{x^* \in X^* : x^*(x) = 0 \quad \forall x \in ker(S)\} \\ &=\{x^* \in X^* : x^*(x) = 0 \quad \forall x \Rightarrow S(x)=0 \} \end{align}
$\mathbf Define:$ $X$ is a space, $S \subset X$ $$Def:S^\perp=\{x^* \in X^* : x^*(x) = 0 \quad \forall x \in S\}$$
I know that $ran(S)$ is closed,But I couldn't find their relationship
Thanks
The following is complement of proof in the Answer
$$(ker(S))^\perp \subset ran(S^*)$$
lemma:$X$ is a Banach space, $S \in \mathscr B(X)$,If $ran(S)$ is closed ,then exists a constant $K$ such that for any $y \in ran(S)$,there is a point $x\in X$ with $S(x)=y$ and $\Vert x \Vert \le K \Vert y \Vert$
Proof of lemma :By open-mapping theorem $S$ maps the unit ball $B_1$ of $X$ onto a set containing a ball in $ran(S)$ with center $0$, i.e. $$SB_1 \supset \{y: y\in ran(S),\Vert y\Vert \lt \delta \}$$ Now let $0 \neq y \in ran(S)$.then $ \frac{ \delta y}{2\Vert y\Vert}$ is in $SB_1$,i.e.$S(z)=\frac{ \delta y}{2\Vert y\Vert}$ for some $z,\Vert z\Vert \lt 1$.Let $x=\frac{ 2\Vert y\Vert z}{\delta}$.then $S(x)=y$ and $\Vert x\Vert \lt (\frac{2}{\delta})\Vert y\Vert$,completed.
Let $x^* \in (ker(S))^\perp.$ Define a linear functional $g$ on $ran(S)$ by $g(S(x))=x^*(x)$.g is well defined.by lemma, there exists a constent $K$ such that for any $y \in ran(S)$,there is a point $x\in X$ with $S(x)=y$ and $\Vert x\Vert \le K \Vert y\Vert$.Hence $$\vert g(y) \vert = \vert x^*(x) \vert \le K \Vert y\Vert \Vert x^*\Vert$$ Thus $g$ is a bounded linear functional on $ran(S)$.By Hahn-Banach theorem we can extend $g$ into a continuous linear functional $y^* \in X^*$.Since,for any $x \in X$, $$S^*(y^*)(x)=y^*(S(x))=g(S(x))=x^*(x)$$ we have $S^*(y^*)=x^*$.Thus $x^* \in ran(S^*)$
If $X$ is an Hilbert space you can prove that relation in this way:
$S^*$ is the adjoint operator of $S$ and so if $y\in ran(S^*)$ then there exists $x\in X$ such that
$S^*(x)=y$
And so you have that for each $z\in X$
$\langle x, S(z) \rangle= \langle y,z \rangle$
So you have that for each $z\in ker(S)$ then
$\langle y,z \rangle=\langle x, S(z) \rangle =\langle x,0 \rangle =0$
So $y\in (ker(S))^\perp$ and you have that
$ran(S^*)\subseteq (ker(S))^\perp$
If you want prove the inverse inclusion you can observe that the perpendicular operation invert the inclusion so you can prove that
$ran(S^*)^\perp \subseteq ker(S)$
Because in this case if you apply $\perp$ you have that
$ker(S)^\perp \subseteq (ran(S^*)^\perp)^\perp=ran(S^*)$
Let $x\in ran(S^*)^\perp$, we want prove that $S(x)=0$. We have that for each y\in X$
$\langle y,S(x)\rangle =\langle S^*(y), x\rangle =0$
because $S^*(y)\in ran(S^*)$.
So you have that
$\langle y,S(x) \rangle=0$ for each $y\in X$ and in particular for $y=S(x)$ you have that
$||S(x)||^2=\langle S(x),S(x) \rangle =0$
Then $S(x)=0$ and so $x\in ker(S)$
In the case in which $X$ is only a Banach space you have that
$S^*: X^*\to X^*$ is the adjoint operator in this sense:
for each $\phi\in X^*$ and $x\in X$ then
$S^*(\phi)(x):=\phi(S(x))$
So if you consider a point $\phi\in ran(S^*)$ then there exist $\psi\in X^*$ such that $S^*(\psi)=\phi$. In this case for each $x\in ker(S)$ you have that
$\phi(x)=S^*(\psi)(x)=\psi(S(x))=\psi(0)=0$
So $\phi\in (ker(S))^\perp$ that means that
$ran(S^*)\subseteq (ker(S))^\perp$
Now we must consider $\phi\in (ker (S))^\perp$ so for each $x\in ker(S)$ you have that $\phi(x)=0$ so it is possible to define
$\phi^\sim :X/ker(S)\to \mathbb{R}$ in the natural way.
Let $\pi : X\to X/ker(S)$ the projection map; we want define a new map $S^\sim: X\to X/ker(S) $ such that $S^\sim \circ S=\pi$.
Infact in this way you have that $\psi:=\phi^\sim\circ S^\sim$ is such that
$S^*(\psi)=\phi$
because $S^*(\psi)=\psi\circ S=\phi^\sim\circ (S^\sim\circ S)=\phi^\sim\circ \pi=\phi$
How it is possible define the map $S^\sim$?
If $S$ is a surjective map is simple because for each $x\in X$ let $y\in S^{-1}(x)$ and define the map
$S^\sim(x):=y+ker(S)$
The map is well defined because if $y,z\in S^{-1}(x)$ then $S(y)=x=S(z)$ so $y-z\in ker(S)$ then $y+ker(S)=z+ker(S)$.
The map is linear because for each $x,y\in X$ and $z\in S^{-1}(x), t\in S^{-1}(y)$ then
$ S(z+t)=S(z)+S(t)=x+y$ so $S^\sim(x+y)=(z+t)+ker(S)=S^\sim(x)+S^\sim(y)$
In the same way you can prove that
$S^\sim(\rho x)=\rho S^\sim(x)$
So you have that $S^*(\psi)=\phi$ and so $\phi\in ran(S^*)$