# Relation between kernel and range

$$X$$ 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 $$x^* \in ran(S^*)$$

• What is your definition of perpendicular of a space? Commented Jul 6, 2019 at 9:46
• Thanks,I added definition in quesition.
– John
Commented Jul 6, 2019 at 10:17
• I suspect that this requires Banach's closed range theorem from which you can conclude that range of $S^{*}$ is also closed. Commented Jul 6, 2019 at 11:41

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 $$\phi\in ran(S^*)$$

• Your proof is in Hilbert space setting but the question is about Banach spaces. Commented Jul 6, 2019 at 11:38
• Now I proved for Banach space when S is a surjective map Commented Jul 6, 2019 at 11:59
• The third row,S^* is map from X^* to X^*, x belongs to X,S^*(x) means what ?
– John
Commented Jul 6, 2019 at 15:34
• @宋智鹏 it is in the case in which H is an Hilbert space where X^*=X up to isomorphism and you can solve the problem in that way Commented Jul 6, 2019 at 15:35
• I added the complement of your proof in the question.
– John
Commented Jul 6, 2019 at 17:57