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Let $X$ and $Y$ be normed spaces whose continuous dual are $X^*,Y^*$, respectively.

Let $T:X\to Y$ be a bounded linear operator. The adjoint $T^*$ of $T$ is defined to be the map $T^*:Y^*\to X^*$ such that $$ T^*g(x):=g\left(Tx\right) $$ for all $g\in Y^*$. Some author call $T^*$ the dual operator of $T$.

So far so good, until I started thinking about what $T^*$ actually is. I was able to prove the following relation:

$$ \begin{matrix} \text{Im}(T^*)\subseteq\ker(T)^\perp, & \ker(T^*)\subseteq\text{Im}(T^*)^\perp \\ \ker(T)=\text{Im}(T^*)_\perp, &\text{Im}(T)=\ker(T)_\perp \end{matrix} $$

where for $A\subseteq X$ we define the annihilator $A^\perp$ by $$ A^\perp :=\{\ f\in X^*\ |\ f(x)=0 \ \ \text{for all}\ x\in A \} $$ and $B\subseteq X^*$ we define the pre-annihilator $B_\perp$ be $$ B_\perp :=\{\ x\in X\ |\ f(x)=0 \ \ \text{for all}\ f\in B \} $$ and likewise for $Y$.

Here comes the stupid question: When I tried to show $\ker(T)^\perp\subseteq\text{Im}(T^*)$ to get the equality sign, it seems that it cannot be done. Why?

I am not asking for counter example to show that $\ker(T)^\perp\subseteq\text{Im}(T^*)$ is not true, I am already convinced of that. What I am trying to understand is why is there an asymmetry there?

I kind of know what an adjoint operator does, i.e. allowing us to "move things around" when trying to get something done, like the integration by parts formula on functions with compact support $$ \int (\partial f)g = -\int f (\partial g). $$ This allow us to extends to distributions and stuff so it's very useful. However, I've realized that although I know what it does, I don't really know what it is. Aside from being able to manipulate $T^*$ algebraically, I am pretty much oblivious to any geometric or any intuitive interpretation of it.

Some related articles that I've founded:

What is an intuitive view of adjoints?

This comes really close to what I am asking, but it is quite Hilbert space oriented and I want to understand general adjoint operators. Mr. Qiaochu Yuan's answer seems very interesting but it is too short and I don't really understand what he meant by "backward transition". Perhaps I'd be really great if anyone could elaborate on that.

Geometric intuition of adjoint

This one doesn't really help me at all. The "geometric view" proposed by one of the answer relies heavily on the Hilbert space structure and even if I were to look over that, I still can't imagine orthogonal complement in that product space.

How to interpret the adjoint?

The top answer is good, but it addresses mostly what I already knew. It doesn't really help me understand the asymmetry there. Other answers are insightful but don't mention anything about the specific relation between the structure of kernel and image of $T,T^*$.

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BigbearZzz
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  • Note $(\ker T)^{\perp} = \bigl((\operatorname{Im} T^{\ast}){\perp}\bigr)^{\perp}$ is the weak$^{\ast}$ closure of $\operatorname{Im} T^{\ast}$. If $X,Y$ are Banach spaces, and $\operatorname{Im} T$ is closed, then $\operatorname{Im} T^{\ast}$ is weak$^{\ast}$-closed, so then you have equality. It's if-and-only-if, by the way (closed range theorem). On the other hand, $\bigl((\ker T)^{\perp}\bigr){\perp}$ is the weak closure of $\ker T$. – Daniel Fischer Oct 26 '16 at 20:18
  • @DanielFischer I noticed that the equality seems to hold when we add the assumption that $X,Y$ are Banach but I didn't quite understand why. Your comment seems to partially explain that, thanks! – BigbearZzz Oct 26 '16 at 20:25
  • We also need that $T$ has closed range, otherwise it doesn't hold. – Daniel Fischer Oct 26 '16 at 20:30

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