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Let $T$ be an operator defined by $$T:X \to Y$$ where X and Y are Banach spaces. The adjoint of $T$, $T^*$ is defined by
$${T^*}:{Y^*} \to {X^*}$$ where ${X^*}$, ${Y^*}$ are the dual spaces. In other words, the adjoint maps linear functionals of $Y$ (elements of $Y^*$) into linear functionals of $X$ (elements of $X^*$). This is a bit hard to wrap one's head around it. Could someone give an example of an operator and its adjoint and how it maps linear functional into other linear functionals, or elaborate on the concepts.

Thanks in advance.

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    $\begingroup$ The definition makes it pretty clear. If $T:U\to V$ and $f:V\to \mathbb R,$ then $f\circ T: U\to \mathbb R.$ This mean $f\mapsto f\circ T$ sends $V^*\to U^*.$ The general way to state this is that the functor $X\to\operatorname{Hom}(X,Z)$ is “contravariant,” which means the arrows are reversed. With real vector spaces, if $Z=\mathbb R,$ you get the dual space. Banach spaces are more complicated, with continuity added to the linear map condition, but intuitively the same. $\endgroup$
    – Thomas Andrews
    Dec 6, 2021 at 21:15

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If $f\in Y^*$ then $f$ is a map from $Y$ to $\Bbb R$ or to $\Bbb C$. We define the map $g=T^*(f)$ from $X$ to $\Bbb R$ or to $\Bbb C$ by letting $g(x)=f(T(x))$ for each $x\in X.$ That is, $T^*(f)(x)=f(T(x)).$

Note that this is not $T^*(f(x)),$ which is meaningless. $T^*(f)$ is to be treated as a single symbol for one member of $X^*$.

For example if $X=Y=\Bbb R^2$ and $T(u,v)=(2v,u)$ for all $(u,v)$ then for any $f\in Y^*$ we have $T^*(f)(u,v)=f(T(u,v))=f(2v,u).$

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  • $\begingroup$ what do you mean by $T^*(f)$ meaningless, and why is it meaningless ? $\endgroup$
    – user830472
    Dec 6, 2021 at 22:02
  • $\begingroup$ Another thing, what guarantees that $f(2v.u) \in X^*$ ? $\endgroup$
    – user830472
    Dec 6, 2021 at 22:04
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    $\begingroup$ $T^*(f)$ is not meaningless. $T^*(f(x))$ is meaningless. The brackets are wrongly placed... Assume $T$ is bounded , that is, $\infty>\|T\|=\sup\{\|T(x)\|:\|x\|=1\}.$ Then $|T^*(f)(x)|=|f(T(x)|\le \|f\|\cdot \|T(x)\|\le\|f\|\cdot \|T\|\cdot \|x\| $....So $T^*(f)$ is bounded for each $f\in Y^*.$. And it is easy to show that if $T$ is linear and if $f $ is linear then $T^*(f)$ is linear. So if $T$ is bounded and linear then $T^*(f)\in X^* $ whenever $f\in Y^*.$ $\endgroup$
    – DanielWainfleet
    Dec 6, 2021 at 23:02
  • $\begingroup$ is $f \in Y^*$ bounded by assumption?. You showed that $T^*$ is bounded and linear, why is this relevant for it to be part of $X^*$ ? $\endgroup$
    – user830472
    Dec 6, 2021 at 23:30
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    $\begingroup$ The usual definition of the symbol $X^*$ is the space of bounded linear functionals on $X$.... And if $T$ is not bounded then $T^*(f)$ could be unbounded for some bounded $f$ . $\endgroup$
    – DanielWainfleet
    Dec 7, 2021 at 21:58

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