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Why if an operator $A$ bounded then $A^{\ast}$ is bounded?

I have already proven that $\|A\|=\|A^{\ast}\|$ but if $A$ is bounded then exists $M$ such that $\|A(f)\|\leq M||f||$ for all $f$ in $H$ Hilbert. How prove $\|A^{\ast} f\|\leq M^{\ast} \|f\|$ for some $M^{\ast}$?

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  • $\begingroup$ I assume by $A^\ast$ you mean the adjoint of $A$, usually denoted $A^\dagger$, which satisfies $\langle A^\ast x, y \rangle = \langle x, Ay \rangle$, $\forall x, y \in \text{your Hilbert space}$? $\endgroup$ – Robert Lewis Oct 23 '18 at 0:43
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    $\begingroup$ Yes. It is correct. $\endgroup$ – eraldcoil Oct 23 '18 at 0:44
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The adjoint has a more general meaning in the context of Banach Spaces, which I think is actually more intuitive than the Hilbert Space case.

Suppose $X$ and $Y$ are Banach spaces and $T : X \to Y$ is bounded and linear. Then, define $$T^* : Y^* \to X^* : f \mapsto f \circ T.$$ Then, because $T$ is bounded and $f$ is bounded, we have $$\|T^* f\| = \|f \circ T\| = \sup_{y \in B_Y} f(Ty) \le \sup_{y \in B_Y} \|f\|\|Ty\| \le \sup_{y \in B_Y} \|f\|\|T\|\|y\| = \|f\|\|T\| \tag{1},$$ implying that $\|T^*\|$ is bounded with $\|T^*\| \le \|T\|$.

In the case of a Hilbert space, we have that $X^* \sim X$ and $Y^* \sim Y$ by the Riesz representation theorem. In particular, the map from $x \in X$ to $\langle \cdot, x \rangle \in X^*$ is a surjective linear isometry; the operator norm of $\langle \cdot, x \rangle$ is precisely $\|x\|$.

So, when we compute $T^* y$, we are really computing $T^*(\langle \cdot, y \rangle)$, which is the bounded linear functional $$X \mapsto \Bbb{F} : v \mapsto \langle Tv, y \rangle.$$ Such a functional must have a unique $x \in X$ associated to it, and that $x$ is what we call $T^* y$. In particular, we must have $$\langle Tv, y \rangle = \langle v, x \rangle$$ for all $v$, which leads us to the defining identity for adjoints: $$\langle Tv, y \rangle = \langle v, T^* y \rangle.$$

So, in short, the reason why the adjoint is bounded is because of (1).

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    $\begingroup$ This is a very beautiful proof, thank you! May I ask you in which context have you encontre such arguments? $\endgroup$ – Surb Oct 23 '18 at 11:06
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    $\begingroup$ @Surb Well, the adjoint is a standard tool in functional analysis, and the Riesz representation theorem is utterly fundamental to the study of Hilbert Spaces in particular, and is what makes the adjoint come into its own. Would that I could say that I did something special here, but the above arguments are standard in these fields. :-) $\endgroup$ – Theo Bendit Oct 23 '18 at 11:13
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    $\begingroup$ @Surb Functional analysis can be very beautiful though. I particularly enjoy the geometric flavour of functional analysis, where geometric properties of the space and its unit ball tell you deep properties about the space itself. $\endgroup$ – Theo Bendit Oct 23 '18 at 11:18
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    $\begingroup$ Thanks for your answer, sorry I wasn't precise enough. I'm particularly interested by the Banach space interpretation of the Adjoint and in particular your first estimate. I know it (quite) well for finite dimensional vector space and the proofs I know, hold for any norm but uses an inner product. $\endgroup$ – Surb Oct 23 '18 at 11:18
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    $\begingroup$ :) me too, I was about to say: "Functional analysis can be very beautiful though. I particularly enjoy the geometric flavour " I agree and actually my favorite is "nonlinear functional analysis" $\endgroup$ – Surb Oct 23 '18 at 11:22

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