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Let $A: B_1 \rightarrow B_2$ be a mapping, where $B_1$, $B_2$ are two Banach Spaces.

Then the adjoint of $A$, $A^*$, is defined by

$A^*:B_2^* \rightarrow B_1^*$ s.t for all $f \in B_1^*, f:B_1 \rightarrow \mathbb{R}$ is linear.

$\langle Ax{,}y\rangle$=$\langle x{,}A^*y\rangle$

My Question is: why when dealing with Hilbert Spaces $H_1$ and $H_2$, we simply say: IF $A:H_1 \rightarrow H_2$, THEN $A^*:H_2 \rightarrow H_1$, without taking into consideration $H_2^*$ and $H_1^*$ ?

sara
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  • Because if $H$ is a Hilbert space, then $H$ and $H^*$ are isometric. – Surb Feb 13 '21 at 12:03
  • Every bounded linear functional is an inner product with a fixed vector in a Hilbert space, this gives a correspondence $H^*\cong H$. – Berci Feb 13 '21 at 12:04

1 Answers1

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The reason behind this is that every Hilbert space is isometrically isomorphic to its dual by the Riesz representation theorem, stating that any functional $\phi$ on a Hilbert space $H$ is of the form $\phi(\cdot)=\langle\cdot,z\rangle$ for some $z\in H$. Using this one-to-one correspondence between $H^*$ and $H$, one can see that if $A:H_1\to H_2$ is a bounded linear operator, then the "Hilbert space adjoint: $A^*:H_2\to H_1$ which is defined by $\langle Ax,y\rangle=\langle x,A^*y\rangle$ for all $x,y$ is "essentially the same" as the "Banach space adjoint: $A^*:H_2^*\to H_1^*$.

If $\sigma:H_1\xrightarrow{\cong}H_1^*$ and $\rho:H_2\xrightarrow{\cong}H_2^*$ are the isometric isomorphisms, then $\sigma^{-1}\circ A^{*,Ban}\circ\rho=A^{*,Hil}$.

The only essential difference of the two adjoints is that in the Hilbert setting we have that $(\lambda A)^*=\bar{\lambda}A^*$ while on the Banach setting we have that $(\lambda A)^*=\lambda\cdot A^*$.

Related: Hilbert space adjoint vs Banach space adjoint?

Just dropped in
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  • Sorry but regarding your comment, 1)are $L^p$ spaces Hilbert spaces? 2)can you talk little bit about Riesz Represntation theorem and where is its point regarding $L^p$ spaces? – sara Feb 18 '21 at 15:59
  • @sara 1) No, the only $L^p$ space that is a Hilbert space is $L^2$, none of the rest are Hilbert spaces. 2) The Riesz rep. theorem is related to this because it says that $H\cong H^*$ (isometrically isomorphic) for any Hilbert space. My point is, that the Hilbert space adjoint is not really different that the adjoint map you are reffering to. It's just the composition of the adjoint together with the isometric isomorphisms that are induced by Riesz's rep. theorem. – Just dropped in Feb 18 '21 at 16:12
  • Sry but why the rest $L^p$ spaces are not Hilbert? How to prove it? can you help me plz – sara Feb 18 '21 at 16:15
  • @sara Use this: https://math.stackexchange.com/questions/3105198/a-banach-space-that-satisfies-parallelogram-law-is-a-hilbert-space/3105282#3105282 – Just dropped in Feb 18 '21 at 16:15
  • @sara also related: https://math.stackexchange.com/questions/3017814/lp-space-is-a-hilbert-space-if-and-only-if-p-2 – Just dropped in Feb 18 '21 at 16:16
  • So as a conclusion, if I can apply the parallelogram inequality to an $L^p$ space, then it should be Hilbert, and so it's Nothing but an $L^2$ space, isn't it! – sara Feb 18 '21 at 16:21
  • @sara exactly. More generally, a normed space has an inner product structure hiding behind the norm if and only if the norm satisfies the parallelogram law. So, in order to show that the $L^p$ spaces for $p\neq 2$ are not Hilbert spaces, find functions $f,g$ in $L^p$ so that the parallelogram law fails for those two. There are definitely such posts on MSE, so there you go! – Just dropped in Feb 18 '21 at 16:23
  • Sorry but what is an MSE? How to search in it! – sara Feb 18 '21 at 16:28
  • @sara MSE= Math Stack Exchange. Just open up google and type "Lp is not a Hilbert space", you will definitely find results that can help you – Just dropped in Feb 18 '21 at 16:29
  • thank you so much! – sara Feb 18 '21 at 16:29