X is a subset of Y if and only if Y* is a subset of X*.

@ Martin Krepla

For m

The relationship is the following: Y* can be canonically identified with the quotient space of X* by the annihilator of Y. See section 9.4.2 of my book https://link.springer.com/book/10.1007%2F978-3-319-92004-7

If Y is dense in X, then any linear continuous form on Y has a unique linear continuous extension to the whole space X. Hence, in this case, Y* is contained in X*. On the other hand, the restriction to Y of any linear continuous form on X is linear and continuous on Y, for any vector subspace Y of X, even in the case when Y is not dense in X. Thus, it seems that always X* is contained in Y*.

The following additional remark also holds: if Y is not dense in X, then two different linear continuous forms on X can have the same restriction to Y. Son we can see that X*=Y* if and only if Y is dense in X.

Octav Olteanu If two different linear continuous functionals f, g on X have the same restriction to Y, then their difference annihilates Y. Conversely, the whole set $\{g \in X^*: g-f \in Y^\bot\}$ generates the same functional on Y. This is exactly the way how Y* can be identified with the quotient space $X^*/Y^\bot$. When Y is dense, then $Y^\bot = \{0\}$, so X^* = Y^*.

Dear Colleague Vladimir Kadets, my answer to the question was: Y*=X* if and only if Y is dense in X. It is possible that this answer be obtained from a more general theory, but my answer is much simpler. Why complicate the problem?

Duality naturally inverts the sense of the arrows. Given a continuous linear map Phi from X to Y, the only natural way to define a map relating X* and Y* is the following: for any continuous linear functional g from Y to the scalars, define f=g.Phi from X to the scalars. The most convenient way to check this is to draw a triangle of maps. One obtains in this way a continuous linear map Psi from Y* to X*. The relation between X* and Y* depends of course on the properties of Phi. For example, in the case of finite dimensional X and Y, it is purely a problem of linear algebra to show that Psi is injective iff Phi is surjective. In the infinite dimensional case, a topological answer must be in the style of Vladimir Kadets.

Octav Olteanu Dear colleague, for sure, your answer is perfectly correct. My comment just develops your argument. Being a mathematics teacher I have a feeling that if a person asks such kind of questions, then for this person it may be useful to make one step more and learn what is the complete description of the object in question, especially if this description is supposed to be a part of standard functional analysis course. Maybe I am wrong ...

Dear Colleague Professor Vladimir Kadets, you are perfectly right, and not wrong. I have learned such things from books on functional analysis too.

There is the natural map X* to Y*, since any continuous functional f on X defines a continuous function on Y (called restriction). If f vanishes on Y, the restriction is zero.

Therefore the natural map as above generates the map p from X*/Y^0 to Y* where Y^0 means the subspace in Y* of functionals vanishing on Y. Map p is surjective if we are ready to apply Hahn-Banach theorem. There are no more of "natural" maps between these "players". In particular there is no natural no map of Y* to X*.

However such a map does exist for any Hilbert X. This map is not "natural" since special properties of Hilbert space are necessary.

Dear colleagues,

I would like to correct a small thing in the first answer of Octav Olteano, Octav asserts that X* is always contained in Y*. This is not true, for in order to identify X* with a subspace of Y*, the restriction map must be one to one. But this is not the case in general. This is true iff Y is dense.

Dear colleague Lahbib Oubbi,

I have corrected/completed the first answer which you refer to in the first part of my second answer, 6 days ago, immediately after my first answer.

I general it is a quotient of the dual space (Y closed, take N all continuous linear forms on X which are zero on Y; then Y* is just X*/N with the quotient norm).

Can anyone please provide me an example which shows that Y* is not a subspace of X*, where Y is a subspace of X?

Hi Udday,

If Y* were a subspace of X*, there would be an into mapping f from Y* into X* , with the restriction of f(y') to Y being y' itself, y' bing in Y*. Now, if X were the space of all bounded séquences, with the sup norm, Y = c_0, and f such a mapping, then the image of the null functional would be an element of X*. But for an arbitrary element a of the Stone-Cech compactification of N not belonging to N, the evaluation at a is zero on c_0. Therefore f is actually not well defined.

Actually, Y* is the quotient of X* on the annihilator Y° of Y in X*, i.e., Y* = X*/Y°.

Lahbib Oubbi Thanks for your answer. I have some doubts. Kindly throw more light on them.

1. Are we considering f:Y*->X* by f(y')=Hahn-Banach extension of y'?

2. Since C(bN) is isomorphic to l^{infty}, therefore any element y in c_0 can be seen as an element of C(bN). Now if x* is the evaluation at a (an element of bN\N), then x*(y)=y(a)=0 and so x* restricted to c_0 is the zero functional. But how to say that f is not well defined?

Dear Uday,

The function you consider in 1. is not well defined, because the Hahn-Banach extension is not unique. For example every continuous functional vanishing on c_0 is an extension of the null functional. In particular all the evaluations at any point of bN/N is an extension of 0-functional. And obviously all such extensions are pairwise distinct.

There is no canonical injection of Y* into X*, although, choosing for any y' one extension x', one obtains an injection. But there infinitely many ways to do it. We can say that Y* is "floating" subspace of X*.

Yes, I got it. Since dual of c_0 is not strictly convex, therefore Hahn-Banach extension can not be unique. Thanks!