On the pillars of Functional Analysis

Many authors consider that the main pillars of Functional Analysis are the Hahn–Banach Theorem, the Uniform Boundedness Principle and the Open Mapping Principle. The first one is derived from Zorn’s Lemma, while the latter two usually are obtained from Baire’s Category Theorem. In this paper we show that these three pillars should be either just two or at least eight, since the Uniform Boundedness Principle, the Open Mapping Principle and another five theorems are equivalent, as we show in a very elemental way. Since one can give an almost trivial proof of the Uniform Boundedness Principle that does not require the Baire’s theorem, we conclude that this is also the case for the other equivalent theorems that, in this way, are simultaneously proved in a simple, brief and concise way that sheds light on their nature.

Theorem 1.1 or, alternatively, the (HBT) and the seven theorems involved in the statement of Theorem 1.1. The proof of Theorem 1.1 is done in a very concise and simple way. Since a direct and elementary proof of the Uniform Boundedness Principle that does not require Baire's Theorem can be given, we conclude that this is also the case for all the results involved in Theorem 1.1. Thus, beyond providing a short and simultaneous proof of all of them, we also show that they have the same relevance because they are logically equivalent as Theorem 1.1 establishes. In fact, the proof of this result reveals how close the seven theorems that it involves are to each other.
As usual, if X and Y are normed spaces over K (= R or C) then, L(X ,Y ) will denote the normed space of all bounded linear operators from X into Y .
We will prove this theorem in Sect. 3.

The background
The aim of this section is to establish Proposition 2.1 (whose proof is straightforward) and Theorem 2.5 (which is a direct consequence of the Bipolar Theorem and, therefore, of the (HBT)).
From now on, the graph of a linear map T will be denoted by G(T ). To show (ii) ⇒ (i), let y ∈ R(T ). Since (0, y) ∈ X × R(T ), by (2.3), there exists (w, T (w)) ∈ G(T ) (which means that w ∈ D(T ) and T w = y) such that

Therefore, T is open.
To prove (i) ⇒ (ii) suppose that T is open and let K > 0 be such that, for every y ∈ R(T ), there exists w ∈ D(T ) with T w = y satisfying w ≤ K y . Assume that K > 1 (replace The next result characterizes this property.

Lemma 2.2
Let M and N be subspaces of a normed space X , and r > 0.
, and the rest is clear.
Next, we characterize again the openness of the map M × N → M + N given by (m, n) → m + n.

Lemma 2.3
Let M and N be subspaces of a normed space X . Then the following assertions are equivalent: Let X * be the topological dual of X . For A ⊆ X and F ⊆ X * let A 0 and 0 F denote the polar of A and the prepolar of F, respectively. Thus For A, B ⊆ X and F, G ⊆ X * basic properties are the following: The well-known Bipolar Theorem states that if A is a subset of a normed space X then, 0 (A 0 ) is the closure of the absolute convex hull of A. This is an immediate (and trivial) consequence of the geometrical version of the Hahn-Banach Theorem (See [12, Theorem 15.5] for details).
On the other hand the orthogonal of A ⊆ X and the preorthogonal of F ⊆ X * are the sets

Lemma 2.4 For closed subspaces M and N of a Banach space X , and r
Theorem 2.5 Let M and N , be closed subspaces of a Banach space X . The following assertions are equivalent: 3, this is equivalent to the existence of r > 0 such that, We conclude this section with some remarks on the dual of a densely defined operator. Let X and Y be Banach spaces and T : D(T ) ⊆ X → Y be a densely defined linear operator. We define Let gT ∈ X * be the unique continuous extension to X that the operator gT has. We define the adjoint (or dual) operator of T as the operator T * : D(T * ) → X * given by T * g = gT , for every g ∈ D(T * ). Note that G(T * ) is a subspace of Y * × X * . Moreover the spaces Y * × X * and (X × Y ) * are canonically identified by defining (g, f ) For the sake of completeness (as we will not need it) note that if, additionally, T is continuous then D(T * ) = Y * , trivially, and T * , is continuous with T * ≤ T . Moreover, for every x ∈ D(T ), there exists g ∈ Y * with g = 1 such that |g(T x)| = T x , so that T x = T * g(x) ≤ T * g x ≤ T * x , and hence T = T * . In fact, we have that T is continuous if and only D(T * ) = Y * and T * : Y * → X * is continuous, in which case T = T * .

The proof of the main theorem
Next we prove Theorem 1.1. Also we will provide a proof of the Uniform Boundedness Principle that does not need to use Baire's Theorem. For the sake of completeness we include the proofs of all the assertions involved being aware of that some of them are well known.

Proof of Theorem 1.1 (i) ⇒ (ii). [(UBP) ⇒ (OMT)].
Let X and Y be Banach spaces, and T ∈ L(X ,Y ) a surjective operator. For n ≥ 1, define Then · n is a norm on Y and y n ≤ n y , for every y ∈ Y (take u = 0 and v = y as T (0) + y = y). Let Z be the space of all the sequences in Y having a finite number of non-zero entries, provided with the norm {z n } := max n∈N z n n , for {z n } ∈ Z .
Let T n : Y → Z be the map T n (y) = {z k } k∈N where z k = 0 if k = n and z n = y. Obviously T n ∈ L(Y ,Z ) (as T n (y) = y n ≤ n y ). In fact, T n ≤ n. On the other hand, {T n : n ∈ N} is pointwise bounded because if y ∈ Y , then there exists x ∈ X such that T x = y, so that T n (y) = y n ≤ x , for every n ≥ 1. By the (UBP) there exists M > 0 such that T n ≤ M, for every n ∈ N. Thus, y n = T n (y) ≤ M y , for every y ∈ Y . Consequently, if y < 1 M then y n < 1, and there exists u n ∈ X , and v n ∈ Y , such that T u n + v n = y and u n + n v n → y n < 1.
Therefore, u n < 1 and n v n < 1. Since v n < 1 n we have T u n → y so that y ∈ T (B X ). This shows that, for We claim now that B Y (0, r 2 ) ⊆ T (B X ). From (3.1), if we rescale then, ). Therefore there exists 3 . Iterating, we obtain x n ∈ B X (0, 1 2 n ) satisfying y − n k=1 T x k < r 2 n+1 , for every n ∈ N. Since X is complete and ∞ k=1 x k < ∞ k=1 1 2 n = 1, we have that x = ∞ k=1 x k ∈ X and x < 1. Moreover, . Let X be a Banach space, Y a normed space and T ∈ L(X ,Y ) a surjective map. If Y is complete then, T is open by the (OMT). Conversely, that T is open means that T −1 is continuous, where T : X / ker T → Y is the canonical factorization of T . Hence T is bicontinuous and the complete norm of X / ker T induces a norm on Y , namely |||y||| = T −1 (y) , which is equivalent to the original one, and consequently Y is complete.
is complete. Since |||·||| ≤ · , by the (NT) we obtain that G(T ) is closed if and only if these norms are equivalent which is nothing but the continuity of T . {T i x} i∈I (note that T is well defined as {T i } i∈I is pointwise bounded). Moreover, G(T ) is closed as we can check directly. Thus, the projection P X : G(T ) → X , is a bijective bounded linear operator (between Banach spaces) whose range is closed. Therefore P X is open, by the (CRT), which means that P −1 X is continuous. Since the projection P Y : G(T ) → Y is obviously continuous we conclude that T := P Y P −1 X is continuous which is nothing but sup i∈I T i < ∞. This is a generalization of the (OMTbis), to the wider class of densely defined operators with closed graph, that can be deduced from the (ST) joint with Proposition 2.1. Moreover, (3.2) is enough to obtain the (UBP) as showed in the assertion (CRT) ⇒ (UBP). Therefore, the role of the Bipolar Theorem is just to show that T is open if and only if T * is open. Consequently, if in Theorem 1.1 we replace the (CRT) statement with (3.2) then, Section 2 can be reduced to Proposition 2.1.
The proof of (UBP) ⇒ (OMT) follows [6], and some ideas from [9] have been useful for our approach in the proof of (ST) ⇒ (CRT).
For the sake of completeness we include a proof of the (UBP) that does not require Baire's Theorem, from [11]. Others proofs of the (UBP) that do not use Baire's Theorem are known; for instance the one given by Hahn applying the gliding hump argument [8,Exercise 1.76].

Proof of the Uniform boundedness principle.
First of all, we establish the following trivial result. Lemma 3.1 Let X and Y be normed spaces and let T ∈ L(X ,Y ). Then, for every x 0 ∈ X and every r > 0, sup x∈B(x 0 ,r ) T (x) ≥ r T .