Ann. Funct. Anal. 4 (2013), no. 1, 109–113
A nnals of F unctional A nalysis
ISSN: 2008-8752 (electronic)
URL:www.emis.de/journals/AFA/
A CHARACTERIZATION OF THE INNER PRODUCT SPACES
INVOLVING TRIGONOMETRY
DAN ŞTEFAN MARINESCU1 , MIHAI MONEA2∗ , MIHAI OPINCARIU3 AND MARIAN
STROE4
Communicated by J. Chmieliński
Abstract. In this paper we will give a new characterization of the inner
product space which use the trigonometry. We conclude that a normed space
(X, k·k) is an inner product space if and only if there exists α ∈ R\πQ so that
2
2
2
2
kx cos α + y sin αk + ky cos α − x sin αk = kxk + kyk ,
for any x, y ∈ X.
1. Introduction
The problem of finding some necessary and sufficient geometric conditions for
a normed space to be an inner product space has been investigated by several
mathematicians for a long time. Some characterizations of inner product spaces
and their generalizations can be found in [1, 2, 4] and references therein.
In [3], Moslehian and Rassias gave a characterization of the inner product
space using an Euler-Lagrange type identity. The result is presented in the next
proposition.
Proposition 1.1. Let be (X, k·k) a normed space. Then the norm is derived
from an inner product space if and only if
kax + byk2 + kbx − ayk2 = a2 + b2 kxk2 + kyk2 ,
for any x, y ∈ X and a, b > 0.
Date: Received: 31 August 2012; Accepted: 20 October 2012.
∗
Corresponding author.
2000 Mathematics Subject Classification. Primary 46C15; Secondary 46B20.
Key words and phrases. Normed spaces, inner product spaces, trigonometry.
109
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D. Ş. MARINESCU, M. MONEA, M. OPINCARIU, M. STROE
Starting from this result, we will obtain another characterization for the inner
product space involving the trigonometry.
2. Some Preliminary Results
The identity from previous proposition could be transformed. After we divide
with a2 + b2 , we obtain
2 2
a
b
b
a
√
+ √
= kxk2 + kyk2 .
√
√
x
+
y
x
−
y
a2 + b 2
2
2
2
2
2
2
a +b
a +b
a +b But, it easy to see that it exists a real number α ∈ 0, π2 for which cos α =
√ a
and sin α = √a2b+b2 . In this context, our identity becomes
a2 +b2
kx cos α + y sin αk2 + kx sin α − y cos αk2 = kxk2 + kyk2 .
In fact, the result is more general and it is exposed and proved in the next
proposition.
Proposition 2.1. Let be (X, k·k) a normed space. Then the norm is derived
from an inner product space if and only if
kx cos α + y sin αk2 + ky cos α − x sin αk2 = kxk2 + kyk2 ,
for any x, y ∈ X and α ∈ R.
Proof. If X is an inner product spaces, we have
kx cos α + y sin αk2 + ky cos α − x sin αk2
= hx cos α + y sin α, x cos α + y sin αi + hy cos α − x sin α, y cos α − x sin αi
= kx cos αk2 + hx cos α, y sin αi + hy sin α, x cos αi + ky sin αk2 +
+ ky cos αk2 − hx sin α, y cos αi − hy cos α, x sin αi + kx sin αk2
= sin2 α + cos2 α kxk2 + sin2 α + cos2 α kyk2
+ (hx, yi + hy, xi − hx, yi − hy, xi) sin α cos α
= kxk2 + kyk2 .
For the second part of the proof, we choose α =
π
4
and identity
kx cos α + y sin αk2 + kx sin α − y cos αk2 = kxk2 + kyk2
becomes
kx + yk2 kx − yk2
+
= kxk2 + kyk2 ,
2
2
which conclude our proof .
A CHARACTERIZATION OF THE INNER PRODUCT SPACES
111
But, our scope is to improve this result. For this we will remind two classical
result from the mathematical analysis. First, we denote C (O, 1) the unit circle
from R2 , and πQ = {πk|k ∈ Q} .
Lemma 2.2. For any α ∈ R\πQ, the set {(cos nα, sin nα) | n ∈ N} is dense in
C (O, 1).
Lemma 2.3. Let be (X, k·k) a normed space. The the function f : X →
R, f (x) = kxk is continuous.
Now we can present and prove the main result of our paper.
3. The Main Result
Using the lemmas reminded in previous paragraph, now we can improve the
results from Proposition 2.1. in the next form:
Theorem 3.1. Let be (X, k·k) a normed space. Then the norm is derived from
an inner product space if and only if it exists α ∈ R\πQ so that
kx cos α + y sin αk2 + ky cos α − x sin αk2 = kxk2 + kyk2 ,
for any x, y ∈ X.
Proof. The only if part of the proof is true for any α ∈ R how we seen in Proposition 2.1., so particullary for an α ∈ R\πQ. For the if part, we consider that
there exists α ∈ R\πQ with
kx cos α + y sin αk2 + ky cos α − x sin αk2 = kxk2 + kyk2 ,
for any x, y ∈ X. We replace x with x cos α + y sin α and y with y cos α − x sin α.
We obtain
kx cos α + y sin αk2 + kx sin α − y cos αk2 =
= k(x cos α + y sin α) cos α + (y cos α − x sin α) sin αk2 +
+ k(y cos α − x sin α) cos α − (x cos α + y sin α) sin αk2
2 2
= x cos2 α − sin2 α + y · 2 sin α cos α +y cos2 α − sin2 α − x · 2 sin α cos α
= kx cos 2α + y sin 2αk2 + ky cos 2α − x sin 2αk2 .
So, we obtain the identity
kx cos 2α + y sin 2αk2 + ky cos α − x sin αk2 = kxk2 + kyk2 ,
for all x, y ∈ X. Further, we will use the mathematical induction and we suppose
that the identity
kx cos kα + y sin kαk2 + ky cos kα − x sin kαk2 = kxk2 + kyk2 ,
112
D. Ş. MARINESCU, M. MONEA, M. OPINCARIU, M. STROE
is true for some k ∈ N and we prove it for k + 1. In the initial identity, we replace
x with x cos kα + y sin kα and y with y cos kα − x sin kα and we have
kx cos kα + y sin kαk2 + ky cos kα − x sin kαk2 =
= k(x cos kα + y sin kα) cos α + (y cos kα − x sin kα) sin αk2 +
+ k(y cos kα − x sin kα) cos α − (x cos kα + y sin kα) sin αk2
= kx (cos kα cos α − sin kα sin α) + y (sin kα cos α + cos kα sin α)k2 +
+ ky (cos kα cos α − sin kα sin α) − x (sin kα cos α + cos kα sin α)k2
= kx cos (k + 1) α + y sin (k + 1) αk2 + ky cos (k + 1) α − x sin (k + 1) αk2 .
So, we have that the identity
kx cos nα + y sin nαk2 + ky cos nα − x sin nαk2 = kxk2 + kyk2 ,
is true for all n ∈ N∗ .
Now, we apply Lemma 2.2. There exists a sequence (an )n∈N of natural numbers
for which
lim an = ∞
n→∞
and
√
lim (cos an α, sin an α) =
n→∞
2
,
2
√ !
2
.
2
Then, we obtain
kx cos an α + y sin an αk2 + ky cos an α − x sin an αk2 = kxk2 + kyk2 .
If we make n → ∞ and use Lemma 2.3., we have
kx + yk2 kx − yk2
+
= kxk2 + kyk2 ,
2
2
for all x, y ∈ X and our proof is ready.
Final remark: Obviously, the set R\πQ can be enlarged. For example, if
√
√
2k+1
2
2
α ∈ 4n π : k ∈ Z, n ∈ N , then (sin nα, cos nα) = ± 2 , ± 2 and from the
identity
kx cos nα + y sin nαk2 + ky cos nα − x sin nαk2 = kxk2 + kyk2 ,
we get the parallelogram identity. The question is: what is the biggest set A,
such that
R\πQ ⊂ A ⊂ R
and for which the theorem holds true.
A CHARACTERIZATION OF THE INNER PRODUCT SPACES
113
References
1. C. Alsina, J. Sikorska and M.S. Tomás, Norm derivatives and characterizations of inner
product spaces, World Scientific Publishing Co. Pte. Ltd., Hackensack, NJ, 2010.
2. D. Amir, Characterizations of inner product spaces, Operator Theory: Advances and Applications, Birkhäuser Verlag, Basel, 1986.
3. M.S. Moslehian and J.M. Rassias, A characterization of inner product spaces concerning an
Euler-Lagrange identity, Commun. Math. Anal. 8 (2010), no. 2, 16–21.
4. K. Nikodem and Z. Páles, Characterizations of inner product spaces by strongly convex functions, Banach J. Math. Anal. 5 (2011), no. 1, 83–87.
1
National College ”IANCU DE HUNEDOARA”, Hunedoara, Romania
E-mail address: marinescuds@gmail.com
2
National College ”DECEBAL”, Deva, Romania
E-mail address: mihaimonea@yahoo.com
3
National College ”AVRAM IANCU”, Brad, Romania
E-mail address: opincariumihai@yahoo.com
4
Economic College ”EMANOIL GOJDU”, Hunedoara, Romania
E-mail address: maricu stroe@yahoo.com