QUADRATIC REVERSES OF THE TRIANGLE INEQUALITY IN INNER PRODUCT SPACES

SEVER S. DRAGOMIR

Abstract. Some sharp quadratic reverses for the generalised triangle inequality in inner product spaces and applications are given.

1. Introduction

In 1966, J.B. Diaz and F.T. Metcalf [1] proved the following reverse of the triangle inequality in the general settings of inner product spaces:

Theorem 1. Let a be a unit vector in the inner product space (H; ⟨·, ·⟩) over the real or complex number field 𝕂. Suppose that the vectors x_{i} ∈ H\ {0} , i ∈ {1, …, n} satisfy

(1.1)  0 ≤ r ≤ \frac{Re ⟨x_{i}, a⟩}{∥x_{i}∥}, i ∈ \{1, …, n\} .

Then

(1.2)  r ∑_{i=1}^{n} ∥x_{i}∥ ≤ ∥∑_{i=1}^{n} x_{i}∥ ,

where equality holds if and only if

(1.3)  ∑_{i=1}^{n} x_{i} = r (∑_{i=1}^{n} ∥x_{i}∥) a.

For some similar results valid for semi-inner products in normed spaces, see [3] and [4].

In the same spirit, but providing a somewhat simpler sufficient condition with a clear geometrical meaning, we note the following result obtained by the author in [2]:

Theorem 2. Let a be as above and ρ ∈ (0, 1) . If x_{i} ∈ H, i ∈ {1, …, n} are such that

(1.4)  ∥x_{i} − a∥ ≤ ρ for each i ∈ {1, …, n} ,

then we have the inequality

(1.5)  \sqrt{1 − ρ^{2}} ∑_{i=1}^{n} ∥x_{i}∥ ≤ ∥∑_{i=1}^{n} x_{i}∥ ,

Date: April 16, 2004.

2000 Mathematics Subject Classification. Primary 46C05; Secondary 26D15.

Key words and phrases. Triangle inequality, Diaz-Metcalf Inequality, Inner products.

with equality if and only if

(1.6)  ∑_{i=1}^{n} x_{i} = \sqrt{1 − ρ^{2}} (∑_{i=1}^{n} ∥x_{i}∥) a.

In a complementary direction providing reverses of the triangle inequality in its additive form, i.e., upper bounds for the nonnegative difference

∑_{i=1}^{n} ∥x_{i}∥ − ∥∑_{i=1}^{n} x_{i}∥ ,

we note the following recent result obtained in [2]:

Theorem 3. Let a be as above and x_{i} ∈ H, k_{i} ≥ 0, i ∈ {1, …, n} such that

(1.7)  ∥x_{i}∥ − Re ⟨a, x_{i}⟩ ≤ k_{i} for each i ∈ {1, …, n} ,

then we have the inequality

(1.8)  0 ≤ ∑_{i=1}^{n} ∥x_{i}∥ − ∥∑_{i=1}^{n} x_{i}∥ ≤ ∑_{i=1}^{n} k_{i}.

The equality holds in (1.8) if and only if

(1.9)  ∑_{i=1}^{n} ∥x_{i}∥ ≥ ∑_{i=1}^{n} k_{i}

and

(1.10)  ∑_{i=1}^{n} x_{i} = (∑_{i=1}^{n} ∥x_{i}∥ − ∑_{i=1}^{n} k_{i}) a.

Another similar result but with a simpler condition, is the following one [2].

Theorem 4. Let a and x_{i}, i ∈ {1, …, n} be as above. If r_{i} > 0, i ∈ {1, …, n} are such that

(1.11)  ∥x_{i} − a∥ ≤ r_{i} for each i ∈ {1, …, n} ,

then we have the inequality

(1.12)  0 ≤ ∑_{i=1}^{n} ∥x_{i}∥ − ∥∑_{i=1}^{n} x_{i}∥ ≤ \frac{1}{2} ∑_{i=1}^{n} r_{i}^{2}.

The equality holds in (1.12) if and only if

(1.13)  ∑_{i=1}^{n} ∥x_{i}∥ ≥ \frac{1}{2} ∑_{i=1}^{n} r_{i}^{2}

and

(1.14)  ∑_{i=1}^{n} x_{i} = (∑_{i=1}^{n} ∥x_{i}∥ − \frac{1}{2} ∑_{i=1}^{n} r_{i}^{2}) a.

For other inequalities related to the triangle inequality, see Chapter XVII of the book [5].

The main aim of the present paper is to point out some quadratic reverses for the generalised triangle inequality, namely, sharp upper bounds for the nonnegative differences

(∑_{i=1}^{n} ∥x_{i}∥)^{2} − ∥∑_{i=1}^{n} x_{i}∥^{2} ,

under various assumptions for the vectors x_{i} ∈ H, i ∈ {1, …, n} involved. Some related results are established. Applications for vector-valued integrals in Hilbert spaces are also given.

2. Quadratic Reverses of the Triangle Inequality

The following lemma holds:

Lemma 1. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂, x_{i} ∈ H, i ∈ {1, …, n} and k_{ij} > 0 for 1 ≤ i < j ≤ n such that

(2.1)  0 ≤ ∥x_{i}∥ ∥x_{j}∥ − Re ⟨x_{i}, x_{j}⟩ ≤ k_{ij}

for 1 ≤ i < j ≤ n. Then we have the following quadratic reverse of the triangle inequality

(2.2)  (∑_{i=1}^{n} ∥x_{i}∥)^{2} ≤ ∥∑_{i=1}^{n} x_{i}∥^{2} + 2 ∑_{1≤i<j≤n} k_{ij}.

The case of equality holds in (2.2) if and only if it holds in (2.1) for each i, j with 1 ≤ i < j ≤ n.

Proof. We observe that the following identity holds:

(∑_{i=1}^{n} ∥x_{i}∥)^{2} − ∥∑_{i=1}^{n} x_{i}∥^{2} (2.3)

= ∑_{i,j=1}^{n} ∥x_{i}∥ ∥x_{j}∥ − ⟨∑_{i=1}^{n} x_{i}, ∑_{j=1}^{n} x_{j}⟩

= ∑_{i,j=1}^{n} ∥x_{i}∥ ∥x_{j}∥ − ∑_{i,j=1}^{n} Re ⟨x_{i}, x_{j}⟩

= ∑_{i,j=1}^{n} [∥x_{i}∥ ∥x_{j}∥ − Re ⟨x_{i}, x_{j}⟩]

= ∑_{1≤i<j≤n} [∥x_{i}∥ ∥x_{j}∥ − Re ⟨x_{i}, x_{j}⟩] + ∑_{1≤j<i≤n} [∥x_{i}∥ ∥x_{j}∥ − Re ⟨x_{i}, x_{j}⟩]

= 2 ∑_{1≤i<j≤n} [∥x_{i}∥ ∥x_{j}∥ − Re ⟨x_{i}, x_{j}⟩] .

Using the condition (2.1), we deduce that

∑_{1≤i<j≤n} [∥x_{i}∥ ∥x_{j}∥ − Re ⟨x_{i}, x_{j}⟩] ≤ ∑_{1≤i<j≤n} k_{ij},

and by (2.3), we deduce the desired inequality (2.2).

The case of equality is obvious by the identity (2.3) and we omit the details.

Remark 1. From (2.2) one may deduce the coarser inequality that might be useful in some applications:

0 ≤ ∑_{i=1}^{n} ∥x_{i}∥ − ∥∑_{i=1}^{n} x_{i}∥ ≤ \sqrt{2} ( ∑_{1≤i<j≤n} k_{ij})^{\frac{1}{2}}  (≤ \sqrt{2} ∑_{1≤i<j≤n} \sqrt{k_{ij}}) .

Remark 2. If the condition (2.1) is replaced with the following refinement of Schwarz’s inequality:

(2.4)  (0 ≤) δ_{ij} ≤ ∥x_{i}∥ ∥x_{j}∥ − Re ⟨x_{i}, x_{j}⟩ for 1 ≤ i < j ≤ n,

then the following refinement of the quadratic generalised triangle inequality is valid:

(2.5)  (∑_{i=1}^{n} ∥x_{i}∥)^{2} ≥ ∥∑_{i=1}^{n} x_{i}∥^{2} + 2 ∑_{1≤i<j≤n} δ_{ij} (≥ ∥∑_{i=1}^{n} x_{i}∥^{2}) .

The equality holds in the first part of (2.5) iff the case of equality holds in (2.4) for each 1 ≤ i < j ≤ n.

The following result holds.

Proposition 1. Let (H; ⟨·, ·⟩) be as above, x_{i} ∈ H, i ∈ {1, …, n} and r > 0 such that

(2.6)  ∥x_{i} − x_{j}∥ ≤ r

for 1 ≤ i < j ≤ n. Then

(2.7)  (∑_{i=1}^{n} ∥x_{i}∥)^{2} ≤ ∥∑_{i=1}^{n} x_{i}∥^{2} + \frac{n (n − 1)}{2}r^{2}.

The case of equality holds in (2.7) if and only if

(2.8)  ∥x_{i}∥ ∥x_{j}∥ − Re ⟨x_{i}, x_{j}⟩ = \frac{1}{2}r^{2}

for each i, j with 1 ≤ i < j ≤ n.

Proof. The inequality (2.6) is obviously equivalent to

∥x_{i}∥^{2} + ∥x_{j}∥^{2} ≤ 2 Re ⟨x_{i}, x_{j}⟩ + r^{2}

for 1 ≤ i < j ≤ n. Since

2 ∥x_{i}∥ ∥x_{j}∥ ≤ ∥x_{i}∥^{2} + ∥x_{j}∥^{2} , 1 ≤ i < j ≤ n;

hence

(2.9)  ∥x_{i}∥ ∥x_{j}∥ − Re ⟨x_{i}, x_{j}⟩ ≤ \frac{1}{2}r^{2}

for any i, j with 1 ≤ i < j ≤ n.

Applying Lemma 1 for k_{ij} := \frac{1}{2}r^{2} and taking into account that

∑_{1≤i<j≤n} k_{ij} = \frac{n (n − 1)}{4}r^{2},

we deduce the desired inequality (2.7). The case of equality is also obvious by the above lemma and we omit the details.

In the same spirit, and if some information about the forward difference Δx_{k} := x_{k+1} −x_{k} (1 ≤ k ≤ n − 1) are available, then the following simple quadratic reverse of the generalised triangle inequality may be stated.

Corollary 1. Let (H; ⟨·, ·⟩) be an inner product space and x_{i} ∈ H, i ∈ {1, …, n} . Then we have the inequality

(2.10)  (∑_{i=1}^{n} ∥x_{i}∥)^{2} ≤ ∥∑_{i=1}^{n} x_{i}∥^{2} + \frac{n (n − 1)}{2} ∑_{k=1}^{n−1} ∥Δx_{k}∥ .

The constant \frac{1}{2} is best possible in the sense that it cannot be replaced in general by a smaller quantity.

Proof. Let 1 ≤ i < j ≤ n. Then, obviously,

∥x_{j} − x_{i}∥ = ∥∑_{k=i}^{j−1} Δx_{k}∥ ≤ ∑_{k=i}^{j−1} ∥Δx_{k}∥ ≤ ∑_{k=1}^{n−1} ∥Δx_{k}∥ .

Applying Proposition 1 for r := ∑_{k=1}^{n−1} ∥Δx_{k}∥ , we deduce the desired result (2.10).

To prove the sharpness of the constant \frac{1}{2}, assume that the inequality (2.10) holds with a constant c > 0, i.e.,

(2.11)  (∑_{i=1}^{n} ∥x_{i}∥)^{2} ≤ ∥∑_{i=1}^{n} x_{i}∥^{2} + cn (n − 1) ∑_{k=1}^{n−1} ∥Δx_{k}∥

for n ≥ 2, x_{i} ∈ H, i ∈ {1, …, n} .

If we choose in (2.11), n = 2, x_{1} = −\frac{1}{2}e, x_{2} = \frac{1}{2}e, e ∈ H, ∥e∥ = 1, then we get 1 ≤ 2c, giving c ≥ \frac{1}{2}.

The following result providing a reverse of the quadratic generalised triangle inequality in terms of the sup-norm of the forward differences also holds.

Proposition 2. Let (H; ⟨·, ·⟩) be an inner product space and x_{i} ∈ H, i ∈ {1, …, n} . Then we have the inequality

(2.12)  (∑_{i=1}^{n} ∥x_{i}∥)^{2} ≤ ∥∑_{i=1}^{n} x_{i}∥^{2} + \frac{n^{2} (n^{2} − 1)}{12} max_{1≤k≤n−1} ∥Δx_{k}∥^{2} .

The constant \frac{1}{12} is best possible in (2.12).

Proof. As above, we have that

∥x_{j} − x_{i}∥ ≤ ∑_{k=i}^{j−1} ∥Δx_{k}∥ ≤ (j − i) max_{1≤k≤n−1} ∥Δx_{k}∥ ,

for 1 ≤ i < j ≤ n.

Squaring the inequality, we get

∥x_{j}∥^{2} + ∥x_{i}∥^{2} ≤ 2 Re ⟨x_{i}, x_{j}⟩ + (j − i)^{2} max_{1≤k≤n−1} ∥Δx_{k}∥^{2}

for any i, j with 1 ≤ i < j ≤ n, and since

2 ∥x_{i}∥ ∥x_{j}∥ ≤ ∥x_{j}∥^{2} + ∥x_{i}∥^{2} ,

hence

(2.13)  0 ≤ ∥x_{i}∥ ∥x_{j}∥ − Re ⟨x_{i}, x_{j}⟩ ≤ \frac{1}{2} (j − i)^{2} max_{1≤k≤n−1} ∥Δx_{k}∥^{2}

for any i, j with 1 ≤ i < j ≤ n.

Applying Lemma 1 for k_{ij} := \frac{1}{2} (j − i)^{2} max_{1≤k≤n−1} ∥Δx_{k}∥^{2} , we can state that

(∑_{i=1}^{n} ∥x_{i}∥)^{2} ≤ ∥∑_{i=1}^{n} x_{i}∥^{2} + ∑_{1≤i<j≤n} (j − i)^{2} max_{1≤k≤n−1} ∥Δx_{k}∥^{2} .

However,

∑_{1≤i<j≤n} (j − i)^{2} = \frac{1}{2} ∑_{i,j=1}^{n} (j − i)^{2} = n ∑_{k=1}^{n} k^{2} − (∑_{k=1}^{n} k)^{2}

= \frac{n^{2} (n^{2} − 1)}{12}

giving the desired inequality.

To prove the sharpness of the constant, assume that (2.12) holds with a constant D > 0, i.e.,

(2.14)  (∑_{i=1}^{n} ∥x_{i}∥)^{2} ≤ ∥∑_{i=1}^{n} x_{i}∥^{2} + Dn^{2} (n^{2} − 1) max_{1≤k≤n−1} ∥Δx_{k}∥^{2}

for n ≥ 2, x_{i} ∈ H, i ∈ {1, …, n} .

If in (2.14) we choose n = 2, x_{1} = −\frac{1}{2}e, x_{2} = \frac{1}{2}e, e ∈ H, ∥e∥ = 1, then we get 1 ≤ 12D giving D ≥ \frac{1}{12}.

The following result may be stated as well.

Proposition 3. Let (H; ⟨·, ·⟩) be an inner product space and x_{i} ∈ H, i ∈ {1, …, n} . Then we have the inequality:

(2.15)  (∑_{i=1}^{n} ∥x_{i}∥)^{2} ≤ ∥∑_{i=1}^{n} x_{i}∥^{2} + ∑_{1≤i<j≤n} (j − i)^{\frac{2}{q}} (∑_{k=1}^{n−1} ∥Δx_{k}∥^{p})^{\frac{2}{p}} ,

where p > 1, \frac{1}{p} + \frac{1}{q} = 1.

The constant E = 1 in front of the double sum cannot generally be replaced by a smaller constant.

Proof. Using Hölder’s inequality, we have

∥x_{j} − x_{i}∥ ≤ ∑_{k=i}^{j−1} ∥Δx_{k}∥ ≤ (j − i)^{\frac{1}{q}} (∑_{k=i}^{j−1} ∥Δx_{k}∥^{p})^{\frac{1}{p}}

≤ (j − i)^{\frac{1}{q}} (∑_{k=1}^{n−1} ∥Δx_{k}∥^{p})^{\frac{1}{p}} ,

for 1 ≤ i < j ≤ n.

Squaring the previous inequality, we get

∥x_{j}∥^{2} + ∥x_{i}∥^{2} ≤ 2 Re ⟨x_{i}, x_{j}⟩ + (j − i)^{\frac{2}{q}} (∑_{k=1}^{n−1} ∥Δx_{k}∥^{p})^{\frac{2}{p}} ,

for 1 ≤ i < j ≤ n.

Utilising the same argument from the proof of Proposition 2, we deduce the desired inequality (2.15).

Now assume that (2.15) holds with a constant E > 0, i.e.,

(∑_{i=1}^{n} ∥x_{i}∥)^{2} ≤ ∥∑_{i=1}^{n} x_{i}∥^{2} + E ∑_{1≤i<j≤n} (j − i)^{\frac{2}{q}} (∑_{k=1}^{n−1} ∥Δx_{k}∥^{p})^{\frac{2}{p}} ,

for n ≥ 2 and x_{i} ∈ H, i ∈ \{1, …, n\} , p > 1, \frac{1}{p} + \frac{1}{q} = 1.

For n = 2, x_{1} = −\frac{1}{2}e, x_{2} = \frac{1}{2}e, ∥e∥ = 1, we get 1 ≤ E, showing the fact that the inequality (2.15) is sharp.

The particular case p = q = 2 is of interest.

Corollary 2. Let (H; ⟨·, ·⟩) be an inner product space and x_{i} ∈ H, i ∈ {1, …, n} . Then we have the inequality:

(2.16)  (∑_{i=1}^{n} ∥x_{i}∥)^{2} ≤ ∥∑_{i=1}^{n} x_{i}∥^{2} + \frac{(n^{2} − 1) n}{6} ∑_{k=1}^{n−1} ∥Δx_{k}∥^{2} .

The constant \frac{1}{6} is best possible in (2.16).

Proof. For p = q = 2, Proposition 3 provides the inequality

(∑_{i=1}^{n} ∥x_{i}∥)^{2} ≤ ∥∑_{i=1}^{n} x_{i}∥^{2} + ∑_{1≤i<j≤n} (j − i) ∑_{k=1}^{n−1} ∥Δx_{k}∥^{2} ,

and since

∑_{1≤i<j≤n} (j − i) = 1 + (1 + 2) + (1 + 2 + 3) + ⋯ + (1 + 2 + ⋯ + n − 1)

= ∑_{k=1}^{n−1} (1 + 2 + ⋯ + k) = ∑_{k=1}^{n−1} \frac{k (k + 1)}{2}

= \frac{1}{2} [\frac{(n − 1) n (2n − 1)}{6} + \frac{n (n − 1)}{2}]

= \frac{n (n^{2} − 1)}{6},

hence the inequality (2.15) is proved. The best constant may be shown in the same way as above but we omit the details.

Finally, we may state and prove the following different result.

Theorem 5. Let (H; ⟨·, ·⟩) be an inner product space, y_{i} ∈ H, i ∈ {1, …, n} and M ≥ m > 0 are such that either

(2.17)  Re ⟨My_{j} − y_{i}, y_{i} − my_{j}⟩ ≥ 0 for 1 ≤ i < j ≤ n,

or, equivalently,

(2.18)  ∥y_{i} − \frac{M + m}{2}y_{j}∥ ≤ \frac{1}{2} (M − m) ∥y_{j}∥ for 1 ≤ i < j ≤ n.

Then we have the inequality

(2.19)  (∑_{i=1}^{n} ∥y_{i}∥)^{2} ≤ ∥∑_{i=1}^{n} y_{i}∥^{2} + \frac{1}{2} · \frac{(M − m)^{2}}{M + m} ∑_{k=1}^{n−1} k ∥y_{k+1}∥^{2} .

The case of equality holds in (2.19) if and only if

(2.20)  ∥y_{i}∥ ∥y_{j}∥ − Re ⟨y_{i}, y_{j}⟩ = \frac{1}{4} · \frac{(M − m)^{2}}{M + m} ∥y_{j}∥^{2}

for each i, j with 1 ≤ i < j ≤ n.

Proof. Firstly, observe that, in an inner product space (H; ⟨·, ·⟩) and for x, z, Z ∈ H, the following statements are equivalent:

(i) Re ⟨Z − x, x − z⟩ ≥ 0 (ii) ∥x − \frac{Z+z}{2} ∥ ≤ \frac{1}{2} ∥Z − z∥ .

This shows that (2.17) and (2.18) are obviously equivalent.

Now, taking the square in (2.18), we get

∥y_{i}∥^{2} + \frac{(M − m)^{2}}{M + m} ∥y_{j}∥^{2} ≤ 2 Re ⟨y_{i}, \frac{M + m}{2}y_{j}⟩ + \frac{1}{n} (M − m)^{2} ∥y_{j}∥^{2}

for 1 ≤ i < j ≤ n, and since, obviously,

2 (\frac{M + m}{2}) ∥y_{i}∥ ∥y_{j}∥ ≤ ∥y_{i}∥^{2} + \frac{(M − m)^{2}}{M + m} ∥y_{j}∥^{2} ,

hence

2 (\frac{M + m}{2}) ∥y_{i}∥ ∥y_{j}∥ ≤ 2 Re ⟨y_{i}, \frac{M + m}{2}y_{j}⟩ + \frac{1}{n} (M − m)^{2} ∥y_{j}∥^{2} ,

giving the much simpler inequality

(2.21)  ∥y_{i}∥ ∥y_{j}∥ − Re ⟨y_{i}, y_{j}⟩ ≤ \frac{1}{4} · \frac{(M − m)^{2}}{M + m} ∥y_{j}∥^{2} ,

for 1 ≤ i < j ≤ n.

Applying Lemma 1 for k_{ij} := \frac{1}{4} · \frac{(M−m)^{2}}{M+m} ∥y_{j}∥^{2} , we deduce

(2.22)  (∑_{i=1}^{n} ∥y_{i}∥)^{2} ≤ ∥∑_{i=1}^{n} y_{i}∥^{2} + \frac{1}{2} \frac{(M − m)^{2}}{M + m} ∑_{1≤i<j≤n} ∥y_{j}∥^{2}

with equality if and only if (2.21) holds for each i, j with 1 ≤ i < j ≤ n.

Since

∑_{1≤i<j≤n} ∥y_{j}∥^{2} = ∑_{1<j≤n} ∥y_{j}∥^{2} + ∑_{2<j≤n} ∥y_{j}∥^{2} + ⋯ + ∑_{n−1<j≤n} ∥y_{j}∥^{2}

= ∑_{j=2}^{n} ∥y_{j}∥^{2} + ∑_{j=3}^{n} ∥y_{j}∥^{2} + ⋯ + ∑_{j=n−1}^{n} ∥y_{j}∥^{2} + ∥y_{n}∥^{2}

= ∑_{j=2}^{n} (j − 1) ∥y_{j}∥^{2} = ∑_{k=1}^{n−1} k ∥y_{k+1}∥^{2} ,

hence the inequality (2.19) is obtained.

3. Further Quadratic Refinements of the Triangle Inequality

The following lemma is of interest in itself as well.

Lemma 2. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂, x_{i} ∈ H, i ∈ {1, …, n} and k ≥ 1 with the property that:

(3.1)  ∥x_{i}∥ ∥x_{j}∥ ≤ k Re ⟨x_{i}, x_{j}⟩ ,

for each i, j with 1 ≤ i < j ≤ n. Then

(3.2)  (∑_{i=1}^{n} ∥x_{i}∥)^{2} + (k − 1) ∑_{i=1}^{n} ∥x_{i}∥^{2} ≤ k ∥∑_{i=1}^{n} x_{i}∥^{2} .

The equality holds in (3.2) if and only if it holds in (3.1) for each i, j with 1 ≤ i < j ≤ n.

Proof. Firstly, let us observe that the following identity holds true:

(∑_{i=1}^{n} ∥x_{i}∥)^{2} − k ∥∑_{i=1}^{n} x_{i}∥^{2} (3.3)

= ∑_{i,j=1}^{n} ∥x_{i}∥ ∥x_{j}∥ − k ⟨∑_{i=1}^{n} x_{i}, ∑_{j=1}^{n} x_{j}⟩

= ∑_{i,j=1}^{n} [∥x_{i}∥ ∥x_{j}∥ − k Re ⟨x_{i}, x_{j}⟩]

= 2 ∑_{1≤i<j≤n} [∥x_{i}∥ ∥x_{j}∥ − k Re ⟨x_{i}, x_{j}⟩] + (1 − k) ∑_{i=1}^{n} ∥x_{i}∥^{2} ,

since, obviously, Re ⟨x_{i}, x_{j}⟩ = Re ⟨x_{j}, x_{i}⟩ for any i, j ∈ {1, …, n} .

Using the assumption (3.1), we obtain

∑_{1≤i<j≤n} [∥x_{i}∥ ∥x_{j}∥ − k Re ⟨x_{i}, x_{j}⟩] ≤ 0

and thus, from (3.3), we deduce the desired inequality (3.2).

The case of equality is obvious by the identity (3.3) and we omit the details.

Remark 3. The inequality (3.2) provides the following reverse of the quadratic generalised triangle inequality:

(3.4)  0 ≤ (∑_{i=1}^{n} ∥x_{i}∥)^{2} − ∑_{i=1}^{n} ∥x_{i}∥^{2} ≤ k [∥∑_{i=1}^{n} x_{i}∥^{2} − ∑_{i=1}^{n} ∥x_{i}∥^{2}] .

Remark 4. Since k = 1 and ∑_{i=1}^{n} ∥x_{i}∥^{2} ≥ 0, hence by (3.2) one may deduce the following reverse of the triangle inequality

(3.5)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ \sqrt{k} ∥∑_{i=1}^{n} x_{i}∥ ,

provided (3.1) holds true for 1 ≤ i < j ≤ n.

The following corollary providing a better bound for ∑_{i=1}^{n} ∥x_{i}∥ , holds.

Corollary 3. With the assumptions in Lemma 2, one has the inequality:

(3.6)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ \sqrt{\frac{nk}{n + k − 1}} ∥∑_{i=1}^{n} x_{i}∥ .

Proof. Using the Cauchy-Bunyakovsky-Schwarz inequality

n ∑_{i=1}^{n} ∥x_{i}∥^{2} ≥ (∑_{i=1}^{n} ∥x_{i}∥)^{2}

we get

(3.7)  (k − 1) ∑_{i=1}^{n} ∥x_{i}∥^{2} + (∑_{i=1}^{n} ∥x_{i}∥)^{2} ≥ (\frac{k − 1}{n} + 1) (∑_{i=1}^{n} ∥x_{i}∥)^{2} .

Consequently, by (3.7) and (3.2) we deduce

k ∥∑_{i=1}^{n} x_{i}∥^{2} ≥ \frac{n + k − 1}{n} (∑_{i=1}^{n} ∥x_{i}∥)^{2}

giving the desired inequality (3.6).

The following result may be stated as well.

Theorem 6. Let (H; ⟨·, ·⟩) be an inner product space and x_{i} ∈ H\ {0} , i ∈ {1, …, n} , ρ ∈ (0, 1) , such that

(3.8)  ∥x_{i} − \frac{x_{j}}{∥x_{j}∥}∥ ≤ ρ for 1 ≤ i < j ≤ n.

Then we have the inequality

(3.9)  \sqrt{1 − ρ^{2}} (∑_{i=1}^{n} ∥x_{i}∥)^{2} + (1 − \sqrt{1 − ρ^{2}}) ∑_{i=1}^{n} ∥x_{i}∥^{2} ≤ ∥∑_{i=1}^{n} x_{i}∥^{2} .

The case of equality holds in (3.9) iff

(3.10)  ∥x_{i}∥ ∥x_{j}∥ = \frac{1}{\sqrt{1 − ρ^{2}}} Re ⟨x_{i}, x_{j}⟩

for any 1 ≤ i < j ≤ n.

Proof. The condition (3.1) is obviously equivalent to

∥x_{i}∥^{2} + 1 − ρ^{2} ≤ 2 Re ⟨x_{i}, \frac{x_{j}}{∥x_{j}∥}⟩

for each 1 ≤ i < j ≤ n.

Dividing by \sqrt{1 − ρ^{2}} > 0, we deduce

(3.11)  \begin{matrix}∥x_{i}∥ & 2 \\ √ & \overline{1 − ρ^{2}} \end{matrix} + \sqrt{1 − ρ^{2}} ≤ \frac{2}{\sqrt{1 − ρ^{2}}} Re ⟨x_{i}, \frac{x_{j}}{∥x_{j}∥}⟩ ,

for 1 ≤ i < j ≤ n.

On the other hand, by the elementary inequality

(3.12)  \frac{p}{α} + qα ≥ 2\sqrt{pq}, p, q ≥ 0, α > 0

we have

(3.13)  2 ∥x_{i}∥ ≤ \begin{matrix}∥x_{i}∥ & 2 \\ √ & \overline{1 − ρ^{2}} \end{matrix} + \sqrt{1 − ρ^{2}}.

Making use of (3.11) and (3.13), we deduce that

∥x_{i}∥ ∥x_{j}∥ ≤ \frac{1}{\sqrt{1 − ρ^{2}}} Re ⟨x_{i}, x_{j}⟩

for 1 ≤ i < j ≤ n.

Now, applying Lemma 1 for k = \frac{1}{\sqrt{1−ρ^{2}}} , we deduce the desired result.

Remark 5. If we assume that ∥x_{i}∥ = 1, i ∈ {1, …, n} , satisfying the simpler condition

(3.14)  ∥x_{j} − x_{i}∥ ≤ ρ for 1 ≤ i < j ≤ n,

then, from (3.9), we deduce the following lower bound for ∥∑_{i=1}^{n} x_{i}∥ , namely

(3.15)  [n + n (n − 1) \sqrt{1 − ρ^{2}}]^{\frac{1}{2}} ≤ ∥∑_{i=1}^{n} x_{i}∥ .

The equality holds in (3.15) iff \sqrt{1 − ρ^{2}} = Re ⟨x_{i}, x_{j}⟩ for 1 ≤ i < j ≤ n.

Remark 6. Under the hypothesis of Proposition 3, we have the coarser but simpler reverse of the triangle inequality

(3.16)  \sqrt[4]{1 − ρ^{2}} ∑_{i=1}^{n} ∥x_{i}∥ ≤ ∥∑_{i=1}^{n} x_{i}∥ .

Also, applying Corollary 3 for k = \frac{1}{\sqrt{1−ρ^{2}}} , we can state that

(3.17)  ∑_{i=1}^{n} ∥x_{i}∥ ≤ \sqrt{\frac{n}{n\sqrt{1 − ρ^{2}} + 1 − \sqrt{1 − ρ^{2}}}} ∥∑_{i=1}^{n} x_{i}∥ ,

provided x_{i} ∈ H satisfy (3.8) for 1 ≤ i < j ≤ n.

In the same manner, we can state and prove the following reverse of the quadratic generalised triangle inequality.

Theorem 7. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂, x_{i} ∈ H, i ∈ {1, …, n} and M ≥ m > 0 such that either

(3.18)  Re ⟨Mx_{j} − x_{i}, x_{i} − mx_{j}⟩ ≥ 0 for 1 ≤ i < j ≤ n,

or, equivalently,

(3.19)  ∥x_{i} − \frac{M + m}{2}x_{j}∥ ≤ \frac{1}{2} (M − m) ∥x_{j}∥ for 1 ≤ i < j ≤ n

hold. Then

(3.20)  \frac{2\sqrt{mM}}{M + m} (∑_{i=1}^{n} ∥x_{i}∥)^{2} + \frac{(\sqrt{M} − \sqrt{m})^{2}}{M + m} ∑_{i=1}^{n} ∥x_{i}∥^{2} ≤ ∥∑_{i=1}^{n} x_{i}∥^{2} .

The case of equality holds in (3.20) if and only if

(3.21)  ∥x_{i}∥ ∥x_{j}∥ = \frac{M + m}{2\sqrt{mM}} Re ⟨x_{i}, x_{j}⟩ for 1 ≤ i < j ≤ n.

Proof. From (3.18), observe that

(3.22)  ∥x_{i}∥^{2} + Mm ∥x_{j}∥^{2} ≤ (M + m) Re ⟨x_{i}, x_{j}⟩ ,

for 1 ≤ i < j ≤ n. Dividing (3.22) by \sqrt{mM} > 0, we deduce

\begin{matrix}∥x_{i}∥ & 2 \\ √ & \overline{mM} \end{matrix} + \sqrt{mM} ∥x_{j}∥^{2} ≤ \frac{M + m}{\sqrt{mM}} Re ⟨x_{i}, x_{j}⟩ ,

and since, obviously

2 ∥x_{i}∥ ∥x_{j}∥ ≤ \begin{matrix}∥x_{i}∥ & 2 \\ √ & \overline{mM} \end{matrix} + \sqrt{mM} ∥x_{j}∥^{2}

hence

∥x_{i}∥ ∥x_{j}∥ ≤ \frac{M + m}{2\sqrt{mM}} Re ⟨x_{i}, x_{j}⟩ , for 1 ≤ i < j ≤ n.

Applying Lemma 2 for k = \frac{M+m}{2\sqrt{mM}} ≥ 1, we deduce the desired result.

Remark 7. We also must note that a simpler but coarser inequality that can be obtained from (3.20) is

(\frac{2\sqrt{mM}}{M + m} )^{\frac{1}{2}} ∑_{i=1}^{n} ∥x_{i}∥ ≤ ∥∑_{i=1}^{n} x_{i}∥ ,

provided (3.18) holds true.

Finally, a different result related to the generalised triangle inequality is incorporated in the following theorem.

Theorem 8. Let (H; ⟨·, ·⟩) be an inner product space over 𝕂, η > 0 and x_{i} ∈ H, i ∈ {1, …, n} with the property that

(3.23)  ∥x_{j} − x_{i}∥ ≤ η < ∥x_{j}∥ for each i, j ∈ {1, …, n} .

Then we have the following reverse of the triangle inequality

(3.24)  ∑_{i=1}^{n} \sqrt{∥x_{i}∥^{2} − η^{2}} ≤ \begin{matrix}∥∑_{i=1}^{n} x_{i}∥ & 2 \\ ∑ & {n \atop i=1} & ∥x_{i}∥ \end{matrix} .

The equality holds in (3.24) iff

(3.25)  ∥x_{i}∥ \sqrt{∥x_{j}∥^{2} − η^{2}} = Re ⟨x_{i}, x_{j}⟩ for each i, j ∈ \{1, …, n\} .

Proof. From (3.23), we have

∥x_{i}∥^{2} − 2 Re ⟨x_{i}, x_{j}⟩ + ∥x_{j}∥^{2} ≤ η^{2},

giving

∥x_{i}∥^{2} + ∥x_{j}∥^{2} − η^{2} ≤ 2 Re ⟨x_{i}, x_{j}⟩ , i, j ∈ {1, …, n} .

On the other hand,

2 ∥x_{i}∥ \sqrt{∥x_{j}∥^{2} − η^{2}} ≤ ∥x_{i}∥^{2} + ∥x_{j}∥^{2} − η^{2}, i, j ∈ \{1, …, n\}

and thus

∥x_{i}∥ \sqrt{∥x_{j}∥^{2} − η^{2}} ≤ Re ⟨x_{i}, x_{j}⟩ , i, j ∈ \{1, …, n\} .

Summing over i, j ∈ {1, …, n} , we deduce the desired inequality (3.24). The case of equality is also obvious from the above, and we omit the details.

4. Applications for Vector-Valued Integral Inequalities

Let (H; ⟨·, ·⟩) be a Hilbert space over the real or complex number field, [a, b] a compact interval in ℝ and η : [a, b] → [0, ∞) a Lebesgue integrable function on [a, b] with the property that ∫_{a}^{b} η (t) dt = 1. If, by L_{η} ([a, b] ; H) we denote the Hilbert space of all Bochner measurable functions f : [a, b] → H with the property that ∫_{a}^{b} η (t) ∥f (t)∥^{2} dt < ∞, then the norm ∥·∥_{η} of this space is generated by the inner product ⟨·, ·⟩_{η} : H × H → 𝕂 defined by

⟨f, g⟩_{η} := ∫_{a}^{b} η (t) ⟨f (t) , g (t)⟩ dt.

The following proposition providing a reverse of the integral generalised triangle inequality may be stated.

Proposition 4. Let (H; ⟨·, ·⟩) be a Hilbert space and η : [a, b] → [0, ∞) as above. If g ∈ L_{η} ([a, b] ; H) is so that ∫_{a}^{b} η (t) ∥g (t)∥^{2} dt = 1 and f_{i} ∈ L_{η} ([a, b] ; H) , i ∈ {1, …, n} , and M ≥ m > 0 are so that either

(4.1)  Re ⟨Mf_{j} (t) − f_{i} (t) , f_{i} (t) − mf_{j} (t)⟩ ≥ 0

or, equivalently,

∥f_{i} (t) − \frac{m + M}{2} f_{j} (t)∥ ≤ \frac{1}{2} (M − m) ∥f_{j} (t)∥

for a.e. t ∈ [a, b] and 1 ≤ i < j ≤ n, then we have the inequality

[∑_{i=1}^{n} (∫_{a}^{b} η (t) ∥f_{i} (t)∥^{2} dt)^{1/2}]^{2} (4.2)

≤ ∫_{a}^{b} η (t) ∥∑_{i=1}^{n} f_{i} (t)∥^{2} dt

+\frac{1}{2} · \frac{(M − m)^{2}}{m + M} ∫_{a}^{b} η (t) (∑_{k=1}^{n−1} k ∥f_{k+1} (t)∥^{2}) dt.

The case of equality holds in (4.2) if and only if

(∫_{a}^{b} η (t) ∥f_{i} (t)∥^{2} dt)^{1/2} (∫_{a}^{b} η (t) ∥f_{j} (t)∥^{2} dt)^{1/2}

− ∫_{a}^{b} η (t) Re ⟨f_{i} (t) , f_{j} (t)⟩ dt

= \frac{1}{4} · \frac{(M − m)^{2}}{m + M} ∫_{a}^{b} η (t) ∥f_{j} (t)∥^{2} dt

for each i, j with 1 ≤ i < j ≤ n.

Proof. We observe that

Re ⟨Mf_{j} − f_{i}, f_{i} − mf_{j}⟩_{η}
= ∫_{a}^{b} η (t) Re ⟨Mf_{j} (t) − f_{i} (t) , f_{i} (t) − mf_{j} (t)⟩ dt ≥ 0

for any i, j with 1 ≤ i < j ≤ n.

Applying Theorem 5 for the Hilbert space L_{η} ([a, b] ; H) and for y_{i} = f_{i}, i ∈ {1, …, n} , we deduce the desired result.

Another integral inequality incorporated in the following proposition holds:

Proposition 5. With the assumptions of Proposition 4, we have

\frac{2\sqrt{mM}}{m + M} [∑_{i=1}^{n} (∫_{a}^{b} η (t) ∥f_{i} (t)∥^{2} dt)^{1/2}]^{2} (4.3)

+\frac{(\sqrt{M} − \sqrt{m})^{2}}{m + M} ∑_{i=1}^{n} ∫_{a}^{b} η (t) ∥f_{i} (t)∥^{2} dt

≤ ∫_{a}^{b} η (t) ∥∑_{i=1}^{n} f_{i} (t)∥^{2} dt.

The case of equality holds in (4.3) if and only if

(∫_{a}^{b} η (t) ∥f_{i} (t)∥^{2} dt)^{1/2} (∫_{a}^{b} η (t) ∥f_{j} (t)∥^{2} dt)^{1/2}

= \frac{M + m}{2\sqrt{mM}} ∫_{a}^{b} η (t) Re ⟨f_{i} (t) , f_{j} (t)⟩ dt

for any i, j with 1 ≤ i < j ≤ n.

The proof is obvious by Theorem 7 and we omit the details.

References

[1] J.B. DIAZ and F.T. METCALF, A complementary triangle inequality in Hilbert and Banach spaces, Proceedings Amer. Math. Soc., 17(1) (1966), 88-97.

[2] S.S. DRAGOMIR, Reverses of the triangle inequality in inner product spaces, RGMIA Res. Rep. Coll., (to appear).

[3] S.M. KHALEELULA, On Diaz-Metcalf’s complementary triangle inequality, Kyungpook Math. J., 15 (1975), 9-11.

[4] P.M. MILIČIĆ, On a complementary inequality of the triangle inequality (French), Mat. Vesnik, 41(1989), No. 2, 83-88.

[5] D.S. MITRINOVIĆ, J.E. PEČARIĆ and A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.

School of Computer Science and Mathematics, Victoria University of Technology, PO Box 14428, MCMC 8001, Victoria, Australia.

   E-mail address: sever@csm.vu.edu.au
   URL: http://rgmia.vu.edu.au/SSDragomirWeb.html