1, otherwise the series is trivially convergent

Problem 12186

(American Mathematical Monthly, Vol.127, May 2020) Proposed by A. Eydelzon (USA).

For v = hx₁, . . . , x_ni in Rⁿ, let kvkp = (Pn

i=1|xi|^p)^1/p and kvk∞ = max_1≤i≤n|xi|; these are the usual p-norm and∞-norm on Rⁿ. For what v does the series

∞

X

p=1

(kvkp− kvk∞)

converge?

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution. We may assume that v is not the zero vector and n > 1, otherwise the series is trivially convergent. Then, we show that the series is convergent if and only if there is exactly one component of maximal absolute value.

(i) If the above condition is satisfied then, without loss of generality, let x₁ be the component of maximal absolute value and let t = _|x¹

1|max_2≤i≤n|xi| ∈ [0, 1). Hence, as p → ∞, 0 ≤ kvkp− kvk∞≤ kvk∞

(1 + (n − 1)t^p)^1/p− 1

= kvk∞



exp ln(1 + (n − 1)t^p) p



− 1



∼ kvk∞(n − 1) ·t^p p and the given series is convergent becauseP∞

p=1 t^p

p <∞.

(ii) If the above condition is not satisfied, then there are at least 2 components of maximal absolute value and therefore

kvkp− kvk∞≥ kvk∞

2^1/p− 1

= kvk∞



exp ln(2) p



− 1



∼ kvk∞ln(2) ·1 p and the given series is not convergent becauseP∞

p=1 1

p = ∞.