TOR VERGATA

UNIVERSITÀ DEGLI STUDI DI ROMA “TOR VERGATA”

DIPARTIMENTO DI MATEMATICA

Perfetti Paolo

Dipartimento di Matematica

Università degli Studi di Roma “Tor Vergata” Via della Ricerca Scientifica 00133 Roma, Italy (http://www.mat.uniroma2.it/˜ perfetti)

Doug Hensley, Monthly Problems Department of Mathematics Texas A&M University College Station, TX 77840

Dear Professor, I would like to submit the included solution of problem 11861

Let n be a natural number and let f be a continuous function from [0, 1] to R such that ∫_{0}^{1} f^{2n+1}dx = 0. Prove that

\frac{(2n + 1)^{2n+1}}{(2n)^{2n}} (∫_{0}^{1} f(x)dx)^{2n} ≤ ∫_{0}^{1}(f(x))^{4n}dx

Proof As stated it is false. Just take λf with a λ ≠ 0 small enough. The r.h.s. goes to zero with λ according to λ^{4n} while the l.h.s. according to λ^{2n}. The correct statement is

\frac{(2n + 1)^{2n+1}}{(2n)^{2n}} (∫_{0}^{1} f(x)dx)^{4n} ≤ ∫_{0}^{1}(f(x))^{4n}dx

By Cauchy–Schwarz

(∫_{0}^{1}(c + f^{2n})fdx)^{2} ≤ ∫_{0}^{1}(c + f^{2n})^{2}dx ∫_{0}^{1} f^{2}dx

thus (use ∫_{0}^{1} f^{2n+1}dx = 0)

(∫_{0}^{1} cfdx)^{2} ≤ ∫_{0}^{1}(c + f^{2n})^{2}dx ∫_{0}^{1} f^{2}dx = (c^{2} + 2c ∫_{0}^{1} f^{2n}dx + ∫_{0}^{1} f^{4n}dx) ∫_{0}^{1} f^{2}dx

or

c^{2} ((∫_{0}^{1} fdx)^{2} − (∫_{0}^{1} f^{2}dx)) − 2c ∫_{0}^{1} f^{2}dx ∫_{0}^{1} f^{2n}dx ≤ ∫_{0}^{1} f^{4n}dx ∫_{0}^{1} f^{2}dx (1)

Via della Ricerca Scientifica, 00133 Roma, Italy – Fax 0672594699 –tel. 0672594722– e-mail: perfetti@mat.uniroma2.it

By Cauchy–Schwarz we have

A ≐ (∫_{0}^{1} fdx)^{2} − ∫_{0}^{1} f^{2}dx ≤ 0

A = 0 means that f ≡ a (constant) but ∫_{0}^{1} f^{2n+1}dx = 0 implies a = 0 and this in turn implies that the inequality we are going to prove trivially holds. So let’s assume A ≠ 0. The maximum of the l.h.s. of (1) occurs for

c = ∫_{0}^{1} f^{2}dx ∫_{0}^{1} f^{2n}dx ((∫_{0}^{1} fdx)^{2} − ∫_{0}^{1} f^{2}dx)^{−1}

and (1) becomes

(∫_{0}^{1} f^{4n}dx) (∫_{0}^{1} f^{2}dx) ≥ (∫_{0}^{1} f^{2}dx)^{2} (∫_{0}^{1} f^{2n}dx)^{2} (∫_{0}^{1} f^{2}dx − (∫_{0}^{1} fdx)^{2})^{−1}

or

∫_{0}^{1} f^{4n}dx ≥ ∫_{0}^{1} f^{2}dx (∫_{0}^{1} f^{2n}dx)^{2} (∫_{0}^{1} f^{2}dx − (∫_{0}^{1} fdx)^{2})^{−1}

By Hölder it implies

∫_{0}^{1} f^{4n}dx ≥ ∫_{0}^{1} f^{2}dx (∫_{0}^{1} f^{2}dx)^{2n} (∫_{0}^{1} f^{2}dx − (∫_{0}^{1} fdx)^{2})^{−1}

Letting ∫_{0}^{1} f^{2}dx ≐ p and (∫_{0}^{1} fdx)^{2} ≐ p_{0} we rewrite it as

∫_{0}^{1} f^{4n}dx ≥ \frac{p^{2n+1}}{p − p_{0}} , p > p_{0}

The function p^{2n+1}/(p − p_{0}), p > p_{0} > 0 is greater than \frac{(2n + 1)^{2n+1}}{(2n)^{2n}} p_{0}^{2n} hence

∫_{0}^{1} f^{4n}dx ≥ \frac{(2n + 1)^{2n+1}}{(2n)^{2n}} (∫_{0}^{1} f^{2}dx)^{2n}

Cauchy–Schwarz yields

∫_{0}^{1} f^{4n}dx ≥ \frac{(2n + 1)^{2n+1}}{(2n)^{2n}} (∫_{0}^{1} f^{2}dx)^{2n} ≥ \frac{(2n + 1)^{2n+1}}{(2n)^{2n}} (∫_{0}^{1} fdx)^{4n}

whence the result.

Roma(Italy) 01/30/2016 Best regards

Paolo Perfetti Perfetti Paolo, Dipartimento di Matematica, Università degli studi di Tor Vergata Roma, via della ricerca scientifica, 00133 Roma, Italy – email: perfetti@mat.uniroma2.it

If you will publish or simply acknowledge my solution, please, mention “Dipartimento di matematica” or “math. dept.” or anything you like with this information. In the Rome–area there are two “Paolo Perfetti” and although one (me) is a mathematician while the second is a physicist, often it is unclear which one is cited.