01st sets of simulated exam for Mathematics Competition Training of Gannan Normal University

 

1. Set $ F, G $ to a continuous function on $ [a, B] $.

(1) for $1 <p <q <\ infty $, $ \ cfrac {1} {p} + \ cfrac {1} {q} = 1,, b> 0 $, test certificate: $ \ Bex AB \ Leq \ cfrac {1} {p} a ^ P + \ cfrac {1} {q} B ^ Q. \ EEx $

(2) set $ \ DPS {\ VSM {n} a_n} $ as the positive series of convergence. test evidence: $ \ DPS {\ VSM {n} a_n ^ {1-\ frac {1} {n }}$ also converges.

(3) for $1 \ Leq p \ Leq \ infty $, define $ \ Bex \ Sen {f} _ p = \ sedd {\ BA {ll} \ DPS {\ sex {\ int_a ^ B | f (x) | ^ p \ RD x} ^ \ frac {1} {p}, & 1 \ Leq P <\ infty, \\\ DPS {\ Max _ {A \ Leq x \ Leq B} | f (x) |}, & P = \ infty. \ EA} \ EEx $ test certificate: If $1 \ Leq P, Q \ Leq \ infty $ \ DPS {\ frac {1} {p} + \ frac {1} {q} = 1} $, $ \ Sen {f} _ p <\ infty $, $ \ Sen {g} _ q <\ infty $, then $ \ Bex \ int_a ^ B | f (x) g (x) | \ RD x \ Leq \ Sen {f} _ p \ Sen {g }_{ Q }. \ EEx $

(4) for $1 \ Leq p \ Leq \ infty $, $ \ Sen {f} _ p <\ infty $, $ \ Sen {g} _ p <\ infty $, test certificate: $ \ Bex \ Sen {F + G} _ p \ Leq \ Sen {f} _ p + \ Sen {g} _ p. \ EEx $

(5) then set $ h $ to the value range of $ F $ (which is an interval) for second-order continuous export, and $ h'' \ Leq 0 $, then $ \ Bex H \ sex {\ frac {1} {B-a} \ int_a ^ B f (x) \ RD x} \ geq \ frac {1} {B-a} \ int_a ^ B H (f (x) \ RD X. \ EEx $

(6) set $ F $ to constant instead of zero. For $ p \ In \ BBR $, define $ \ Bex a_p (f) = \ sex {\ frac {1} {B-a} \ int_a ^ B | f (x) | ^ p \ RD x} ^ \ frac {1} {p }. \ EEx $ test certificate: $ \ beex \ Bea \ lim _ {P \ To-\ infty} a_p (f) & =\ min _ {A \ Leq x \ Leq B} | f (x) |, \ lim _ {P \ to 0} a_p (f) & =\ exp \ SEZ {\ frac {1} {B-a} \ int_a ^ B \ ln | f (x) | \ RD x }, \\\ LiM _ {P \ to + \ infty} a_p (f) &=\ Max _ {A \ Leq x \ Leq B} | f (x) |. \ EEA \ eeex $

(7) Test Certificate: $ \ Xi \ In (a, B) $ exists, so that $ \ DPS {\ int_a ^ B f (x) \ rd x = F (\ xi) (B-a)} $.

(8) If you set $ F $ to strictly increase the number of non-negative values, then (7) knows $ \ forall \ P> 0 $, $ \ Bex \ exists | \ x_p \ In (a, B), \ st f ^ P (x_p) = \ frac {1} {B-a} \ int_a ^ B f ^ p (x) \ RD X. \ EEx $ test certificate: $ \ vlm {p} x_p $ exists, and request it.

 

Proof: (1) $ \ beex \ Bea & \ quad AB \ Leq \ frac {1} {p} a ^ P + \ frac {1} {q} B ^ q = \ frac {1} {p} a ^ P + \ sex {1-\ frac {1} {p} B ^ \ frac {p} {P-1} \ & \ LRA \ frac {a} {B ^ \ frac {1} {P-1} \ Leq \ frac {1} {p} \ sex {\ frac {A} {B ^ \ frac {1} {P-1 }}^ p + 1-\ frac {1} {P }\\& \ LRA x \ Leq \ frac {1} {p} x ^ p + 1- \ frac {1} {p} \ & \ LRA X-1 \ Leq \ frac {1} {p} (x ^ P-1) \ & \ La \ frac {X-1} {\ frac {1} {p} (x ^ P-1 )} =\ frac {1} {\ Xi ^ {P-1} \ sedd {\ BA {ll}> 1, & 0 \ Leq x <1, \ x <\ xi <1, \ <1, & 1 <x <\ infty, \ 1 <\ xi <X. \ EA} \ EEA \ eeex $ can also be proved as follows: Set $ \ Bex f (x) =\ frac {1} {p} x ^ p-x + 1-\ frac {1} {p}, \ EEx $ then $ F (1) = 0 $, $ \ Bex f' (x) = x ^ {P-1}-1 \ sedd {\ BA {ll} <0, & 0 <x <1, \> 0, & amp; X & gt; 1. \ EA} \ EEx $ and $ f (x) \ geq F (1) = 0, \ x \ in [0, \ infty) $.

 

(2) by (1) zhi $ \ beex \ Bea \ VSM {n} a_n ^ {1-\ frac {1} {n }}& = 2 \ VSM {n} a_n ^ {1 -\ frac {1} {n }}\ cdot \ frac {1} {2} \ & \ Leq 2 \ VSM {n} \ sex {\ frac {n-1} {n} a_n + \ frac {1} {n} \ frac {1} {2 ^ n }}\& <\ infty. \ EEA \ eeex $

 

(3) set $ \ Sen {f} _ p \ NEQ 0 $, $ \ Sen {g} _ Q \ NEQ 0 $, $ p \ NEQ 1 $, $ Q \ NEQ \ infty $, and $ \ beex \ Bea \ int_a ^ B \ sev {\ frac {f (x )} {\ Sen {f }_p} \ cdot \ frac {g (x )} {\ Sen {g }_q }}\ RD x \ Leq \ int_a ^ B \ SEZ {\ frac {1} {p} \ sex {\ frac {f (x)} {\ Sen {f }_p }}^ P + \ frac {1} {q} \ sex {\ frac {g (x )} {\ Sen {g }_q }}^ Q} \ RD x = \ frac {1} {p} + \ frac {1} {q} = 1. \ EEA \ eeex $

(4) set $ p \ NEQ 1 $, $ p \ NEQ \ infty $, $ \ beex \ Bea \ Sen {F + G} _ P ^ P & = \ int_a ^ B | f (x) + g (x) | ^ p \ RD x \ & \ Leq \ int_a ^ B | f (x) + g (x) | \ cdot | f (x) + g (x) | ^ P-1} \ RD x \ & \ Leq \ int_a ^ B (| f (x) | + | G (x) |) \ cdot | f (x) + g (x) | ^ P-1} \ RD x \ & \ Leq \ int_a ^ B | f (x) | \ cdot | f (x) + g (x) | ^ {pm} \ RD x + \ int_a ^ B | G (x) | \ cdot | f (x) + g (x) | ^ {P-1} \ RD x \ & \ Leq \ sex {\ int_a ^ B | f (x) | ^ p \ RD x} ^ \ frac {1} {p} \ cdot \ sex {\ int_a ^ B | f (x) + g (x) | ^ {P-1) \ cdot \ frac {p} P-1} \ RD x} ^ \ frac {p1} {p} \ & \ quad + \ sex {\ int_a ^ B | G (x) | ^ p \ RD x} ^ \ frac {1} {p} \ cdot \ sex {\ int_a ^ B | f (x) + g (x) | ^ {P-1) \ cdot \ frac {p} P-1 }\ RD x} ^ \ frac {P-1 {P }\\\&=\ Sen {f }_p \ cdot \ Sen {F + G} _ P ^ {P-1} + \ Sen {g} _ p \ cdot \ Sen {F + G} _ P ^ {P-1 }. \ EEA \ eeex $

(5) set $ \ DPS {A = \ frac {1} {B-a} \ int_a ^ B f (x) \ RD x }$, by $ h'' \ Leq 0 $ $ \ beex \ Bea & \ quad H (y) \ Leq H (A) + H' () (Y-a) \ & \ rA H (f (x) \ Leq H (A) + H' (A) (f (x)-) \ & \ Ra \ frac {1} {B-a} \ int_a ^ B H (f (x) \ RD x \ Leq H (). \ EEA \ eeex $

(6) Verify $ \ DPS {\ lim _ {P \ to + \ infty} a_p (f) = \ Max _ {A \ Leq x \ Leq B} | f (x) |}$. obviously, $ \ DPS {\ limsup _ {P \ to + \ infty} a_p (f) \ Leq \ Max _ {A \ Leq x \ Leq B} | f (x) |}$, forward certificate $ \ Bex \ liminf _ {P \ to + \ infty} a_p (f) \ geq \ Max _ {A \ Leq x \ Leq B} | f (x) |, \ EEx $. set $ \ Bex \ Xi \ in [a, B], \ st | f (\ XI) | =\ Max _ {A \ Leq x \ Leq B} | f (x) |. \ EEx $ and the number-preserving feature of continuous functions (Note: $ f (x) \ NEQ 0, \ forall \ x $ ), $ \ Bex \ exists \ [a, B] \ supset [c, d] \ Ni \ Xi, \ st x \ in [C, d] \ rA | f (x) |> | f (x) |-\ ve. \ EEx $ then $ \ beex \ Bea a_p (f) & \ geq \ sex {\ frac {1} {B-a} \ int_c ^ d | f (x) | ^ p \ RD x} ^ \ frac {1} {p} \ & \ geq \ SEZ {\ frac {1} {B-a} \ int_c ^ d (| F (\ XI) |-\ ve) ^ p \ RD x} ^ \ frac {1} {p} \\\\=\ sex {\ frac {d-c} {B-a }}^ \ frac {1} {p} (| f (\ XI) |-\ ve), \\\ liminf _ {P \ to + \ infty} a_p (f) & \ geq | f (\ xi) |-\ ve. \ EEA \ eeex $ further evidence $ \ DPS {\ lim _ {P \ To-\ infty} a_p (f) =\min _ {A \ Leq x \ Leq B} | f (x) |}$: $ \ beex \ Bea \ lim _ {P \ To-\ infty} a_p (f) & =\ LiM _ {P \ To-\ infty} \ sed {\ frac {1} {B-a} \ SEZ {\ frac {1} {| f (x) |}}^ {-p} \ RD x} ^ {\ frac {1} {-p} \ cdot (-1 )} \ & =\ LiM _ {Q \ to + \ infty} \ sed {\ frac {1} {B-a} \ SEZ {\ frac {1} {| f (x) |}}^ Q \ RD x }^{\ frac {1} {q} \ cdot (-1 )} \\& =\ SEZ {\ lim _ {Q \ to + \ infty} a_q \ sex {\ frac {1} {| f (x) |}}^{-1 }\\\&=\ SEZ {\ Max _ {A \ Leq x \ Leq B }\ frac {1 }{| f (x) |}}^ {-1 }\\\& =\ min _ {A \ Leq x \ Leq B} | f (x) |. \ EEA \ eeex $ final proof $ \ DPS {\ lim _ {P \ to 0} a_p (f) =\ exp \ SEZ {\ frac {1} {B-a} \ int_a ^ B \ ln | f (x) | \ RD x }}$. on the one hand, $ \ beex \ Bea \ ln a_p (f) & =\ frac {1} {p} \ ln \ SEZ {\ frac {1} {B-a} \ int_a ^ B | f (x) | ^ p \ RD x }\\\\geq \ frac {1} {p} \ frac {1} {B-a} \ int_a ^ B \ ln | f (x) | ^ p \ RD x \ quad \ sex {(\ ln x) ''<0, \ mbox {By} (5 )} \\<=\ frac {1} {B-a} \ int_a ^ B \ ln | f (x) | \ RD X. \ EEA \ eeex $ on the other hand, $ \ beex \ Bea \ ln a_p (f) & =\ ln \ SEZ {\ frac {1} {B-a} \ int_a ^ B | f (x) | ^ p \ RD x} ^ \ frac {1} {p} \ & \ Leq \ cfrac {\ DPS {\ frac {1} {B-a} \ int_a ^ B | f (x) | ^ p \ RD X-1 }}{ P }\\\& \ quad \ sex {\ ln x \ Leq X-1 \ (x> 0) \ Ra \ ln x ^ p \ Leq x ^ P-1 \ ln x \ Leq \ frac {x ^ pm} {p} \ (P> 0 )} \\& =\ frac {1} {B-a} \ int_a ^ B \ frac {| f (x) | ^ P-1} {p} \ RD X, \\\ liminf _ {P \ to 0} a_p (f) & \ Leq \ frac {1} {B-a} \ int_a ^ B \ lim _ {P \ to 0} \ frac {| f (x) | ^ P-1} {p} \ RD x = \ frac {1} {B-a} \ int_a ^ B \ ln | f (x) | \ RD X. \ EEA \ eeex $

(7) set $ \ DPS {f (x) = \ int_a ^ x F (t) \ RD t} $, which is based on the mean value theorem of the Laplace, $ \ Bex \ exists \ Xi \ In (a, B), \ st F (\ xi) = f' (\ xi) = \ frac {F (B) -F (a)} {B-a }=\ frac {1} {B-a} \ int_a ^ B f (t) \ RD T. \ EEx $

(8) by (1) zhi $ \ beex \ Bea f ^ P (x_p) & =\ cfrac {1} {B-a} \ int_a ^ B f ^ p (t) \ cdot 1 \ RD t \ & \ Leq \ cfrac {1} {B-a} \ sex {\ int_a ^ B f ^ {P \ cdot \ frac {p + 1} {p} (t) \ RD t} ^ \ frac {p} {p + 1} \ sex {\ int_a ^ B 1 ^ {p + 1} \ RD t} ^ {\ frac {1} {p + 1 }}\\\&=\ cfrac {1} {B-a} \ sex {\ int_a ^ B f ^ {p + 1} (t) \ RD t} ^ {\ frac {p} {p + 1} (B-) ^ {\ frac {1} {p + 1 }}\\\&=\ sex {\ cfrac {1} {B-a} \ int_a ^ B f ^ {p + 1} (t) \ RD t} ^ \ frac {p} {p + 1} \\& = f ^ p (x _ {p + 1 }). \ EEA \ eeex $ $ F $ strictly incrementing. We have $ x_p \ Leq X _ {p + 1} $. in this case, the increment of $ x_p $ has an upper bound. from the monotonic bounded theorem, $ \ DPS {\ vlm {p} x_p = X _ \ infty} $ exists. in addition, by (6) $ \ beex \ Bea F (x_p) & =\ SEZ {\ cfrac {1} {B-a} \ int_a ^ B f ^ p (t) \ RD t} ^ {\ frac {1} {p}, \ f (x _ \ infty) & =\ vlm {p} \ SEZ {\ cfrac {1} {B-a} \ int_a ^ B f ^ p (t) \ RD t} ^ {\ frac {1} {P }}=\ Max _ {A \ Leq t \ Leq B} f (t) = F (B ), \ x _ \ infty & = B, \ EEA \ eeex $