PLANE TRIGONOMETRY AND TABLES
ANXN |AN|2 IT IS PLAIN THAT ANY COMPLETE ORTHONORMAL SYSTEM IN A SEPARABLE HILBERT SPACE IS A RIESZ BASIS EXAMPLES OF SUCH SYSTEMS ARE THE SEQUENCE OF TRIGONOMETRIC FUNCTIONS (EIN?X)N?Q IN L2(?1,1) AND THE SEQUENCE OF STANDARD UNIT VECTORS IN L2 OUTSIDE THE WORLD OF HILBERT SPACES WITH ITS STRONG GEOMETRICAL ?AVOUR PROVIDED BY THE NOTION OF ORTHOGONALITY, MORE EFFORT IS OFTEN NEEDED TO PRODUCE EXAMPLES OF BASES FOR EXAMPLE, WHEN 1 < P < ?A BASIS OF LP(?1,1) IS GIVEN BY (EIN?X)N?Z THIS FOLLOWS FROM A RESULT DUE TO M RIESZ, NAMELY THAT (SEE [56], CHAP 12, SECT 10, P 106) LIM N?? F ?? |N|?N FNEIN?X P = 0, (1 3) FOR ALL F ?LP(?1,1), WHERE FN = X,X? ?(N) X?(N) CONVERGES WHENEVER ? IS A PERMUTATION OF THE NATURAL NUMBERS CLEARLY EVERY RIESZ BASIS IS AN UNCONDITIONAL BASIS IT IS OBVIOUS THAT THE STANDARD UNIT VECTORS IN LP (1 ?P < ?) FORM AN UNCONDI- TIONAL BASIS HOWEVER, THE TRIGONOMETRIC SYSTEM (EIN?X)N?Z IS AN UNCONDITIONAL BASIS OF LP(?1,1) ONLY WHEN P = 2, ALTHOUGH LP(?1,1) DOES HAVE AN UNCONDITIONAL BASIS IF 1 < P < ?AND FAILS TO HAVE ONE IF P = 1 FOR THESE RESULTS WE REFER TO [122], IID NOTE ALSO THAT IF (XN)N?N IS A BASIS OF A RE?EXIVE SPACE X, THEN THE CORRESPONDING BIORTHOGONAL FUNCTIONALS X? N FORM A BASIS OF X?(SEE [33], CHAP IV, SECT 3, LEMMA AND THEOREM 3); IF IN ADDITION ?XN?= 1 FOR ALL N ?N, THEN FROM (1 4) WE SEE THAT (?X? N?)N?N ?L? 1 2 STRICT AND UNIFORM CONVEXITY STRICT AND UNIFORM CONVEXITY LET X BE A BANACH SPACE WITH DUAL X?; THE VALUE OF X??X?AT X ?X IS DENOTED BY ?X,X??X OR ?X,X?? WE RECALL THAT X IS SAID TO BE STRICTLY CONVEX IF WHENEVER X,Y ?X ARE SUCH THAT X ?= Y AND ?X?= ?Y?= 1, AND ? ?(0,1), THEN ??X+(1 ??)Y?< 1 THIS SIMPLY MEANS THAT THE UNIT SPHERE IN X DOES NOT CONTAIN ANY LINE SEGMENT NOTE THAT AN EQUIVALENT CONDITION ON X IS THAT NO SPHERE OF ANY RADIUS AND ANY CENTRE CONTAINS A LINE SEGMENT FOR EXAMPLE, IF R2 IS EQUIPPED WITH THE LP NORM IT IS STRICTLY CONVEX IF 1 < P < ?BUT NOT IF P = 1 OR ? PROPOSITION 1 1 THE SPACE X IS STRICTLY CONVEX IF AND ONLY IF IT IS THE CASE THAT WHENEVER X,Y ?X ARE SUCH THAT ?X+Y?= ?X?+?Y?THEN EITHER Y = 0 OR X = ?Y FOR SOME ? ?0 PROOF SUPPOSE THAT X IS STRICTLY CONVEX AND X,Y ?X\{0} ARE SUCH THAT X ?= Y AND ?X+Y?= ?X?+ ?Y? THEN ?X??= ?Y?, FOR OTHERWISE X+Y = ?X?, CONTRADICTING THE STRICT CONVEXITY OF X HOWEVER, IF ?Y?< ?X?, WE PUT ? = ?Y?/?X?AND OBSERVE THAT 1 ? X ?X?+? Y ?Y??X ?X? ??X+Y??? ?X? ?X? = ?X?+?Y??? ?X? ?X? = 1, WHICH MEANS THAT X = Y/? THE CONVERSE IS OBVIOUS ?? IT IS PLAIN THAT EVERY LINEAR SUBSPACE OF A STRICTLY CONVEX SPACE IS ITSELF STRICTLY CONVEX, WITH THE INHERITED NORM HERE ARE SOME OF THE LESS TRIVIAL PROPERTIES OF SUCH SPACES PROPOSITION 1 2 LET X BE A BANACH SPACE WITH DUAL X? THEN X IS STRICTLY CONVEX IF AND ONLY IF GIVEN ANY X??X?\{0}, THERE IS AT MOST ONE X ?X SUCH THAT ?X?= AND ?X,X??= ?X??; IF X IS RE?EXIVE, SUCH AN X EXISTS PROOF LET X BE STRICTLY CONVEX AND X??X?\{0} SUPPOSE THERE ARE TWO DISTINCT SUCH POINTS X, SAY X1 AND X2 THEN IF 0 < ? < 1, ?X??= ? ?X1,X??+(1 ??)?X2,X??= ??X1 +(1 ??)X2,X?? ??X????X1 +(1 ??)X2?< ?X??, WHICH IS ABSURD CONVERSELY, SUPPOSE THAT ?X+ ?(Y?X)?= 1 FOR SOME X,Y ?X WITH ?X?= ?Y?= 1 AND SOME ? ?(0,1) BY THE HAHN–BANACH THEOREM, THERE EXISTS X??X?SUCH THAT ?X+ ?(Y?X),X??= 1 AND ?X??= 1 THEN (1 ??)?X,X??+ ? ?Y,X??= 1, AND SINCE |?X,X??|, |?Y,X??| ?1 WE MUST HAVE ?X,X??= ?Y,X??= 1 BY HYPOTHESIS, THIS IMPLIES THAT X = Y, AND SO X IS STRICTLY CONVEX NOW SUPPOSE THAT X IS RE?EXIVE AND LET (XK) BE A SEQUENCE IN X SUCH THAT ?XK?= FOR ALL K ?N AND ?X??= LIMK???XK,X?? SINCE X IS RE?EXIVE, THERE IS A WEAKLY CON- VERGENT SUBSEQUENCE OF (XK), STILL DENOTED BY (XK) FOR SIMPLICITY, WITH WEAK LIMIT X, SAY THEN ?X??1 AND ?X,X??= LIMK???XK,X??= ?X?? ?? 2 ?X+Y?: X,Y ?X,?X??1,?Y??1,?X?Y??? (1 5) THIS FUNCTION IS INTRODUCED SO THAT, GIVEN ANY TWO DISTINCT POINTS X AND Y IN THE CLOSED UNIT BALL B IN X, WE SHALL HAVE AN IDEA OF HOW FAR INSIDE B IS THE MIDPOINT OF THE LINE SEGMENT JOINING X TO Y NOTE THAT THE SAME FUNCTION IS OBTAINED IF THE IN?MUM IS TAKEN OVER ALL X,Y ?X WITH ?X?= ?Y?= 1 AND ?X?Y?= ? : SEE [88], VOL II, P 60 THE BANACH SPACE X IS CALLED UNIFORMLY CONVEX IF FOR ALL ? ?(0,2], ?X(?) > 0 THIS MEANS THAT X IS UNIFORMLY CONVEX IF, GIVEN ANY ? ?(0,2], ?X+Y??2(1??X(?)) WHENEVER X,Y ?X ARE SUCH THAT ?X?= ?Y?= 1 AND ?X?Y??? IT IS EASY TO SEE THAT AN EQUIVALENT CONDITION IS THAT WHENEVER (XK) AND (YK) ARE SEQUENCES IN X SUCH THAT ?XK??1,?YK??1 (K ?N) AND LIMK???XK +YK?= 2, THEN LIMK???XK ?YK?= 0 NOTE THAT ?X IS INCREASING ON [0,2], CONTINUOUS ON [0,2) (NOT NECESSARILY AT 2) AND ?X(0) = 0 MOREOVER, ?X(2) = 1 IF AND ONLY IF X IS STRICTLY CONVEX PLAINLY EVERY CLOSED LINEAR SUBSPACE OF A UNIFORMLY CONVEX SPACE IS UNIFORMLY CONVEX WHEN GIVEN THE INHERITED NORM MOREOVER, IT IS CLEAR THAT EVERY UNIFORMLY 1 2 STRICT AND UNIFORM CONVEXITY 2 ?X+Y?> 0 WHENEVER ?X?= ?Y?= 1 AND ?X?Y?= ? > 0 IMPLIES THAT ?X(?) > 0 SINCE CLOSED BOUNDED SETS ARE COMPACT IN ?NITE-DIMENSIONAL SPACES THE SIMPLEST EXAMPLE OF A UNIFORMLY CONVEX SPACE IS ANY HILBERT SPACE H, FOR THE PARALLELOGRAM LAW ENABLES US TO SEE THAT ?H(?) = 1 ?(1 ??2/4)1/2, WHICH IS CLEARLY POSITIVE FOR ALL ? ?(0,2]; IN FACT, SINCE (1 ?XQ)1/Q ?1 ?XQ/Q (0 ?X ?1,1 < Q < ?), WE HAVE ?H(?) ??2/8 NOTE THAT IN THE OPPOSITE DIRECTION, FOR EVERY BANACH SPACE X OF DIMENSION AT LEAST 2, IT IS KNOWN THAT ?X(?) ??H(?) = 1 ?(1 ??2/4)1/2 ?C?2; SEE [88], VOL II, P 63 IN THIS SENSE, HILBERT SPACES ARE THE ‘MOST’ UNIFORMLY CONVEX SPACES IF 1 < P < ?, THE SEQUENCE SPACE LP AND THE LEBESGUE SPACE LP (OVER ANY SET AND WITH ANY MEASURE) ARE UNIFORMLY CONVEX THIS WAS ESTABLISHED BY CLARKSON [25], WHO INTRODUCED THE NOTION OF UNIFORM CONVEXITY AND PROVED THE FOLLOWING INEQUAL- ITIES (NOW KNOWN AS THE CLARKSON INEQUALITIES) THAT HOLD FOR ARBITRARY ELEMENTS F,G ?LP : ?(F +G)/2?P? P +?(F ?G)/2?P? P 1/P? ? ?F?P P +?G?P P 1/P IF 1 ?P ?2, (1 6) AND ?(F +G)/2?P P +?(F ?G)/2?P P 1/P ? ?F?P P +?G?P P 1/P IF 2 ?P ?? (1 7) HERE ? ?P DENOTES THE NORM ON LP IN THE CASES 2 ?P ??AND 1 ?P ?2 THE INEQUALITIES IN (1 6) AND (1 7) RESPECTIVELY HOLD IN THE REVERSED SENSE HE USED THESE TO OBTAIN LOWER BOUNDS FOR ?P := ?LP OF THE FORM ?P(?) ?(?/C)R, (1 8) THUS F ?(T) ?0 IF 0 < T < Y, F ?(T) ?0 IF Y < T < ? IT FOLLOWS THAT THE MAXIMUM VALUE OF F ON (0,?) OCCURS WHEN T = Y, AND THE DESIRED INEQUALITY FOLLOWS THE PROOF WHEN 2 ?P < ?IS SIMILAR ?? IN [3] IT IS SHOWN THAT CLARKSON’S INEQUALITIES MAY BE OBTAINED FROM THIS RESULT, WHICH ALSO GIVES RISE TO THE FOLLOWING THEOREM, DUE ORIGINALLY TO HANNER [71] AND FROM WHICH THE MODULUS OF CONVEXITY ?P CAN BE ESTIMATED THEOREM 1 1 SUPPOSE THAT 1 < P ?2 AND LET F,G ?LP THEN ?F +G?P P +?F ?G?P P ? ?F?P +?G?P P + ?F?P ??G?P P IF 2 ?P < ?THE INEQUALITY IS REVERSED PROOF SUPPOSE THAT 1 < P ?2 THEN BY LEMMA 1 1, 1 2 STRICT AND UNIFORM CONVEXITY , THE ?NAL INEQUALITY FOLLOWING FROM (1 9) THIS GIVES THE THEOREM WHEN 1 < P ?2; SIMILAR ARGUMENTS WORK WHEN 2 ?P < ? TO PROVE (1 9) IT IS ENOUGH TO SHOW THAT H(T) := (1 +T)P +(1 ?T)P ?2{1 +(P ?1)T2}P/2 ?0, 0 ?T ?1 (1 10) (3 ?P)(2N ?1 ?P) (2N)! T2N +I, WHERE I = U +V P +2(P ?1) U ?V