故由f(x)在x?0连续性,令n??,得f(x)?f(0),x????,???. k?例2.设f(x)在R满足f(x)?fx???k?2,k?N?,且在x?1连续,证明:f(x) ?k1nf??x??, ???是常数. 证明:?x?R,因为 11??12?k???1?k???kkkk2????f(x)?f?x??f(x)?f?x????f??x???...???????????????故由f(x)在x?1连续性,令n??,得f(x)?f(1),x??0,???. ? 例3.设?x,y?R,均有f?x?y??f?x??f?y?,且f?x?在x?0点连续,证明: f?x??f?1?x. 证明:(1).f?0?0??f?0??f?0?,故f?0??0,故由f(x)在x?0处连续性,得 limf?x??0.于是?x0?R,得 x?0x?x0limf?x??limf?x0??x??lim?f?x0??f??x?? ?x?0?x?0?x?0?f?x0??limf??x??f?x0?. 所以f?x?在x0?R连续,因此f?x?在R连续. (2).?m,n?Z,?x?R,f?mx??f?x?x?...?x??mf?x?,于是 ??x??x??x?1f?x??f?n??nf??,f???f?x?, ?n??n??n?n?m??x?mf?x??mf???f(x). ?n??n?n?m?m令x?1,得f???f?1?,故f?r??rf?1?,r?Q?. ?n?n?r?Q?,f?r??f??r??f?r???r???f?0??0,得 f?r???f??r?????r?f?1??rf?1?, 于是,?x?Q,有f?x??xf?1?. ?x?R\\Q,则存在?rn??Q,使得limrn?x,故由f?x?的连续性,得 n??f?x??flimrn?limf?rn??limrnf?1??xf?1?. n??n??n???? 故f?x??xf?1?,x?R. 例4.设f?x??f?解:因为 f??1??f??x??,x?1,且f?x?在x?0处连续性,求f?x?. ?1?x???1??1??f????, ?1?1??2? 37
?1???2??1??1?f????f??f????, 12???3??1???2??1???3??1??1?f????f??f????, 1?3??4??1???3?……………………………………….. ?1???n??1??1?f????f??f????, 1?n??n?1??1???n??1??1?故f??1??f????...?f???,于是由f?x?在x?0处连续性,f??1??f?0?. ?2??n?xt?t??t,则x?t?xt,故x?令,于是f?t??f??.因此当t??1时 1?x1?t?1?t??t??1?t??t??t?f??f?f?????. t?1?t??1?2t??1???1?t?t???1?2t??t??t?f??f?f?????. t?1?2t??1?3t??1???1?2t?………………………………………………….. ?t??t??t??t?f?t??f??f?f?...?f???????. ?1?t??1?2t??1?3t??1?nt?于是由f?x?在x?0处连续性,f?t??f?0?. 例5.设f?x?在R连续,且关于有理数单调,即对任意的r1,r2?Q,当r1?r2?r1?r2?时,R单调. 有f?r1??f?r2?f?r1??f?r2?,证明:f?x?在证明:设f?x?在R连续,且关于有理数单增.任给x1,x2?R,且x1?x2. (1).若x1,x2?Q,则f?x1??f?x2?. (2).若x1?Q,x2?Q,则存在?rn??Q??x1,x2?,使得limrn?x2,且可使f?x1? n?????f?rn?,故f?x2??flimrn?limf?rn??f?x1?. n??n????(3).若x1?Q,x2?Q,同(2)一样可证. (4).若x1?Q,x2?Q,则存在r?Q??x1,x2?,故f?x1??f?r??f?x2?. 故f?x?在R单增. 例6.设f?x?在R连续且为非常数的周期函数,证明:f?x?必存在最小正周期. 证明:令A?TT是f?x?的正周期,则infA存在,设infA?T0.
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??由下确界的定义,存在Tn使得T0?Tn?T0?1Tn?T0.于是由 ?n?1,2,...?,故limn??nf?x?的连续性,得 f?x?T0??fx?limTn?limf?x?Tn??f?x?, n??n????即T0是f?x?的周期. 由Tn?0,得T0?limTn?0.若T0?0,则limTn?0,于是?x,有 n??n??x?knTn??n?n?1,2,...?, 其中kn是整数,0??n?Tn?n?1,2,...?.令xn?knTn?n?1,2,...?,则xn?x,故 f?x??flimxn?limf?xn??limf?0?xn??limf?0??f?0?, n??n??n??n????即f?x??f?0?,x?R,矛盾. §2.2闭区间上连续函数性质的应用 一.零点存在定理的应用 定理1(零点存在定理).若f?x?在?a,b?连续,且f?a?f?b??0,则至少存在 ???a,b?，使得f????0,即f?x??0在?a,b?至少有一根. y a o ? ? ? b x 图 2.2.1 定理2(零点存在定理).若f?x?在?a,b?连续,且f?a?f?b??0,则至少存在 ???a,b?，使得f????0,即f?x??0在?a,b?至少有一根. 例1.设f?x?,g?x?在?a,b?连续,且 f?a??g(a),f?b??g(b), 证明:方程f?x??g(x)在?a,b?至少有一个根.
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证明:令 F?x??f?x??g(x),x??a,b?, 则F?x?在?a,b?连续,且 F?a??f?a??g(a)?0,F?b??f?b??g(b)?0, 故由零点存在定理,方程f?x??g(x)在?a,b?至少有一个根. 例2.设f?x?在?a,a?2b?连续,证明:至少存在???a,a?b?，使得 f???b??f????证明:令 1f?a?2b??f?a??. ?2F?x??f?x?b??f?x??则F?x?在?a,a?b?连续,且 1?f?a?2b??f?a??,x??a,a?b?, 2F?a??f?a?b??f?a??1?f?a?2b??f?a?? 2?f?a?b??1?f?a?2b??f?a??; 21F?a?b??f?a?2b??f?a?b???f?a?2b??f?a?? 21?????f?a?b???f?a?2b??f?a???, 2??故F?a?F?b??0,于是由零点存在定理,至少存在???a,a?b?，使得 1?f?a?2b??f?a??. 2?3a?b??3a?b?例3.设f?x?在R连续,a,b?R,a?b,f??f???,f?a?b??f?a?, ?2??2?b?a?b?a???证明:存在???a,b?使得f????f?????. 22????f???b??f????证明:令 b?a??F?x??f?x???2??则F?x?在?a,b?a??f?x??,x?R, 2????a?b?连续,且 ?2?b?a?b?a????3a?b??3a?b?F?a??f?a??fa??f?f????????0; 2?2????2??2??b?a??b?ab?a??b?ab?a?F??f??f???????f?a?b??f?a??0, 2?2??2??2?2
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故由零点存在定理,存在???a, ??a?b????a,b?，使得2?b?a?b?a???f????f?????. 2?2???例4.设f?x?在?k,k?2?连续,f?k??f?k?1?,证明:至少存在???k,k?1?，使得 1??f?????f????n?1,2,...?. n??1?? 证明:当n?1时,取??k,则f?????f???. n??当n?1时,令 1??F?x??f?x???f?x?,x??k,k?1?, n??则 1?n?1???F?k??F?k???...?F?k?? n?n?????1????2???f?k???f?k????f?k???fn?n???????f?k?1??f?k??0. 若F?k??0,或F?k?1????k??...?????fn????n??k????n??n?1???f?k??? n???1?n?1??,…,或?0Fk?????0,则取??k,或取 n?n??1n?1??k?,…,或取??k?即可. nn??若 1?n?1???F?k??0,F?k???0,…,F?k???0, nn????1??1?则必存在i,j??0,1,2,...,n?1k??Fk??，使得F???0,于是故由零点存在定理,至???ij????少存在???k?,k?1??f?????f????n?1,2,...?. n???1?? 故至少存在???k,k?1?，使得f?????f????n?1,2,...?. n?? 例5.设f?x?在R连续,且 ?1i1??，使得j?fffff?f?x??证明:至少存在??R，使得f证明:令 ?????????x, ?f?f???????. ??F?x??ff?f?x???x, 若F?x?在R不变号,不妨设 F?x??ff?f?x???x?0,x?R, 则f???f?f?x????x,x?R.于是 41