14 thoughts on “Phương trình hàm Cauchy trong lớp đơn điệu”

  1. Chung ta chac deu da biet ket qua sau day:

    Dinh li 1. Neu f:\mathbb{R}\to\mathbb{R} thoa man f(x+y)=f(x)+f(y)\forall x,y\in \mathbb{R} (*) va no lien tuc tren \mathbb{R} thi ton tai so thuc a sao cho f(x)=ax\forall x\in \mathbb{R}.
    Chung minh. Tu (*) bang quy nap va cac tinh toan don gian ta duoc f(q)=aq\forall q\in\mathbb{Q}, o day a=f(1). Nhung \overline{\mathbb{Q}}=\mathbb{R} va f(x) lien tuc tren \mathbb{R} nen chung ta co f(x)=ax\forall x\in \mathbb{R}.\Box

    Ro rang ham so f(x)=ax\forall x\in\mathbb{R} cung thoa man dieu kien (*), le tu nhien chung ta dat cau hoi: Neu f(x) chi thoa man (*) nhung khong thoa man dieu kien lien tuc thi lieu co nhat thiet phai co f(x)=ax\forall x\in\mathbb{R} voi mot so a\in\mathbb{R}? Cau tra loi la khong😀 Nhu vay can dieu kien bo sung thi chung ta moi co ket qua do. Co nhieu dieu kien nhu vay, nhung trong tiet nay chung ta chi quan tam den ket qua sau day:

    Dinh li 2. Neu f:\mathbb{R}\to\mathbb{R} thoa man f(x+y)=f(x)+f(y)\forall x,y\in \mathbb{R} (*) va no don dieu tren \mathbb{R} thi ton tai so thuc a sao cho f(x)=ax\forall x\in \mathbb{R}.

    Chung minh. Tu (*) bang quy nap va cac tinh toan don gian ta duoc f(q)=aq\forall q\in\mathbb{Q}, o day a=f(1). Chung ta chi can chung minh cho truong hop f don dieu tang la du( neu f don dieu giam thi -f don dieu tang). Voi moi so thuc x, lay hai day so huu ti (u_n),(v_n) sao cho chung thoa man dong thoi cac dieu kien:
    a)\lim_{n\to\infty}u_n=\lim_{n\to\infty}v_n=x.
    b)u_n\leq x\leq v_n\forall n.
    Tu a) ta co f(u_n)\leq f(x)\leq f(v_n)\forall n hay au_n\leq f(x) \leq av_n\forall n, o day cho n\to\infty va chu y den a) chung ta co dieu phai chung minh.\Box

    Trong mot vai bai toan , dieu kien don dieu thuong duoc ”giau di”😛 , day la mot vi du:

    He qua 1. Neu f:\mathbb{R}\to\mathbb{R} thoa man f(x+y)=f(x)+f(y)\forall x,y\in \mathbb{R} (*) va f(xy)=f(x)f(y)\forall x,y\in\mathbb{R} thi f(x)=0\forall x\in\mathbb{R} hoac f(x)=x\forall x\in\mathbb{R}.

    He qua 2. Neu mot ham f:(0,\infty)\to\mathbb{R} thoa man f(xy)=f(x)f(y)\forall x,y>0 va no la don dieu thi f(x)=0\forall x>0 hoac co so c sao cho f(x)=x^c\forall x>0.

    Chung minh hai he qua nay de lai cho nguoi doc xem nhu bai tap, cac ban hay post chung minh cho no ngay trong topic nay.

    Sau day la vai bai tap ap dung.

    Bai 1(IMO 1992).
    Tim tat ca cac ham f:\mathbb{R}\to\mathbb{R} sao cho f(x^2+f(y))=f^2(x)+y\forall x,y\in\mathbb{R}.

    Bai 2(Italy 1999).
    a)Tim tat ca cac ham f:\mathbb{R}\to\mathbb{R} sao cho f don dieu ngat tren \mathbb{R} va f(x+f(y))=f(x)+y\forall x,y\in\mathbb{R}.
    b)Chung minh rang voi moi so nguyen n>1, khong ton tai ham f:\mathbb{R}\to\mathbb{R} sao cho f don dieu ngat tren \mathbb{R} va f(x+f(y))=f(x)+y^n\forall x,y\in\mathbb{R}.

    Bai 3(Vietnam TST 2004).
    Tim tat ca cac so thuc a sao cho ton tai duy nhat ham f:\mathbb{R}\to\mathbb{R} thoa man f(x^2+y+f(y))=f^2(x)+ay\forall x,y\in\mathbb{R}.

    Bai 4(China TST 2003 quizzes).
    Tim tat ca cac ham f,g:\mathbb{R}\to\mathbb{R} sao cho f(x+yg(x))=g(x)+xf(y)\forall x,y\in\mathbb{R}.

    Cac ban hay giai no va post loi giai ngay trong topic nay.
    (Moi cac ban cung dong gop cac vi du😀 )

  2. Bai 1 thi qua quen thuoc roi .Chac khong can post loi giai lam gi nua😀

    Bai 2 :a,
    Cho x =0 ta co f(f(y))=y \forall y\in \mathbb{R} .
    Thay x boi f(x) ta co f(f(x)+f(y))=x+y=f(f(y)).
    Suy ra : f(x)+f(y)=f(x+y)
    Do f(x) don dieu va cong tinh nen f(x)=kx \forall x\in\mathbb{R}
    Thay vao bieu thuc ta co ngay k=1 .
    Vay f(x)=x \forall x\in\mathbb{R}

    b,
    Tuong tu cau a, ta co f(x)+f(y)=f(x+y) va f(x)=kx.
    Thay vao bieu thuc ta co k(x+ky)=kx+y^n suy ra
    k^2=y^{n-1} \forall y\in \mathbb{R} vo ly .

  3. Chet that em chu quan qua .Edit lai ngay day .
    Ta chi can chung minh f(f(x)+f(y))=f((f(x+y)) la duoc.
    Thay y=0, ta co f(x+f(0))=f(x)
    Lai thay x=0 ta co f(f(y))=y+f(0) .
    Thay x boi f(x) ta co :
    f(f(x)+f(y))=x+y+f(0)=f(f(x+y))
    Xong roi nha .:D.Xin loi moi nguoi .:”>

    Con bai 1, em noi no quen thuoc chu co noi no de dau. no cung binh thuong thoi ma anh😛

  4. Minh biet la chi can lien tuc tai mot diem la du, nhung topic nay chi can quan tam den dieu do. Cam on Cuong.

    Gui Quang: Post loi giai Bai 2 b) di , hay la chua giai duoc, khong giai duoc thi cu nhan di , khong phai xau ho dau, khong lam duoc mot bai Toan la chuyen binh thuong thoi ma .😀

  5. Anh Tuan kich deu em phai khong.:-P.
    Loi giai day nay:
    Cho x=y=0 ta co f(0)=0
    Cho x=0 duoc f(f(y))=y^n,\forall y\in\mathbb{R}
    Suy ra :
    f(x+y^n=f(x)+f(f(x))^n
    Lai cho x=0 ta co f(y^n)=f(f(y))^n
    \Rightarrow f(x+y^n)=f(x)+f(y^n)
    Do do:
    (x+y^n)^n=f(f(x+y^n))=f(f(x)+f(y^n))=f(f(x))+(y^n)^n=x^n+y^{n^2}
    vo ly cho ta ket luan:voi n\ge 2 khong ton tai ham sao thoa man.

  6. Bai 5 day😀
    Thay x=y=0 ta duoc:f(0)=0
    Thay x=1,y=-1 ta duoc -f(1)=f(1).f(-1)
    Ta xet hai truong hop:
    TH 1.f(-1)=0.Khi do thay y=1 ta duoc :
    f(x+1)=0 \forall x\in\mathbb{R} hay f(x)=0\forall x\in\mathbb{R}ta duoc 1 ham so thoa man.

    TH 2.f(-1)=-1.
    Thay y=-1 ta duoc f(-x)=-f(x-1)-1
    Thay xboi -x, y=-1\Rightarrow f(-x-1)=-f(x)-1(*)
    Thay y=-1\Rightarrow f(x)=1+f(x-1)
    Tu (*) ta co f(-x-1)=-f(x+1) hay f(x)=-f(-x)
    Trong gia thiet thay x boi -x, y boi -y
    \Rightarrow f(-x-y)+f(xy)=f(-x).f(-y)+f(-x)+f(-y)
    Hay :
    -f(x+y)+f(xy)=f(x).f(y)-f(x)-f(y)
    Suy ra :f(xy)=f(x).f(y) va f(x+y)=f(x)+f(y)
    Do do :f(x)=k.x thay vao PT ta co ngay f(x)=x
    done.

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