Metric and Hilbert Spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 4 November 2014

Assignment 1 Solutions

  1. Let A and B be bounded subsets of a metric space (X,d) such that AB. Show that diam(AB)diam (A)+diam(B). What can you say if A and B are disjoint?

    Solution.
    The definition of diam(A) is diam(A)=sup {d(x,y)|x,yA}. Assume AX and BX and AB= and diam(A)< and diam(B)<. To show: diam(AB)diam(A)+diam(B).
    To show: diam(A)+diam(B) is an upper bound of {d(x,y)|x,yAB}.
    To show: If x,yAB then d(x,y)diam(A)+diam(B).
    Assume x,yAB.
    Case 1: x,yA. Then d(x,y) diam(A) diam(A)+ diam(B). Case 2: x,yB. Then d(x,y) diam(B) diam(A)+ diam(B). Case 3: xA and yB. Let zAB. Then d(x,y) d(x,z)+ d(z,y) diam(A)+ diam(B). Case 4: xB and yA. Let zAB. Then d(x,y) d(x,z)+ d(z,y) diam(A)+ diam(B). So diam(A)+diam(B) is an upper bound of {(x,y)|x,yAB}.
    So diam(AB)diam(A)+diam(B).

  2. Let X=C[0,1]={f:[0,1]|fis continuous}. The supremum metric d:X×X0 and the L1 metric d1:X×X0 are defined by d(f,g) = sup { |f(x)-g(x)| |x[0,1] } and d1(f,g) = 01f(x)-g(x) dx. Consider the sequence {f1,f2,f3,} in X where fn(x)=nxn(1-x) for 0x1.
    1. Determine whether {fn} converges in (X,d1).
    2. Determine whether {fn} converges in (X,d).
    (You may use any standard results about limits of real sequences.)

    Solution.
    The graph of y=x-x2 1 1 and y=xn-1 1 1 help to determine the graph of y=nxn(1-x)=nxn-1(x-x2). 1 1 4 Since limnnxn-1=0 for x[0,1) and nxn-1=n for x=1 and x-x2=0 for x=1, then limnnxn-1(x-x2)=0 for x[0,1].
    So the pointwise limit of {fn} is the zero function f:[0,1] given by f(x)=0.
    {fn} converges in (X,d1) if limnd1(fn,f)=0.
    {fn} converges in (X,d) if limnd(fn,f)=0.
    Compute: d1(fn,f). d1(fn,f) = 01 fn(x)-f(x)dx = 01 (nxn(1-x)-0)dx = 01 (nxn-nxn+1)dx = ( nxn+1n+1- nxn+2n+2 ) |x=0x=1 = ( nn+1- nn+2 ) = n(n+1)(n+2). So limn d1(fn,f) = limn n(n+1)(n+2) = limn 1(1+1n) 1(n+2) = 1·0 = 0. So {fn} converges in (X,d1).
    Compute d(fn,f):
    To compute d(fn,f)=sup{nxn(1-x)-0|x[0,1]} find the maximum of fn(x)=nxn(1-x) on the interval [0,1].
    This maximum occurs at x=0 or x=1 or at a critical point.
    Since dfndx= dnxn(1-x)dx= n2xn-1-n (n+1)xn=n xn-1 (n-nx-x) the critical points are at x=0 and x=nn+1.
    Since fn(0)=0 and fn(1)=0 and fn(nn+1)=n(nn+1)n(1-nn+1) the maximum of fn is (nn+1)n+1= (1-1n+1)n+1. So d(fn,f) = sup{fn(x)-f(x)|x[0,1]} = sup{nxn(1-x)|x[0,1]} = (1-1n+1)n+1 and limn (1-1n+1)n+1 = limn elog(1-1n+1)n+1 = limn e(n+1)log(1-1n+1) = limn e-(n+1)(1n+1+12(1n+1)2+13(1n+1)3+) since -log(1-x)=11-xdx=(1+x+x2+)dx=x+12x2+13x3+.
    So limn d(fn,f)= limn (1-1n+1)n+1= e-(1+12·0+13·02+)= e-1. So {fn} does not converge in (X,d).

  3. Let X and Y be topological spaces. Let AX and BY. Show that A×B= A×B.

    Solution.
    To show:
    (a) A×BA×B.
    (b) A×BA×B.
    (a) Assume (x,y)A×B.
    To show: (x,y)A×B.
    To show: (x,y) is a close point of A×B.
    Let N be a neighbourhood of (x,y) in X×Y.
    By the definition of the product topology on X×Y there exist NX, a neighbourhood of x in X, and Ny, a neighbourhood of y in Y, such that Nx×NyN.
    Since xA there exists aA with aNx.
    Since yB there exists bB with bNy.
    So (a,b)Nx×NyN and (a,b)A×B.
    So (x,y) is a close point of A×B.
    So (x,y)A×B.
    So A×BA×B.
    (b) To show: A×BA×B.
    Assume (x,y)A×B.
    To show: (x,y)A×B.
    To show: xA and yB.
    Let Nx be a neighbourhood of xX and let Ny be a neighbourhood of yY.
    Then Nx×Ny is a neighbourhood of (x,y)X×Y.
    Since (x,y) is a close point of A×B, there exists (a,b)A×B with (a,b)Nx×Ny.
    So aNx and bNy and aA and bB.
    So x is a close point of A and y is a close point of B.
    So xA and yB.
    So (x,y)A×B.

  4. Let (X,d) be a metric space and let A be a non-empty subset of X. Recall that for each xX, the distance from x to A is d(x,A)=inf {d(x,a)|aA}.
    1. Prove that A={xX|d(x,A)=0}.
    2. Prove that |d(x,A)-d(y,A)|d(x,y) for all x,yX. [Hint: first show that d(x,A)d(x,y)+d(y,A).]
    3. Deduce the function f:X defined by f(x)=d(x,A) is continuous.
    4. Show that if xA then U={yX|d(y,A)<d(x,A)} is an open set in X such that AU and xU.

    Solution.
    (a)
    To show:
    (aa) {xX|d(x,A)=0}A.
    (ab) A{xX|d(x,A)=0}.
    (aa) Assume xX and d(x,A)=0.
    To show: xA.
    Let N be a neighbourhood of x in X.
    Then there exists ε>0 such that B(x,ε)N.
    Since d(x,A)=inf{d(x,a)|aA}=0, there exists aA such that d(x,a)<ε.
    Then aB(x,ε)N and aA.
    So x is a close point of A.
    So {xX|d(x,A)=0}A.
    (ab) To show: A{xX|d(x,A)=0}.
    Let xA.
    So x is a close point of A.
    To show: d(x,A)=0.
    Let ε>0.
    Then B(x,ε) is a neighbourhood of x in X.
    Since x is a close point of A there exists aA such that aB(x,ε).
    So d(x,a)<ε.
    So d(x,A)<ε for all ε>0.
    So d(x,A)=0.
    So x{xX|d(x,A)=0}.
    So A{xX|d(x,A)=0}.
    Thus A={xX|d(x,A)=0}.
    (b)
    Assume x,yX.
    To show:
    (ba) d(x,A)-d(y,A)d(x,y).
    (bb) -(d(x,A)-d(y,A))d(x,y).
    (ba) Since d(x,A) is a lower bound of {d(x,a)|aA}, if aA then d(x,A)d(x,a).
    Using d(x,a)d(x,y)+d(y,a), if aA then d(x,A)d(x,y)+d(y,a).
    So d(x,A) is a lower bound of {d(x,y)+d(y,a)|aA}.
    Since d(x,y)+d(y,A) is the greatest lower bound of {d(x,y)+d(y,a)|aA} then d(x,A) d(x,y)+ d(y,A). So d(x,A)-d(y,A)d(x,y).
    So d(y,A)-d(x,A)d(y,x)=d(x,y).
    So -(d(x,A)-d(y,A))d(x,y).
    So d(x,A)-d(y,A)d(x,y) and -(d(x,A)-d(y,A))d(x,y).
    So d(x,A)-d(y,A)d(x,y).
    (c) To show: If ε>0 and xX then there exists δ>0 such that if yX and d(x,y)<δ then d(f(x),f(y))<ε. Assume ε>0 and xX.
    To show: There exists δ>0 such that if yX and d(x,y)<δ then d(f(x),f(y))<ε. Let δ=ε.
    To show: If yX and d(x,y)<δ then d(f(x),f(y))<ε.
    Assume yX and d(x,y)<δ.
    To show: d(f(x),f(y))<ε.
    By part (b), d(f(x),f(y))= d(y,A)-d(x,A) d(x,y)<δ=ε. So f is continuous.
    (d)
    Assume xA and let U={yX|d(y,A)<d(x,A)}.
    To show:
    (da) xU.
    (db) U is open.
    (dc) AU.
    (da) Let D=d(x,A).
    Since xA and, by part (a), A={yX|d(y,A)=0} then d(x,A)0.
    So D0.
    We know U={yX|d(y,A)<D}.
    Since d(x,A)=D, xU.
    (db) Since U=f-1(<D)=f-1((-,D)) and f is continuous, then U is open.
    (dc) By part (a), A= { yX| d(y,A)=0 } {yX|d(y,A)<D} =U. So AU.

  5. Determine whether the following sequences of functions converge uniformly.
    1. fn=e-nx2,x[0,1];
    2. gn=e-x2/n,x[0,1].
    3. gn=e-x2/n,x.

    Solution.
    Let (X,ρ) be a metric space and let fn:X, n>0, be a sequence of functions from X to .
    Assume that f:X defined by f(x)=limn fn(x) is well defined.
    The sequence f1,f2,f3, converges uniformly to f if limn ( sup {ρ(fn(x),f(x))|xX} ) =0.
    (a) Define fn:[0,1] by fn(x)= e-nx2= e-(nx)2, for n>0.
    Then limn e-n·02= e-0=1 and limn e-n·12= e-=0 and if x(0,1) then limn e-nx2= e-=0. 1 Let f:[0,1] be given by f(x)= { 1, ifx=0, 0, ifx(0,1]. Let n>0. Then sup{ρ(fn(x),f(x))|xX} = sup ( { e-nx2-0 |x(0,1] } {1-1} ) = sup{e-nx2|x(0,1]} = 1. So limn ( sup {e-nx2|x[0,1]} ) =limn1=1. So f1,f2,f3, is not uniformly convergent.
    (b) Define gn:[0,1] by gn(x)= e-x2/n= e-(x/n)2, for n>0.
    Then limn e-02/n= e-0=1and limn e-12/n= e-1/= e-0=1 and if x(0,1) then limn e-x2/n= e-0=1. 1 Let g:[0,1] by given by g(x)=1.
    Let n>0.
    Then sup{ρ(gn(x),g(x))|xX} = sup{e-x2/n-1|x[0,1]} = e-12/n-1 = 1-e-1/n. So limn (sup{e-x2/n-1|x[0,1]}) = limn1-e-1n = 1-e-1/ = 1-e0 = 1-1 = 0. So g1,g2,g3, uniformly converges to g.
    (c) Define gn: by gn(x)=e-x2/n= e-(x/n)2, for n>0.
    Let x.
    Then limn e-x2/n= e-x2/= e-0=1. Let g: be given by g(x)=1.
    Let n>0.
    Then sup{ρ(gn(x),g(x))|xX} = sup{e-x2/n-1|x} = e-2/n-1 = e--1 = 0-1 = 1. So limnsup {ρ(gn(x),g(x))|xX} =limn1=1. So g1,g2, is not uniformly convergent.

  6. Let X be the set of all real sequences with finitely many non-zero terms with the supremum metric: if x=(xi) and y=(yi) then d(x,y)=sup{|xi-yi||i>0}.
    For each n, let xn=(1,1/2,1/3,,1/n,0,0,).
    1. Show that {xn} is a Cauchy sequence in X.
    2. Show that {xn} does not converge to a point in X. (So X is not complete.)

    Solution.
    Let X= { (x1,x2,) |xi and all but a finite number ofxiare zero } and define d:X×X>0 by d(x,y)=sup {xi-yi|i>0}. Let fn=(1,12,13,,1n,0,0,,0).
    (a) Show that {fn} is a Cauchy sequence in X.
    To show: If ε>0 then there exists N>0 such that if m,n>0 and m>N and n>N then d(fm,fn)<ε.
    Assume ε>0.
    To show: There exists N>0 such that if m,n>0 and m>N and n>N then d(fm,fn)<ε.
    Let N=1ε.
    Assume m,n>0, m>N and n>N and m<n.
    To show: d(fn,fm)<ε. d(fn,fm) = sup { 1-1, 12-12,, 1m-1m, 1m+1, 1m+2,, 1n,0-0,0-0, } = 1m+1 < 1N+1 = 11ε+1 = εε+1 < ε1 = ε. So {fn} is a Cauchy sequence in X.
    (b) The limit of {f1,f2,} is the sequence f=(1,12,13,14,). If k>0 the kth entry of f is 1k which is not equal to 0.
    So all entries of f are nonzero.
    So fX.
    So {f1,f2,} is a Cauchy sequence in X which does not converge to a point in X.
    So X is not complete.

  7. Let X be a nonempty set and let (Y,d) be a complete metric space. Let f:XY be an injective function and define df(x,y)=d (f(x),f(y)) for x,yX.
    1. Explain briefly why df is a metric on X.
    2. Show that (X,df) is a complete metric space if f(X) is a closed subset of Y.

    Solution.
    (a)
    To show:
    (aa) If x,yX then df(x,y)=df(y,x).
    (ab) If xX then df(x,x)=0.
    (ac) If x,yX and df(x,y)=0 then x=y.
    (ad) If x,y,zX then df(x,y)df(x,z)+df(z,y).
    (aa) Assume x,yX.
    To show: df(x,y)=df(y,x). df(x,y)= d(f(x),f(y))= d(f(y),d(x))= df(y,x).
    (ab) Assume xX.
    To show: df(x,x)=0. df(x,x)= d(f(x),f(x))=0.
    (ac) Assume x,yX and df(x,y)=0.
    To show: x=y.
    Since 0=df(x,y)=d(f(x),f(y)) and d is a metric, then f(x)=f(y).
    Since f:XY is injective and f(x)=f(y) then x=y.
    (ad) Assume x,y,zX.
    To show: df(x,y)df(x,z)+df(z,y). df(x,y) = d(f(x),f(y)) d(f(x),f(z))+ d(f(z),f(y)) = df(x,z)+ df(z,y).
    (b) Assume f(x) is a closed subset of Y.
    Since f:XY is injective and f:Xf(X) is surjective, then f:Xf(X) is bijective.
    Since df(x,y)=d(f(x),f(y)) for x,yX then f:(X,df) (f(X),d) is an isometry.
    To show: (X,df) is complete.
    To show: (f(X),d) is complete.
    To show: If z1,z2, is a Cauchy sequence in f(X) then z1,z2, converges with limnzn in f(X).
    Assume z1,z2, is a Cauchy sequence in f(X).
    Then z1,z2, is a Cauchy sequence in Y.
    Since Y is complete z=limnzn exists with zY.
    Since z is a close point of z1,z2,, then z is a close point of f(X).
    So zf(X).
    So f(X) is complete.
    So (X,df) is complete.

  8. Let f:00 be given by f(x)=22+x.
    1. Show that f defines a contraction mapping f:00.
    2. Fix x00 and xn+1=f(xn) for all n0. Show that the sequence {xn} converges and find its limit with respect to the usual metric on .

    Solution.
    (a) A function f:XX is a contraction mapping if there exists α(0,1) such that if x,yX then d(f(x),f(y))αd(x,y). Let f:00 be given by f(x)=22+x.
    Let α=12.
    To show: If x,y>0 then d(f(x),f(y))αd(x,y).
    Assume x,y>0.
    To show: f(y)-f(x)αy-x. f(y)-f(x) = 22+x-22+y = 22+y-(2+x)(2+x)(2+y) = 2(2+x)(2+y) y-x 24y-x = 12y-x. So f is a contraction mapping.
    (b) The Banach fixed point theorem gives that the sequence xn+1=f(xn) for x0X converges to the (unique) fixed point of f.
    In our case the fixed point is p>0 such that p=f(p)=22+p. So p2+2p=2 and p=-2±4+4·22=-1±3.
    Since p>0 then p=-1+3.

  9. Let X be a connected topological space. Let f:X be continuous with f(X). Show that f is a constant function.

    Solution.
    To show: If f is not a constant function then X is not connected.
    Assume f is not a constant function.
    Let a,bf(X) with a<b.
    Let z, z with a<z<b.
    Let A=f-1((-,z)) and B=f-1((z,)).
    Then AsinceaA, BsincebB, AB=since (-,z) (z,)=. Since z then zf(X) and since (-,z)(z,)=-{z} then AB=X.
    So X is not connected.

  10. Show that X={(x,y)2|xy=0} is not homeomorphic to .

    Solution.
    Let X = {(x,y)2|xy=0} = {(x,y)2|x=0} {(x,y)2|y=0}. y = 0 x = 0 Assume that f:X is a homeomorphism.
    Let a=f((0,0)).
    Then f:X-{(0,0)}-{a} is a homeomorphism.
    So X-{(0,0)} and -{a} have the same number of connected components.
    Since X-{(0,0)} has 4 connected components and -{a} has 2 connected components, this is a contradiction.
    So X is not homeomorphic to .

Notes and References

These are a typed copy of Assignment 1 Solutions from a series of handwritten lecture notes for the class Metric and Hilbert Spaces.

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