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Linear Operators on Hilbert Spaces

  1. Apr 1, 2005 #1
    Let [tex]U, V, W[/tex] be inner product spaces. Suppose that [tex]T:U\rightarrow V[/tex] and [tex]S:V\rightarrow W[/tex] are bounded linear operators. Prove that the composition [tex]S \circ T:U\rightarrow W[/tex] is bounded with [tex]\|S\circ T\| \leq \|S\|\|T\|[/tex]
     
    Last edited: Apr 1, 2005
  2. jcsd
  3. Apr 1, 2005 #2
    We know that [tex]T[/tex] and [tex]S[/tex] are bounded linear operators, and that they 'operate' on three inner product spaces [tex]U,V,W[/tex]. The composition is bounded if and only if it continuous. So does it make sense for me to prove that [tex]S \circ T[/tex] is continuous which implies that it is bounded?

    But from what I have seen the composition of two bounded linear operators is a bounded linear operator. So what I'm trying to prove here is an obvious definition, obvious in the sense that it seems intuitive.
     
  4. Apr 1, 2005 #3
    We know that [tex]S[/tex] is bounded. By definition it has a norm [tex]\|S\|[/tex] which is the lower bound of all the constants [tex]K[/tex] that make [tex]S[/tex] a bounded operator. ie

    [tex]\|S\| = \inf\{K:\|Sx\| \leq K\|x\|, \, \forall \, x\in X\}[/tex]

    So I want to prove that [tex]S[/tex] is uniformly continuous if and only if [tex]S[/tex] is a bounded linear operator, say [tex]\|S\|= K[/tex].

    [tex]proof[/tex]

    Take any two points [tex]v \in V[/tex] and [tex]w \in W[/tex], let [tex]d = v - w[/tex]. Now [tex]\|S(d)\|[/tex] is bounded by [tex]K\|d\|[/tex]. By linearity, [tex]S(d) = S(v) - S(w)[/tex]. Thus distance is scaled by at most [tex]K[/tex], everywhere, and [tex]S[/tex] is uniformly continuous.

    Conversely, take an [tex]r[/tex] such that [tex]\|u\| < r [/tex] implies [tex]\|S(u)\| < 1[/tex]. The norm of the image of the ball of radius [tex]r[/tex] is at most 1, hence [tex]1/r[/tex] acts as a bound for [tex]S[/tex].

    Hence [tex]S[/tex] is uniformly continuous, and by the same argument, so is [tex]T[/tex].
     
  5. Apr 1, 2005 #4
    If I am on the right track, then since [tex]S[/tex] and [tex]T[/tex] are uniformly continuous linear operators, then [tex]S\circ T[/tex] is uniformly continuous (the composition of two continuous linear operators is again continuous and linear).

    Hence [tex]S \circ T[/tex] is bounded.
     
  6. Apr 1, 2005 #5
    Prove that if [tex]S,T : V \rightarrow W[/tex] are bounded linear operators between inner product spaces, then the map [tex]S + T : V \rightarrow W[/tex] , defined by [tex](S+T)(v):= S(v) + T(v), \, \forall \, v \in V[/tex] , is a bounded linear operator with [tex]\|S+T\| \leq \|S\| + \|T\|[/tex].
     
  7. Apr 2, 2005 #6
    Does anyone know if the answer to my first question is right or not?

    Any pointers for my second question?
     
  8. Apr 3, 2005 #7
    Oops, missed this! Silly physics labs.

    While your strategy for the first question may work out (I haven't looked at it too hard), there is a much easier (and as you noted, intuitive!) way to go about the proof.

    Say [itex]\|T\| = K[/itex], and [itex]\|S\| = C[/itex]. Let [itex]u \in U[/itex]. Then

    [tex] T(u) = v \in V \ \mbox{with} \ \|v\| \leq K\|u\|[/tex]

    and

    [tex] S(v) = w \in W \ \mbox{with} \ \|w\| \leq C\|v\|[/tex]

    combine the results, and see what you get :smile:

    Now for the second question. Again let [itex]\|S\| = C, \ \|T\| = K[/itex].

    Then say [itex] S(v) = w_1 \in W, \ T(v) = w_2 \in W[/itex], so that [itex]\|w_1\| \leq C\|v\|[/itex] and [itex]\|w_2\| \leq K\|v\|[/itex]. Look at

    [tex](S+T)(v) = S(v)+T(v) = w_1 + w_2[/tex]

    What can be said about the magnitude of the right side (as you're probably tired of me saying, you might need the triangle inequality! :smile:)?
     
  9. Apr 3, 2005 #8
    Thanks Data (I was wondering where you got to?). I will never tire of you reminding me to use the triangle inequality :smile:

    Anyway, I'll give it a go and type back later.
     
  10. Apr 3, 2005 #9
    Why can't [tex]\|v\|[/tex] be greater than [tex]K\|u\|[/tex] is that because we are free to choose any [tex]K[/tex] we want, and we can always choose a [tex]K[/tex] such that [tex]\|v\| \leq K\|u\|[/tex].
     
  11. Apr 3, 2005 #10
    Let [tex]\|T\| = K[/tex] and [tex]\|S\| = C[/tex], where [tex]C, K \in \mathbb{R}[/tex].

    Take any [tex]u \in U[/tex], [tex]v \in V[/tex], and [tex]w \in W[/tex]. Then.

    [tex]T(u) = v \in V[/tex] with [tex]\|v\| \leq K\|u\|[/tex]
    [tex]S(v) = w \in W[/tex] with [tex]\|w\| \leq C\|v\|[/tex]

    Combining the results gives

    [tex].\,\, \frac{1}{C}\|w\| \leq \|v\| \leq K\|u\|[/tex]
    [tex]\rightarrow \|w\| \leq M\|u\|[/tex]

    where [tex]M = CK \in \mathbb{R}[/tex].

    That is, [tex]S\circ T[/tex] is bounded. Because, by the definition of boundedness, [tex]S \circ T[/tex] is bounded if there exists an [tex]M > 0 \in \mathbb{R}[/tex] such that [tex]\|(S\circ T)x\| \leq M\|x\|[/tex] for all [tex]x \in X[/tex].

    In this case [tex]M = CK[/tex]
     
  12. Apr 3, 2005 #11
    That looks (almost - look at the next paragraph~) fine. Do you understand why [itex]\|v\| \leq K \|u\|[/itex] now? You can't choose [itex]K[/itex]. It's just the order of [itex]T[/itex], which you know has the property that

    [tex] \|T\| \|u\| = K\|u\| \geq \|T(u)\| = \|v\|[/tex]

    since [itex]T(u)=v[/itex] :smile:

    Your proof looks fine, except one thing. Taking any [itex]u \in U[/itex] was right, but you do not get to choose [itex]v[/itex] and [itex]w[/itex] arbitrarily. Since [itex]T : U \longrightarrow V[/itex], [itex]v[/itex] is just whatever vector [itex]T[/itex] takes [itex]u[/itex] to. Similarly, [itex]w[/itex] is whatever vector [itex]S[/itex] takes [itex]v[/itex] to. They are effectively determined by your choice of [itex]u[/itex]. Then, you proved that [itex]CK\|u\| \geq \|w\| = \|(S \circ T)(u)\|[/itex], and since you could choose any [itex]u[/itex] to start with, you're finished!

    You've even proved the last part of the question, that [itex]\|S \circ T \| \leq \|S\|\|T\|[/itex]. Do you see why?
     
  13. Apr 4, 2005 #12
    Thanks Data for that help and clarification. I will post back with an attempt of the second question in a couple of minutes.
     
  14. Apr 4, 2005 #13
    By the way, regarding the first question, why does [itex]\|S\circ T\|\leq \|S\|\|T\|[/itex]? Could you explain it please. I am really interested. In all the proofs I have seen, I actually haven't seen THIS proved to me. (don't tell me it requires the triangle inequality).

    edit - please disregard this post (see below)
     
    Last edited: Apr 4, 2005
  15. Apr 4, 2005 #14
    I know why it wasn't proved to me! Because it is bloody obvious!

    [tex]\|S\|\|T\| = CK = M[/tex]

    Hence

    [tex]\|S\circ T\| \leq M = CK = \|S\|\|T\| [/tex]

    So [itex]\|S\circ T\| \leq \|S\|\|T\|[/itex]
     
  16. Apr 4, 2005 #15
    Second Question

    Let [itex]\|S\| = B[/itex] and [itex]\|T\| = C[/itex], where [itex]B, C \in \mathbb{R}[/itex]. Then take any [itex]v \in V[/itex] we have

    [tex]S(v) = w_1 \in W[/tex] such that [itex]\|w_1\| \leq B\|v\|[/itex]
    [tex]T(v) = w_2 \in W[/tex] such that [itex]\|w_2\| \leq C\|v\|[/itex]

    Now [itex](S+T)(v) = S(v) + T(v)[/itex] by linearity, and hence

    [tex](S+T)(v) = S(v) + T(v) [/tex]
    [tex]\quad = \|w_1\| + \|w_2\|[/tex]
    [tex]\leq B\|v\| + C\|v\|[/tex]
    [tex]= \|B+C\|\|v\|[/tex]
    [tex]= M\|v\|[/tex]

    where [itex]M = B+C \in \mathbb{R}[/itex]. Hence by definition the map [itex]S + T[/itex] is bounded.

    Also

    [tex](S+T)(v) = \|w_1\| + \|w_2\| \leq B\|v\| + C\|v\| = \|B+C\|\|v\| = \left( \|S\|+\|T\|\right)\|v\|[/tex]

    Hence

    [tex]\|S+T\| \leq \|S\| + \|T\|[/tex]
     
    Last edited: Apr 4, 2005
  17. Apr 5, 2005 #16
    Pretty close.

    What doe you mean by

    [tex]S(v) + T(v) = \|w_1\| + \|w_2\|,[/tex]

    though? On one side there are vectors and on the other real numbers!
     
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