Hey guys! I've been studying with the problems from my textbook and have been stuck for quite a while now. The problem asks to prove that if H is a real Hilbert space and T linear operator on H such that (Tx,x)>=0 for all x in H, where (-,-) denotes the inner product. I found the same question on Stack Exchange (https://math.stackexchange.com/questions/803293/show-that-t-is-continuous-with-langle-x-tx-rangle-geq-0?noredirect=1&lq=1), where somebody says that the problem is a special case of a more general theorem for Banach spaces (https://math.stackexchange.com/questions/216858/positive-operator-is-bounded?noredirect=1&lq=1) and mentions the Riesz Representation Theorem. I understand the proof for Banach spaces, what I don't get is how one would use the boundedness of A (using the notation of the second question) to conclude that T is bounded. Any hints are appreciated! Thanks!
Hello
Does anyone have some example for a bounded self-adjoint operator or an application of the spectral theorem for bounded self-adjoint operators?
Right now I can only find self-adjoin linear operator on a finite dimensional Hilbert space, self-adjoint compact operators, multiplicate operator and the discrete laplacian operator on l²(Z).
Let A and B be linear operators defined over a Euclidean space such that AB-BA=I where I is the identity operator. Show that both operators cannot be bounded. Pls help, I have tried using triangle inequality but I’m stuck rn
I'm trying to work out a solution to a differential equation whose solutions are the banach space-valued functions on [0,/infty), that is f € C(R+, X). where X is the Banach space of scalar-valued functions. I am trying to prove its weak solution via lax milgram theorem. But to apply lax milgram, I need my vector space to be a Hilbert space. Also, if someone has the book "Banach and Hilbert space of vector-valued functions: their general theory and applications to holomorphy", please post it here, I can't find its free pdf.
Hello,
It is well known that a bounded linear operator A in L^1 (R) or L^2 (R) that commutes with translations is a convolution operator (Au = f*u for a given function f).
I am looking for the most general setting for this result. I am thinking about something like a "continuous" linear mapping in D'(R) (distributions) that commutes with translations, but I cannot find any good references on the subject and I am not sure what continuous means in this context.
Thanks for your help !
Hi everyone, I'm trying to prove that any self-adjoint linear operator A:H->H on a Hilbert space is bounded. I figured the best way would be to show that A is continuous, and therefore bounded. I followed a path of the form:
take x,y close to eachother.
|A(x-y)|^2 = <A(x-y),A(x-y)> = <(x-y),(A^2)(x-y)> <= |x-y||A^2(x-y)|
But this didn't get me anywhere. Any tips would be appreciated
Let T be a linear operator on a Hilbert space H such that
<Tx,y>=i <x,Ty> ∀x,y∈H and i is the imaginary root.
Show that T is bounded
I know that I have to use the Closed Graph Theorem but i'm not sure about the mechanism to do this.
Hey r/learnmath! I have a question on functional analysis! Given is the complex sequence space l^(2)_C. (I will denote it l^(2) from now on). I have to show that a linear, bounded operator A that maps from l^(2) to l^(2) with a finite dimensional range also has an adjoint with a finite dimensional range.
My first thought was this: If I take any sequence a = (a1, a2, a3, ...) from l^(2) and plug it into A (which has finite dimensional range), I get a 'deformed' version of each entry of the sequence: A(a) = (b1, b2, ..., bn, 0, 0, 0,...). If I then work out the requirement for A^(*) (A(a), b) = (a, A^(*)(b)) it is easy to find that the range of the adjoint of A must also be finite dimensional and I am done.
But what if A is the projection of some vector on a infinitely long other vector in l^(2)? Then I still have a finite dimensional range. I think there are more exceptions to the finite operator A just being zeros from some point on, and therefore I would like to give a more general proof. I think there must be some sort of connection between the ranges en nullspaces of A and A^(*) but I can't really grasp it unfortunately. I hope the problem is a bit clear any help would be greatly appreciated!
I only know that ||T|| = ||T* || but this equality could hold for a smaller subspace D(T*) of H. Could someone clarify this? Thanks!
Let P[0,1] be the space of polynomials on [0,1], and suppose we equip the norm ||u|| := |u(0)| + max_[0≤t≤1] |u'(t)|.
Is f(u) = u''(1/2) a bounded or unbounded functional?
Don't really have any intuition. I have tried to construct a counterexample, but to no avail. And I can't see a way for a direct proof . . .
Can I ever use renown to buy an operator again or should I start splashing out on skins or hoarding for limited edition alpha packs?
