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PostPosted: Tue, 22 Nov 2011 17:10:02 UTC 
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Say H is a separable Hilbert space and S\in H^* and T=T^*\in C(H,H) is strictly positive.

Then what is the minimal value for F(x)=\langle Tx,x\rangle - STx?

I had this on a recent exam and I just loved the solution so much, I thought I would share it. Enjoy.

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PostPosted: Tue, 22 Nov 2011 17:59:59 UTC 
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Shadow wrote:
Say H is a separable Hilbert space and S\in H^* and T=T^*\in C(H,H) is strictly positive.

Then what is the minimal value for F(x)=\langle Tx,x\rangle - STx?

I had this on a recent exam and I just loved the solution so much, I thought I would share it. Enjoy.


C(H,H) means B(H), the set of bounded linear operators on H? I don't think you need the separability of H.

Spoiler:
Since S\in H', by Riesz there exists a unique y\in H such that Sz=\langle z,y\rangle for all z\in H. So F(x)=\langle Tx,x-y\rangle. Now since T is positive self-adjoint, we get a unique positive self-adjoint square-root R, and F(x)=\langle Rx,R(x-y)\rangle=\lVert R(x-\frac{1}{2}y)\rVert^2-\lVert R(\frac{1}{2}y)\rVert^2. So the minimum value is -\frac{1}{4}\langle Ry,Ry\rangle=-\frac{1}{4}\langle Ty,y\rangle=-\frac{1}{4}STy=-\frac{1}{4}STS^*(1), where S^*\colon\mathbb{F}\to H is the adjoint to S\colon H\to\mathbb{F}.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Tue, 22 Nov 2011 19:29:08 UTC 
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outermeasure wrote:
Shadow wrote:
Say H is a separable Hilbert space and S\in H^* and T=T^*\in C(H,H) is strictly positive.

Then what is the minimal value for F(x)=\langle Tx,x\rangle - STx?

I had this on a recent exam and I just loved the solution so much, I thought I would share it. Enjoy.


C(H,H) means B(H), the set of bounded linear operators on H? I don't think you need the separability of H.

Spoiler:
Since S\in H', by Riesz there exists a unique y\in H such that Sz=\langle z,y\rangle for all z\in H. So F(x)=\langle Tx,x-y\rangle. Now since T is positive self-adjoint, we get a unique positive self-adjoint square-root R, and F(x)=\langle Rx,R(x-y)\rangle=\lVert R(x-\frac{1}{2}y)\rVert^2-\lVert R(\frac{1}{2}y)\rVert^2. So the minimum value is -\frac{1}{4}\langle Ry,Ry\rangle=-\frac{1}{4}\langle Ty,y\rangle=-\frac{1}{4}STy=-\frac{1}{4}STS^*(1), where S^*\colon\mathbb{F}\to H is the adjoint to S\colon H\to\mathbb{F}.


C(H,H) means compact operators, and though your solution works, there's a particularly cute one when you use that T is compact.

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PostPosted: Wed, 23 Nov 2011 05:28:56 UTC 
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Shadow wrote:
C(H,H) means compact operators, and though your solution works, there's a particularly cute one when you use that T is compact.


Oh, you mean \mathcal{K}(H) (I think the K comes from German "Kompakter").

It isn't one of those "true for finite rank, so true for all compact" proof, is it?

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Wed, 23 Nov 2011 06:00:34 UTC 
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outermeasure wrote:
Shadow wrote:
C(H,H) means compact operators, and though your solution works, there's a particularly cute one when you use that T is compact.


Oh, you mean \mathcal{K}(H), the Calkin algebra on H (I think the K comes from German "Kompakter").

It isn't one of those "true for finite rank, so true for all compact" proof, is it?


No, MUCH cuter.

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PostPosted: Thu, 24 Nov 2011 19:35:58 UTC 
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Here's a hint for those that want one:

Hint 1:
Spoiler:
Try using Hilbert Schmidt


Hint 2:
Spoiler:
By "cute" I mean once you see it in the right light, you can do it using more or less 8th grade information

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PostPosted: Sun, 4 Dec 2011 19:52:57 UTC 
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And it's been long enough, I'll post the solution, because it's so darn cute.

Spoiler:
By the properties listed for T, and since H is separable, we have an orthonormal basis of eigenvectors, $\{u_i\}_{i=1}^\infty with associated positive eigenvalues \lambda_i,\; i=1,2,3,\ldots. Now by Riesz, Sx=\langle x,y\rangle for some fixed y\in H.

So we use linearity of \langle\cdot,\cdot\rangle and orthonormality of the basis, to write

$F(x)=\sum_{i=1}^\infty (\lambda_ix_i^2-\lambda_ix_iy_i)

Now the cute part is that each of these expressions is an upward opening parabola, so the minimum is achieved -- by the quadratic formula -- at $x_i=-{(-\lambda_iy_i)\over 2\lambda_i}={y_i\over 2}, so the minimum occurs at x={1\over 2}y.

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PostPosted: Sun, 4 Dec 2011 20:41:50 UTC 
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A fun follow-up question I think:

Why is it important that we have STx and not just Sx?

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PostPosted: Sun, 4 Dec 2011 21:46:58 UTC 
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Shadow wrote:
And it's been long enough, I'll post the solution, because it's so darn cute.

Spoiler:
By the properties listed for T, and since H is separable, we have an orthonormal basis of eigenvectors, $\{u_i\}_{i=1}^\infty with associated positive eigenvalues \lambda_i,\; i=1,2,3,\ldots. Now by Riesz, Sx=\langle x,y\rangle for some fixed y\in H.

So we use linearity of \langle\cdot,\cdot\rangle and orthonormality of the basis, to write

$F(x)=\sum_{i=1}^\infty (\lambda_ix_i^2-\lambda_ix_iy_i)

Now the cute part is that each of these expressions is an upward opening parabola, so the minimum is achieved -- by the quadratic formula -- at $x_i=-{(-\lambda_iy_i)\over 2\lambda_i}={y_i\over 2}, so the minimum occurs at x={1\over 2}y.


I was thinking of how to avoid Riesz and to come up with a natural S^* appearing one, but couldn't find any.

Shadow wrote:
A fun follow-up question I think:

Why is it important that we have STx and not just Sx?


Spoiler:
Because you don't get a minimum that way. For example, T=\mathop{\mathrm{diag}}(1,2^{-1},3^{-1},4^{-1},\dots), and y=(1,2^{-1},3^{-1},4^{-1},\dots). The resulting supposed minimum would need n^{-1}x_n^2-n^{-1}x_n minimum, i.e. x_n=\frac{1}{2} for all n, and that x is not in \ell^2 (and the infimum would be -\infty).

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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