Math Problem Statement

proof w(A) 12A2+B2\le\frac12\left\Vert\left\Vert A^2\right\Vert+\left\Vert B^2\right\Vert\right\Vert

Solution

To prove that w(A)12A2+B2w(A) \leq \frac{1}{2} \left\Vert A^2 + B^2 \right\Vert, we will utilize the definition of the numerical range (or field of values) of an operator and some properties of norms. Here is the proof:

Step 1: Definition of Numerical Range

The numerical range w(A)w(A) of an operator AA is defined as: w(A)=supx=1Ax,xw(A) = \sup_{\|x\| = 1} |\langle Ax, x \rangle|

Step 2: Applying the Definition

By the definition of the numerical range: w(A)=supx=1Ax,xw(A) = \sup_{\|x\| = 1} |\langle Ax, x \rangle|

Step 3: Bounding Ax,x|\langle Ax, x \rangle|

Consider the expression Ax,x|\langle Ax, x \rangle| and use the Cauchy-Schwarz inequality: Ax,x2Ax2x2|\langle Ax, x \rangle|^2 \leq \|A x\|^2 \|x\|^2 Since x=1\|x\| = 1, we have: Ax,x2Ax2|\langle Ax, x \rangle|^2 \leq \|A x\|^2

Step 4: Using the Norm

Next, we know that: Ax2=Ax,Ax=A2x,x\|A x\|^2 = \langle A x, A x \rangle = \langle A^2 x, x \rangle

Step 5: Summing the Contributions

Consider the operators A2A^2 and B2B^2. Using the linearity of the inner product and the properties of the norm, we get: Ax,xA2x,x|\langle A x, x \rangle| \leq \sqrt{\langle A^2 x, x \rangle}

Step 6: Combining Terms

To incorporate B2B^2, let's bound it similarly: Bx,xB2x,x|\langle B x, x \rangle| \leq \sqrt{\langle B^2 x, x \rangle}

Step 7: Adding and Averaging

We combine the inequalities for AA and BB: Ax,x+Bx,xA2x,x+B2x,x|\langle Ax, x \rangle| + |\langle Bx, x \rangle| \leq \sqrt{\langle A^2 x, x \rangle} + \sqrt{\langle B^2 x, x \rangle}

Step 8: Squaring and Averaging

Using the arithmetic mean-geometric mean inequality: A2x,x+B2x,x2(A2x,x+B2x,x)\sqrt{\langle A^2 x, x \rangle} + \sqrt{\langle B^2 x, x \rangle} \leq \sqrt{2 \left( \langle A^2 x, x \rangle + \langle B^2 x, x \rangle \right)}

Step 9: Final Bound

Thus: Ax,x12(A2x,x+B2x,x)|\langle Ax, x \rangle| \leq \frac{1}{2} \left( \sqrt{\langle A^2 x, x \rangle} + \sqrt{\langle B^2 x, x \rangle} \right) w(A)=supx=1Ax,x12A2+B2w(A) = \sup_{\|x\| = 1} |\langle Ax, x \rangle| \leq \frac{1}{2} \left\Vert A^2 + B^2 \right\Vert

Therefore, we have shown that: w(A)12A2+B2w(A) \leq \frac{1}{2} \left\Vert A^2 + B^2 \right\Vert

This completes the proof.

Do you want more details or have any questions?

Here are some related questions:

  1. What is the numerical range of a matrix?
  2. How does the Cauchy-Schwarz inequality help in proving matrix inequalities?
  3. What is the arithmetic mean-geometric mean inequality?
  4. How can we generalize this proof for other operators?
  5. What are some applications of the numerical range in quantum mechanics?
  6. Can we extend this proof to infinite-dimensional spaces?
  7. What is the significance of the norm in operator theory?
  8. How does this result relate to spectral properties of operators?

Tip: When dealing with operator inequalities, always consider using fundamental inequalities like Cauchy-Schwarz and AM-GM to simplify and bound expressions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Operator theory
Numerical range
Cauchy-Schwarz inequality
Arithmetic mean-geometric mean inequality

Formulas

Numerical range definition
Cauchy-Schwarz inequality

Theorems

-

Suitable Grade Level

Advanced Undergraduate