proof w(AB) $$ \le\frac12\left\Vert\left\Vert A^2\right\Vert+\left\Vert B^2\right\Vert\right\Vert $$

To prove that $ w(AB) \leq \frac{1}{2} \left\Vert A^2 \right\Vert + \frac{1}{2} \left\Vert 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 and inequalities.

**Step 1: Definition of Numerical Range**

The numerical range $ w(AB) $ of an operator $ AB $ is defined as:
\[ w(AB) = \sup_{\|x\| = 1} |\langle ABx, x \rangle| \]

**Step 2: Applying the Definition**

By the definition of the numerical range:
\[ w(AB) = \sup_{\|x\| = 1} |\langle ABx, x \rangle| \]

**Step 3: Decomposing the Inner Product**

Consider the expression $ |\langle ABx, x \rangle| $:
\[ |\langle ABx, x \rangle| = |\langle Bx, A^*x \rangle| \]
Here, $ A^* $ is the adjoint of $ A $.

**Step 4: Using the Cauchy-Schwarz Inequality**

Apply the Cauchy-Schwarz inequality to the inner product:
\[ |\langle Bx, A^*x \rangle| \leq \|Bx\| \|A^*x\| \]

**Step 5: Bounding the Norms**

Next, we use the sub-multiplicative property of the norm:
\[ \|Bx\| \leq \|B\| \|x\| \]
\[ \|A^*x\| \leq \|A^*\| \|x\| \]

Since $\|x\| = 1$, these simplify to:
\[ \|Bx\| \leq \|B\| \]
\[ \|A^*x\| \leq \|A^*\| \]

**Step 6: Relationship Between Norms of $ A $ and $ A^* $**

For any operator $ A $, $\|A^*\| = \|A\|$, so we have:
\[ |\langle ABx, x \rangle| \leq \|B\| \|A\| \]

**Step 7: Combining the Inequalities**

Since we want to incorporate $ A^2 $ and $ B^2 $, we use the fact that:
\[ \|A\| \leq \sqrt{\|A^2\|} \]
\[ \|B\| \leq \sqrt{\|B^2\|} \]

Thus:
\[ \|A\| \|B\| \leq \sqrt{\|A^2\|} \sqrt{\|B^2\|} \]

**Step 8: Using the Arithmetic Mean-Geometric Mean Inequality**

Using the arithmetic mean-geometric mean (AM-GM) inequality:
\[ \sqrt{\|A^2\|} \sqrt{\|B^2\|} \leq \frac{1}{2} \left( \|A^2\| + \|B^2\| \right) \]

**Step 9: Final Bound**

Putting it all together:
\[ w(AB) \leq \sup_{\|x\| = 1} |\langle ABx, x \rangle| \leq \|A\| \|B\| \leq \sqrt{\|A^2\|} \sqrt{\|B^2\|} \leq \frac{1}{2} \left( \|A^2\| + \|B^2\| \right) \]

Therefore, we have shown that:
\[ w(AB) \leq \frac{1}{2} \left( \|A^2\| + \|B^2\| \right) \]

This completes the proof.

Do you want more details or have any questions?

Here are some related questions:

1. What is the adjoint of an operator, and how is it used in numerical range problems?
2. How does the Cauchy-Schwarz inequality help in proving operator inequalities?
3. What is the significance of the sub-multiplicative property of norms?
4. Can we apply this result to non-square matrices?
5. What is the relationship between the norm of an operator and its adjoint?
6. How does the arithmetic mean-geometric mean inequality help in bounding expressions?
7. Are there other inequalities that can be used to bound numerical ranges?
8. What are some practical applications of numerical range inequalities in quantum mechanics or control theory?

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

Explore a detailed proof of the operator inequality $ w(AB) \leq \frac{1}{2} \left\Vert A^2 \right\Vert + \frac{1}{2} \left\Vert B^2 \right\Vert $, utilizing concepts from numerical range, norms, and inequalities. This problem covers advanced topics in operator theory and is suitable for advanced undergraduate and graduate students.

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