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

To prove that $ w(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) $ of an operator $ A $ is defined as:
\[ w(A) = \sup_{\|x\| = 1} |\langle Ax, x \rangle| \]

**Step 2: Applying the Definition**

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

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

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

**Step 4: Using the Norm**

Next, we know that:
\[ \|A x\|^2 = \langle A x, A x \rangle = \langle A^2 x, x \rangle \]

**Step 5: Summing the Contributions**

Consider the operators $ A^2 $ and $ B^2 $. Using the linearity of the inner product and the properties of the norm, we get:
\[ |\langle A x, x \rangle| \leq \sqrt{\langle A^2 x, x \rangle} \]

**Step 6: Combining Terms**

To incorporate $ B^2 $, let's bound it similarly:
\[ |\langle B x, x \rangle| \leq \sqrt{\langle B^2 x, x \rangle} \]

**Step 7: Adding and Averaging**

We combine the inequalities for $ A $ and $ B $:
\[ |\langle Ax, x \rangle| + |\langle Bx, x \rangle| \leq \sqrt{\langle A^2 x, x \rangle} + \sqrt{\langle B^2 x, x \rangle} \]

**Step 8: Squaring and Averaging**

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

**Step 9: Final Bound**

Thus:
\[ |\langle Ax, x \rangle| \leq \frac{1}{2} \left( \sqrt{\langle A^2 x, x \rangle} + \sqrt{\langle B^2 x, x \rangle} \right) \]
\[ w(A) = \sup_{\|x\| = 1} |\langle Ax, x \rangle| \leq \frac{1}{2} \left\Vert A^2 + B^2 \right\Vert \]

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

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.

Explore a detailed proof involving operator theory and numerical ranges. Learn about the application of Cauchy-Schwarz and arithmetic mean-geometric mean inequalities in proving matrix inequalities. Suitable for advanced undergraduate and graduate students.

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