Question

Suppose that \(p\) is a homogeneous polynomial of degree \(r\) in \(\mathbf{Y}\) and \(p(\mathbf{Y})>0\) for all nonzero \(\mathbf{Y}\) in \(\mathbb{R}^{n}\). Show that there is a \(\rho>0\) such that \(p(\mathbf{Y}) \geq \rho|\mathbf{Y}|^{r}\) for all \(\mathbf{Y}\) in \(\mathbb{R}^{n}\). HINT: \(p\) assumes a minimum on the set \(\\{\mathbf{Y}|| \mathbf{Y} \mid=1\\}\). Use this to establish the inequality in Eqn. (5.4.41).

Short answer

Question: Given a homogeneous polynomial of degree \(r\) in \(\mathbf{Y}\), denoted by \(p(\mathbf{Y})\), with \(p(\mathbf{Y})>0\) for all nonzero \(\mathbf{Y}\) in \(\mathbb{R}^{n}\), show that there exists a positive constant \(\rho > 0\) such that \(p(\mathbf{Y}) \geq \rho|\mathbf{Y}|^{r}\). Answer: To prove the given inequality, first find the minimum value of the polynomial on the set \(\\{\mathbf{Y}|| \mathbf{Y} \mid=1\\}\) and denote it as \(\rho_1\). Then, construct a new vector \(\mathbf{Y}_1\) by dividing \(\mathbf{Y}\) by its magnitude. Since \(p\) is a homogeneous polynomial, we have \(p(\mathbf{Y}) = |\mathbf{Y}|^{r}p(\mathbf{Y}_1)\). Now, using the minimum value of \(p(\mathbf{Y}_1)\), we obtain the inequality \(p(\mathbf{Y}) \geq \rho_1 |\mathbf{Y}|^{r}\), where \(\rho = \rho_1 > 0\).

Step-by-step solution

Compute the minimum of the polynomial on the set

Since \(p(\mathbf{Y})>0\) for all nonzero \(\mathbf{Y}\) in \(\mathbb{R}^{n}\), we know that \(p\) will have a positive minimum on the set \(\\{\mathbf{Y}|| \mathbf{Y} \mid=1\\}\). We denote this minimum value as \(\rho_1 > 0\). This means that: $$p(\mathbf{Y}) \geq \rho_1 \textrm{ for all } \mathbf{Y} \textrm{ such that } |\mathbf{Y}| = 1 $$ Now, we will use this minimum value to show and construct the required inequality.

Use the minimum value to show the inequality

Consider any \(\mathbf{Y}\) in \(\mathbb{R}^{n}\). Let's define a new vector \(\mathbf{Y}_1\) as follows: $$\mathbf{Y}_1 = \frac{\mathbf{Y}}{|\mathbf{Y}|}$$ Note that, for this new vector \(\mathbf{Y}_1\), we have \(|\mathbf{Y}_1| = 1\). Now, since \(p\) is a homogeneous polynomial of degree \(r\), we have: $$p(\mathbf{Y})=|\mathbf{Y}|^{r}p\left( \frac{\mathbf{Y}}{|\mathbf{Y}|}\right)=|\mathbf{Y}|^{r}p(\mathbf{Y}_1)$$ Now, we know that \(p(\mathbf{Y}_1) \geq \rho_1\), from Step 1. Therefore, we have: $$p(\mathbf{Y}) \geq \rho_1 |\mathbf{Y}|^{r}$$ Finally, we have the inequality \(p(\mathbf{Y}) \geq \rho |\mathbf{Y}|^{r}\), with \(\rho = \rho_1 > 0\). This concludes the proof.

Key concepts

Real Analysis

Real Analysis is a branch of mathematical analysis dealing with the set of real numbers and the analytic properties of real-valued sequences and functions. Central to this field is the study of limits, continuity, differentiation, integration, and sequences of functions. When we talk about properties, such as continuity or differentiability for functions with real variables, we're navigating the realm of real analysis.

In the context of finding the minimum of a polynomial, real analysis provides the tools to investigate the behavior of polynomials on the set of real numbers. For instance, the fact that a continuous function achieves a minimum and maximum on a closed and bounded set is a consequence of the extreme value theorem in real analysis. This foundational concept helps us understand why a homogeneous polynomial, like the one given in the exercise, would exhibit a minimum value on the unit sphere in the real vector space \( \mathbb{R}^n \).

Minimum of Polynomial

In a polynomial function's graph, the minimum points are where the value of the function is lower than at any nearby points. To find the exact minimum of a polynomial with real coefficients, one typically sets the derivative of the polynomial equal to zero and solves for the roots to find critical points, then analyzes the second derivative or uses other criteria to identify whether these points are minima.

For the special case of homogeneous polynomials, as seen in the exercise, the minimum will be investigated on the unit sphere because the property of homogeneity allows us to scale the input while proportionally scaling the output by the degree of the polynomial. This scaling property simplifies the search for the minimum to a manageable set. The solution confirms the existence of a positive minimum \( \rho_1 \) by utilizing this strategy, which then extends to the whole space through the homogeneous property of the polynomial.

Homogeneous Polynomial Properties

Homogeneous polynomials have some captivating properties that make them significant in various areas of mathematics. These are special kinds of polynomials that, for any scalar \( \lambda \) and any vector \( \mathbf{Y} \) in \( \mathbb{R}^n \) satisfy the property:

\[ p(\lambda \mathbf{Y}) = \lambda^r p(\mathbf{Y}) \]
where \( r \) is the degree of the polynomial. This means that all terms in the polynomial have the same total degree.

Due to this property, the value of the polynomial over the unit sphere can reveal much about its behavior over the entire space. If \( p \) is positive on the unit sphere, then it will be positive everywhere except at the origin because scaling the input by any non-zero real number scales the output by the corresponding power of that number. This scaling feature is what ultimately facilitates the establishment of an inequality like the one in the exercise, where the polynomial is bounded from below by \( \rho |\mathbf{Y}|^r \) for some \( \rho > 0 \).