Question

Let \(\Pi\) be the space of all polynomials on \([0,1]\) with the supremum norm. For \(p \in \Pi\) let \(D p\) denote the polynomial which is the derivative of \(p\). Show that \(D\) defines a linear mapping on \(\Pi\), and that \(D\) is discontinuous.

Short answer

Short answer: The derivative operator, \(D\), defines a linear mapping on the space of all polynomials on the interval \([0,1]\) since it preserves both the addition and scalar multiplication operations. However, the mapping is discontinuous because there exists a sequence of polynomials \((p_n)\) such that \(p_n \rightarrow 0\) but \(Dp_n\) does not converge to \(D0 = 0\). Specifically, we can consider the sequence of polynomials \(p_n(x) = \frac{x^n}{n}\) for \(x \in [0, 1]\), which converges to 0, but its derivatives \(Dp_n = x^{n-1}\) do not converge to 0 since their supremum norm remains constant at 1 for all values of \(n\).

Step-by-step solution

Prove that \(D\) is a linear mapping

Let \(p, q \in \Pi\) be arbitrary polynomials and let \(\alpha\) be an arbitrary scalar. We need to show that \(D\) preserves the addition and scalar multiplication operations. First, we will show that \(D\) preserves addition: $$ D(p + q) = \frac{d(p+q)}{dx} = \frac{dp}{dx} + \frac{dq}{dx} = Dp + Dq $$ Next, we will show that \(D\) preserves scalar multiplication: $$ D(\alpha p) = \frac{d(\alpha p)}{dx} = \alpha \frac{dp}{dx} = \alpha Dp $$ Since \(D\) preserves both addition and scalar multiplication, it defines a linear mapping on \(\Pi\).

Prove discontinuity

To show that \(D\) is discontinuous, we need to find a sequence of polynomials \((p_n)\) in \(\Pi\) such that \(p_n \rightarrow p\) but \(Dp_n\) does not converge to \(Dp\). Consider the sequence of polynomials \((p_n)\) defined by \(p_n(x) = \frac{x^n}{n}\) for \(x \in [0, 1]\). Notice that \(p_n \rightarrow 0\) as \(n \rightarrow \infty\) with respect to the supremum norm, since \(p_n(x) \leq \frac{1}{n}\) for any \(x \in [0, 1]\). Now, let's find the derivatives of \(p_n\): $$ Dp_n = \frac{d}{dx} \left( \frac{x^n}{n} \right) = \frac{nx^{n-1}}{n} = x^{n-1} $$ Now, consider the supremum norm of \(Dp_n\) for every \(n \in \mathbb{N}\): $$ \| Dp_n \|_\infty = \sup_{x \in [0, 1]} |x^{n-1}| $$ We can easily see that \(\| Dp_n \|_\infty\) does not converge to 0, because the supremum is 1 for all values of \(n\). Since \(p_n \rightarrow 0\) and \(Dp_n\) does not converge to \(D0 = 0\), we conclude that \(D\) is discontinuous.

Key concepts

Polynomial Derivatives

Polynomial derivatives represent the change in a polynomial function. To find the derivative of a polynomial, we apply basic calculus rules. For a polynomial \( p(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \), the derivative \( Dp(x) \) is obtained by differentiating each term individually. This means:

• Constant terms, such as \( a_0 \), disappear since they don't change.
• For each term \( a_i x^i \), the derivative is \( i a_i x^{i-1} \).

Given these rules, polynomial differentiation is a straightforward process involving multiplying coefficients by their corresponding exponents and decreasing the power of \( x \) by one. This concise method allows us to understand how polynomial functions change as their input varies.
In the context of the exercise, we examined how the derivative operation acts linearly across polynomial spaces. Specifically, the derivative of a sum of polynomials or a scaled polynomial results in simple operations on the derivatives themselves. This consistent behavior is key in establishing the linearity of differentiation.

Supremum Norm

The supremum norm, or infinity norm, measures the "largest" value a function can achieve over a specified interval. For function \( f \) defined over \([0,1]\), the supremum norm is expressed as \( ||f||_\infty = \sup_{x \in [0,1]} |f(x)| \). This provides the maximum absolute value attained by \( f \) on that interval.
Using this concept, one can evaluate how the output of a function varies across its entire domain. Although this measure is especially useful for ensuring certain types of continuity, it can also highlight limits to continuity in operator functions, as seen in the exercise.

• The concept helps quantify how close sequences of functions come to zero or any other function.
• It helps in testing continuity of operators such as in polynomial differentiation.

Evaluating \( Dp_n \) using the supremum norm proved that although \( p_n(x) \) approaches zero, \( Dp_n(x) \) does not due to its unchanging supremum. This demonstrates properties of the supremum norm in continuity analysis.

Discontinuity of Operators

Operators map functions from one space to another. Their continuity is crucial for the stability and reliability of operations like differentiation. An operator is continuous if small changes in input lead to small changes in output.
In our exercise, we showed discontinuity in the differentiation operator \( D \). We examined polynomials \( p_n(x) = \frac{x^n}{n} \) converging to zero, yet their derivatives \( Dp_n(x) = x^{n-1} \) maintain a constant supremum norm.

• This indicates that although the original functions \( p_n(x) \) get closer to zero, the derivatives do not.
• The differentiation operator doesn't satisfy the condition that function sequences and their derivatives converge together.

This example clarifies how the differentiation operator is discontinuous under the supremum norm. Recognizing such discontinuities aids in understanding where particular mathematical methods may fall short in practical applications.