Question

Let \(x \in C[0,1] .\) Let \(B_{n} x\) denote the \(n^{\text {th }}\) Bernstein polynomial associated with \(x\) (see chapter \(\mathbf{5})\). Show that the sequence \(\left\\{B_{n}\right\\}\) is bounded, when \(C[0,1]\) has the supremum norm.

Short answer

Question: Prove that the sequence of Bernstein polynomials associated with a continuous function on the interval [0,1] is bounded when equipped with the supremum norm. Answer: We have shown that the supremum norm of the \(n^{\text{th}}\) Bernstein polynomial \(B_nx\) is bounded by a constant multiple of the supremum norm of the function \(x\): \(\lVert B_nx \rVert_\infty \leq 2^n \lVert x \rVert_\infty\). This implies that the sequence of Bernstein polynomials \(\{B_n\}\) is bounded when equipped with the supremum norm.

Step-by-step solution

Define Bernstein polynomials

The \(n^{\text{th}}\) Bernstein polynomial associated with a function \(x\) on the interval \([0,1]\) is defined as follows: $$ B_nx(t) = \sum_{k=0}^{n} \binom{n}{k} x\left(\frac{k}{n}\right) t^k (1-t)^{n-k} $$ for \(t \in [0,1]\).

Define the supremum norm

The supremum norm for a function \(f\) in the function space \(C[0,1]\) is defined as: $$ \lVert f \rVert_\infty = \sup_{t\in[0,1]}|f(t)| $$

Bounding the supremum norm of the \(n^{\text{th}}\) Bernstein polynomial

Now we will find an upper bound for the supremum norm of the \(n^{\text{th}}\) Bernstein polynomial \(B_nx(t)\): $$ \begin{aligned} \lVert B_nx \rVert_\infty &= \sup_{t\in[0,1]} \left|\sum_{k=0}^{n} \binom{n}{k} x\left(\frac{k}{n}\right) t^k (1-t)^{n-k}\right| \\ &\leq \sup_{t\in[0,1]} \sum_{k=0}^{n} \binom{n}{k} \left|x\left(\frac{k}{n}\right)\right| \left|t^k (1-t)^{n-k}\right|\\ \end{aligned} $$ Since \(0\leq t\leq 1\) and \(0\leq 1-t\leq 1\), we have \(0\leq t^k(1-t)^{n-k} \leq 1\). Thus, $$ \begin{aligned} \lVert B_nx \rVert_\infty &\leq \sup_{t\in[0,1]} \sum_{k=0}^{n} \binom{n}{k} \left|x\left(\frac{k}{n}\right)\right| \\ &\leq \sum_{k=0}^{n} \binom{n}{k} \sup_{t\in[0,1]} \left|x(t)\right| \\ &= (2^n) \lVert x \rVert_\infty. \end{aligned} $$ We have found that \(\lVert B_nx \rVert_\infty \leq 2^n \lVert x \rVert_\infty\). Since the right-hand side of the inequality does not depend on \(t\), we see that the supremum norm of the \(n^{\text{th}}\) Bernstein polynomial is bounded by a constant multiple of the supremum norm of the function \(x\), which implies that the sequence \(\{B_n\}\) is bounded.

Key concepts

Supremum Norm

When studying Bernstein polynomials or other mathematical functions, understanding the supremum norm is essential. The supremum norm, often denoted \( \| f \|_\infty \), of a function \( f \) in the space of continuous functions \( C[0,1] \) is calculated by finding the maximum absolute value that \( f \) assumes on the interval \( [0,1] \). This corresponds to the greatest 'peak' in magnitude, no matter whether it lies above or below the x-axis.

Using the supremum norm, it becomes possible to measure the 'size' of functions in a uniform manner across the interval. In the context of Bernstein polynomials, we can use this measure to determine how 'big' the polynomial gets without pinpointing the exact location where the function reaches its peak. This norm helps us to work out various properties of the functions such as uniform convergence and boundedness in a theoretically robust way.

Suppose that \( x(t) \) is a continuous function on \( [0,1] \) and \( \| x(t) \|_\infty \) is its supremum norm. By taking the supremum of the absolute value across the interval, we ensure that no matter how \( x(t) \) may behave in between, we have a fixed upper limit, making the supremum norm particularly useful for analyzing bounded sequences of functions.

Bounded Sequence

In mathematics, a sequence is considered bounded if there exists a real number \( M \) such that every term in the sequence is less than or equal to \( M \) in absolute value. In the scenario of Bernstein polynomials used to approximate continuous functions, demonstrating that the sequence of polynomials is bounded means showing that there exists a supremum norm that serves as an upper limit for all polynomials in the sequence.

For a given continuous function \( x \) over the interval \( [0,1] \) and its associated Bernstein polynomials \( B_nx \), proving boundedness involves showing that \( \| B_nx \|_\infty \) does not grow beyond a certain threshold as \( n \) increases. In practical terms, this suggests that as we generate more Bernstein polynomials for \( x \) by increasing \( n \) (which could be viewed as increasing the 'resolution' of our approximation), we are guaranteed that these polynomials won't have peaks shooting off to infinity. The grooviness of the polynomials is contained, making analysis and convergence behavior predictable and manageable.

Continuous Functions

Continuous functions are the bread and butter of calculus and mathematical analysis. Speaking informally, a function is continuous if you can draw it without lifting your pencil from the paper. More formally, a function \( f \) on an interval is continuous if, for every point in that interval, the limit of \( f \) as it approaches the point is equal to the function's value at that point.

Regarding Bernstein polynomials, these polynomials approximate continuous functions on the interval \( [0,1] \) as \( n \) becomes large. This offers a powerful tool in numerical analysis because continuous functions cover a vast spectrum of practical applications, from describing the trajectory of a moving object to the change in stock market prices over time.

Continuity ensures that Bernstein polynomials, which themselves are continuous, closely follow the function they are approximating without abrupt jumps or breaks. This aligns with the Weierstrass approximation theorem, which states that every continuous function defined on a closed interval can be uniformly approximated as closely as desired by a polynomial function, a role perfectly played by Bernstein polynomials.