Question

Let \(T \in \mathcal{B}(X)\). Define \(T^{n}\) by \(T^{n} x=T\left(T^{n-1} x\right)\), for \(n=2,3, \ldots\) Show that \(\left\|T^{n}\right\| \leq\|T\|^{n}\).

Short answer

Question: Prove that for any bounded linear operator T on a normed space X, and for each positive integer n, the inequality ∥𝑇ⁿ∥≤∥𝑇∥ⁿ is satisfied. Solution: By mathematical induction, we proved that for each positive integer n, the inequality ∥𝑇ⁿ∥≤∥𝑇∥ⁿ holds true for any bounded linear operator T on a normed space X. We first established the base case where the expression was true when n=1. Then, we assumed the statement was true for a positive integer k and used this assumption to prove the inequality for the case n=k+1. Therefore, the inequality is true for all positive integers n.

Step-by-step solution

Base Case

We shall start by noting that the base case, where \(n=1\), is already true. This is because for \(n=1\), the expression becomes: \(\|T^1\| \leq \|T|^1\) Since both sides are equal, the inequality is satisfied for the base case.

Inductive Hypothesis

Next, let's assume that the statement is true for some positive integer \(k\), i.e. we will assume the following: \(\|T^k\| \leq \|T\|^k\) We will use this assumption to prove the inequality for the case where \(n = k + 1\).

Inductive Step

Now, let's prove the inequality for \(n = k + 1\). Observe that: \(\|T^{k+1}\| = \|T(T^k)\|\) Using the properties of norms and operator multiplication, we have: \(\|T(T^k)\| \leq \|T\|\|T^k\|\) Now using the inductive hypothesis, we can write: \(\|T\|\|T^k\| \leq \|T\|\|T\|^k\) By simplification, we get: \(\|T\|\|T\|^k = \|T\|^{k+1}\) Thus, we have proven that: \(\|T^{k+1}\| \leq \|T\|^{k+1}\)

Conclusion

Since our inductive hypothesis assumed the inequality to be true for a positive integer \(k\), and we showed that it is also true for the case \(n = k +1\), we can conclude that the inequality is true for all positive integers \(n\). Therefore, the given statement: \(\|T^n\| \leq \|T\|^n\) is proven to be true by mathematical induction.

Key concepts

Inductive Hypothesis

The inductive hypothesis is a crucial part of mathematical induction.
It involves assuming a statement or proposition is true for a specific case, say for some integer \(k\).
This allows us to use this assumption to prove the statement is true for the next integer, \(k+1\). By doing this, we can establish the truth for all integers starting from a base case. In the context of our problem, the induction process involves assuming that the inequality \(\|T^k\| \leq \|T\|^k\) holds true. This assumption lets us proceed to verify the inequality \(\|T^{k+1}\| \leq \|T\|^{k+1}\).
This step relies heavily on believing the hypothesis for \(k\) holds, allowing a logical bridge to \(k+1\). This method is powerful because once we prove it for \(k+1\), the principle of mathematical induction tells us that it holds for \(k+2\), \(k+3\), and so forth.

Thus, the inductive hypothesis acts like a stepping stone in a domino effect, ensuring each step supports the next.

Operator Multiplication

Operator multiplication refers to the process of applying one linear operator after another.
In mathematical terms, if \(A\) and \(B\) are two operators, then the product \(AB\) represents applying \(B\) first and then \(A\).
The critical aspect of operator multiplication in the context of norms is how it affects the size, or the maximum stretching of vectors in normed spaces.Norms provide a measure of the 'size' of operators, capturing their influence on vector spaces. For any operators \(A\) and \(B\), an important property is that \(\|AB\| \leq \|A\|\|B\|\), meaning the norm of their product is no greater than the product of their norms.
This allows us to control how much the composed operation can "stretch" a vector in space.In our problem, when we consider \(\|T^{k+1}\| = \|T(T^k)\|\), the inequality \(\|T(T^k)\| \leq \|T\|\|T^k\|\) is key. It showcases how by multiplying operators, we can still maintain control over their norm using the norms of the individual operators.

Mathematical Induction

Mathematical induction is a methodical way to prove statements that are claimed to be true for all natural numbers.
It leverages two essential steps: a base case and an inductive step, creating a chain of truth for statements over natural numbers.The base case is your starting point; it's where you establish the statement is true for a particular initial value, often \(n=1\).
In the exercise, confirming \(\|T^1\| \leq \|T\|^1\) sets the basis.The inductive step is where the power of induction shines. Here, you assume the statement is true for \(n=k\), your inductive hypothesis, and then prove it's true for \(n=k+1\). Successfully completing this step is akin to proving all subsequent steps are true.
In our provided induction, after assuming \(\|T^k\| \leq \|T\|^k\), we demonstrate \(\|T^{k+1}\| \leq \|T\|^{k+1}\), thus establishing a continuous proof for all natural numbers.By demonstrating both these steps, mathematical induction grants us a powerful tool to confirm propositions for infinite scenarios.
This method is not only logical but also proves essential in many fields of mathematics and its applications.