Show that the sequence $f_n(x) = x^n$ converges pointwise but not uniformly on [0,1].

Step 1: Determine Pointwise Limit

To find the pointwise limit of the sequence $$f_n(x) = x^n$$ on the interval $$[0{,}1]$$ , we need to evaluate the limit of $$x^n$$ as $$n$$ approaches infinity for each $$x$$ in the interval. We consider three cases for $$x$$ : $$x=0$$ , $$0 < x < 1$$ , and $$x=1$$ .

Step 2: Calculate Pointwise Limit for Cases

For $$x=0$$ , $$\lim_{n \to \infty} 0^n = 0$$ . For $$0 < x < 1$$ , $$\lim_{n \to \infty} x^n = 0$$ . For $$x=1$$ , $$\lim_{n \to \infty} 1^n = 1$$ . Thus, the pointwise limit function f(x) is $$0$$ for $$0 \le x < 1$$ and $$1$$ for $$x=1$$ .

Step 3: Check for Uniform Convergence

To check for uniform convergence, we need to evaluate $$M_n = \sup_{x \in [0{,}1]} |f_n(x) - f(x)|$$ . If $$\lim_{n \to \infty} M_n = 0$$ , then the convergence is uniform. Otherwise, it is not uniform.

Step 4: Calculate $$M_n$$

For $$0 \le x < 1$$ , $$f(x) = 0$$ , so $$|f_n(x) - f(x)| = |x^n - 0| = x^n$$ . For $$x=1$$ , $$f_n(1) = 1^n = 1$$ and $$f(1) = 1$$ , so $$|f_n(1) - f(1)| = |1 - 1| = 0$$ . Therefore, we need to find the supremum of $$x^n$$ on the interval $$[0{,}1)$$ .

Step 5: Evaluate $$\lim_{n \to \infty} M_n$$

As $$x$$ approaches $$1$$ from the left, $$x^n$$ approaches $$1$$ . Since $$x^n$$ is an increasing function on $$[0{,}1)$$ , the supremum of $$x^n$$ on $$[0{,}1)$$ is $$1$$ . Therefore, $$M_n = 1$$ for all $$n$$ .

Step 6: Conclusion

Since $$\lim_{n \to \infty} M_n = 1 \ne 0$$ , the sequence $$f_n(x) = x^n$$ does not converge uniformly on $$[0{,}1]$$ . However, we have already shown that it converges pointwise to $$f(x) = \begin{cases} 0 & \text{if } 0 \le x < 1 \\ 1 & \text{if } x = 1 \end{cases}$$ .

More from this chapter

• 2. Consider the function f(x) = 3cos(e^(5x-3)) + (4)/(x) Find f'(x) [(d)/(dx)(cos x) = -sin x]
• **प्रश्नावली 5.2** प्रश्न 1 से 8 में x के सापेक्ष निम्नलिखित फलनों का अवकलन कीजिए: 1. sin(x² + 5) 2. cos(sin x) 3. sin(ax + b) 4. sec(tan(√x)) 5. (sin(ax + b))/
• 1. Prove that the function f(x) = 2x² - 3x + 2 is continuous at x = 1. 2. The function f(x) is defined as follows: f(x) = 2x - 3, & x < 2 x - 1, & x ≥ 2 Prove t
• Find dy/dx if y = (ln(x))^(cos(x)).
• Differentiate the following w.r.t. x. (i) cos^-1(sin x) (ii) tan^-1(sin x / (1 + cos x)) (iii) sin^-1(2^(x+1) / (1 + 4^x))
• If x = t² and y = t³, then (d² y)/(dx²) is equal to:
• Find (d y)/(d x), if x=a(θ+sin θ), y=a(1-cos θ).
• Differentiate the following w.r.t. x. (i) cos ^(-1)(sin x) (ii) tan ^(-1)((sin x)/(1+cos x)) (iii) sin ^(-1)((2^(x+1))/(1+4^x))

All Continuity and Differentiability solutions →