I have to find out if the functions $f: \mathbb{R} \to \mathbb{R}$ converge uniformly on $M$
$$\text{(a) }f_n(x) := \frac{1}{1+n^2x^2}, M = [0, \infty)$$ $$\text{(b) }f_n(x) := \frac{1}{1+n^2x^2}, M = (0, \infty)$$ $$\text{(c) }f_n(x) := \frac{1}{1+n^2x^2}, M = [\delta, \infty) \text{ for } \delta > 0$$ $$\text{(d) }f_n(x) := \sum_{k=0}^n \frac{(x \sin x)^k}{k!}, M = [0, 1]$$
Regarding $\text{(a)}$:
- For $x = 0$ we get a constant function, it converges towards $1$.
- For $x != 0$ the denominator converges towards $\infty$, so the function converges towards $0$
- Since the function is continuous at $0$ (because $M = [0, \infty))$, it converges uniformly
Regarding $\text{(b)}$:
- For $x = 0$ we get a constant function, it converges towards $1$.
- For $x != 0$ the denominator converges towards $\infty$, so the function converges towards $0$
- Since the function is not continuous at $0$ (because $M = (0, \infty)) $, it does not converge uniformly
Regarding $\text{(c)}$:
- Same reasoning as $\text{(b)}$. --> Not converging uniformly
Regarding $\text{(d)}$:
I think that it converges uniformly (because the sine function's derivative has absolute value of at most one to see that $|\sin(x) - \sin(y)| \le |x - y|.$ But I don't think that explanation works here.
Can someone tell me if these are correct/incorrect?
a) Τhe function does not converge uniformly since its a sequence of continuous function and its pointwise limit is discontinuous
b)$f_n \to 0$ pointwise and $f_n(\frac{1}{n^2})=\frac{1}{2}$ which does not go to zero. So the convergence is not uniform.
c) whe have that $\sup_{x \in [\delta,\infty)}|f_n(x)| \leq \frac{1}{1+n^2\delta^2} \to 0$ so it converges uniformly.
d) Use the Weierstrass $M-$Test to show that the series of functions converges uniformly.