Which one of the given sequences of functions is uniformly convergent $?$

Question

Which one of the given sequences of functions is uniformly convergent $?$ $$A.\ \ f_n(x)=x^n;x\in[0,1].$$ $$B.\ \ f_n(x)=1-x^n;x\in\left[{1\over2},1\right].$$ $$C.\ \ f_n(x)={{1}\over{1+nx^2}};x\in\left[0,{1\over 2}\right].$$ $$D.\ \ f_n(x)={{1}\over{1+nx^2}};x\in\left[{1\over2},1\right].$$

I think option $D.$ is correct . For if we take $\lim_{n\rightarrow \infty}f_n=f$ then for

$A$ $f=1$ at $1$ and $0$ elsewhere.

$B$ $f=0$ at $1$ and $1$ elsewhere.

$C$ $f=1$ at $1$ and $0$ elsewhere.

Did I got things right $?$

Thank you.

Answer 1

You are right that the only possible case is D. Then $$ \frac{1}{1+nx^2}\geq\frac{1}{1+(n+1)x^2} $$ and you can use Dini's theorem.

Answer 2

Noting that the uniform limit of continuous functions must be continuous too, only $D$ is possible (as your (correct) calculations have shown).

For the reason why D is continuous on the give interval, you can use Dini's theorem as is indicated in the previous answer. Or more directly, note that $$|f_n(x)-f_m(x)|=\frac1{(1+mx^2)(1+nx^2)}|m-n|x^2\le\frac1{16mn}|m-n|\le \frac1{16mn}(m+n)\le \frac1{16}(\frac1m+\frac1n)$$ Then use Cauchy's criterion for uniform convergence.