Answer to Question #310728 in Real Analysis for Nikhil rawat

Question

Let fn(x)= nx/(1+nx) is not uniformly convergent on [0,1]

Answer 1

Solution

$\%\; MathType!Translator!2!1!LaTeX.tdl!LaTeX\; 2.09\; and\; later!\; \%\; MathType!MTEF!2!1!+-\; \%\; feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn\; \%\; hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr\; \%\; 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4HqGqFfpeea0xe9Lq-Jc9\; \%\; vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x\; \%\; fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaaBa\; \%\; aaleaacaWGUbaabeaakmaabmaabaGaamiEaaGaayjkaiaawMcaaiab\; \%\; g2da9maalaaabaGaamOBaiaadIhaaeaacaaIXaGaey4kaSIaamOBai\; \%\; aadIhaaaGaaCzcaiaaxMaadaWadaqaaiaaicdacaGGSaGaaGymaaGa\; \%\; ay$ ${f_n}\left( x \right) = \frac{{nx}}{{1 + nx}}$ for the interval $[0, 1]$

For $x=0$ ,

$\mathop {\lim }\limits_{n \to \infty } {f_n}\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{nx}}{{1 + nx}}\\ \mathop {\lim }\limits_{n \to \infty } {f_n}\left( 0 \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{n\left( 0 \right)}}{{1 + n\left( 0 \right)}}\\ = 0$

For $0 < x \le 1\$

$\mathop {\lim }\limits_{n \to \infty } {f_n}\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{nx}}{{1 + nx}}\\ \mathop {\lim }\limits_{n \to \infty } {f_n}\left( x \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{nx}}{{n\left( {\frac{1}{n} + x} \right)}}\\ = \mathop {\lim }\limits_{n \to \infty } \frac{x}{{\frac{1}{n} + x}}\\ = 1$

Since we can see that the two limits are not equal, therefore ${f_n}\left( x \right) = \frac{{nx}}{{1 + nx}}$ , is not uniformly convergent over the interval $[0, 1]$ .