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$$g_n(x)=\frac{\frac{x}{n\:}}{1+\frac{x^2}{n^2}\:}$$

I have to show that it converges uniformly in $[0,\infty)$, I already got that it converges pointwise to $\frac{1}{2}$ but I am just stuck on uniformly proving it.

I have tried using Weierstrass M-Test and Abel's Uniform Convergence Test, but I always reach something not logical.

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  • $\begingroup$ You should revise your calculations regarding pointwise convergence. Note that for example, $g_n(0) = 0 \ (\forall n)$ so we do not have pointwise convergence to $\frac{1}{2}$. $\endgroup$ – qualcuno May 27 '18 at 12:04
  • $\begingroup$ $g_n(\infty)=0$, so something is fishy here $\endgroup$ – tired May 27 '18 at 12:05
  • $\begingroup$ To clarify . . . converges as $n\to\infty$ or as $x\to\infty$? $\endgroup$ – gen-z ready to perish May 27 '18 at 12:10
  • $\begingroup$ I used the inequality of $a^2 + b^2 \ge 2ab$ and so it all got canceled neatly, can you pls explain why it is wrong? $\endgroup$ – David May 27 '18 at 12:32
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$g_n$ does not converge uniformly. It converges pointwise to 0 and $g_n(n)=\frac 1 2$ so it does not converge uniformly.

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  • $\begingroup$ Thank for the answer, but now I have hit a wall in the last part of the question. we are given another function in the beginning of the question that is: $h_n(x) = \frac{xn}{1+x^2n^2}$ and it goes like this. define a new function $k_n(x) = g_n(x) * h_n(x)$, does it uniformly converge? I tried multiple algebraical ways to solve and simplify it but it just wont work. $\endgroup$ – David May 27 '18 at 18:16
  • $\begingroup$ @David what does * stand for? (Convolution or just the product?). $\endgroup$ – Kavi Rama Murthy May 28 '18 at 5:36
  • $\begingroup$ the product of both functions, probably should have left out the * $\endgroup$ – David May 28 '18 at 8:10
  • $\begingroup$ @David You can easily see that $|h_n(x)| \leq \frac 1 {n^{2}}$ so $h_n \to 0$ uniformly. $\endgroup$ – Kavi Rama Murthy May 28 '18 at 9:00
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    $\begingroup$ @David Sorry, there was a typo in the last comment. I meant $|k_n(x)| \leq \frac 1 {n^{2}}$. So $k_n \to 0$ uniformly. $\endgroup$ – Kavi Rama Murthy May 28 '18 at 9:50

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