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I have prove the pointwise convergence in $(0,1)$ to the null function. For the uniformly convergence I can calculate the derivative $f'_n(x)$?

If I prove $f_n$ uniformly converge, $\lim_{n\rightarrow+\infty}\int_{0}^{1} f_n(x) dx=\int_{0}^{1}\lim_{n\rightarrow+\infty} f_n(x) dx$?

$\forall n, \forall x \in (0,1) , x^{(n-\frac{x}{n})}>x^n$ so can I say Sup $x^{(n-\frac{x}{n})}$>Sup $x^n=1$ so there isn't uniformly convergence in (0,1)?

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    $\begingroup$ Is that exponent $\;n-\frac xn\;$ or is it $\;\frac{n-x}n\;$ ? $\endgroup$ – DonAntonio Feb 13 '18 at 10:56
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For any $\;x\in (0,1)\;$ we have

$$x^{n-\frac xn}=\frac{x^n}{\left(x^{1/n}\right)^x}\xrightarrow[n\to\infty]{}\frac0{1}=0$$

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  • $\begingroup$ Your first equality sign does not hold. $\endgroup$ – Christian Blatter Feb 13 '18 at 16:44
  • $\begingroup$ @ChristianBlatter True, thanks. Editing. $\endgroup$ – DonAntonio Feb 13 '18 at 21:25
  • $\begingroup$ @DonAntonio and what do you want to show with it concerning uniform convergence? $\endgroup$ – Gono Feb 14 '18 at 9:40
  • $\begingroup$ @Gono So far, nothing: you already did something (didn't check, just read). What do you want to do concerning pointwise convergence to zero? $\endgroup$ – DonAntonio Feb 14 '18 at 12:18
  • $\begingroup$ Should I do something? The OP already stated "I have prove the pointwise convergence in (0,1) to the null function." So nothing to be done… $\endgroup$ – Gono Feb 14 '18 at 14:34
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Take $$x_n = \left(1-\frac{1}{n}\right) \in (0,1)$$

Then $$f_n(x_n) \to \frac{1}{e}$$ hence $$||f_n||_\infty \ge f_n(x_n) \to \frac{1}{e}$$so the convergence is not uniform.

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  • $\begingroup$ If there isn't uniformly convergence can I calculate two limits? $\endgroup$ – Giulia B. Feb 14 '18 at 9:56
  • $\begingroup$ two limits for what? A limit is always unique… so I don't get what you mean. $\endgroup$ – Gono Feb 14 '18 at 14:34
  • $\begingroup$ The two limits in the initial question $\endgroup$ – Giulia B. Feb 14 '18 at 14:47
  • $\begingroup$ OFC you can calculate both limits… both are $0$. For the RHS you already had $$\int_{0}^{1}\lim_{n\rightarrow+\infty} f_n(x) dx = \int_0^1 0 dx = 0$$ For the LHS consider that hold $$0 \le \int_{0}^{1} f_n(x) dx \le \int_0^1 x^{n - \frac{1}{n}} dx = \frac{n}{n^2 + n - 1}$$ Taking limits leads to $$\lim_{n\to\infty} \int_{0}^{1} f_n(x) dx = 0$$ So LHS and RHS are both $0$ hence equal. $\endgroup$ – Gono Feb 14 '18 at 15:00

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