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)?

  • 1
    $\begingroup$ Is that exponent $\;n-\frac xn\;$ or is it $\;\frac{n-x}n\;$ ? $\endgroup$ – DonAntonio Feb 13 '18 at 10:56

For any $\;x\in (0,1)\;$ we have

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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.