I calculted the Fourier series for $f(x)= x^2$ and I get that it is:

$$\frac{ \pi^{2}}{3} + 4 \sum_{m = 1}^{\infty} \frac{(-1)^m}{m^2} \cos (mx).$$

But the rest of the question asks me to show directly (without using the theorem below) that the series converges uniformly.

Let $f$ be a continuous, piecewise differentiable function in ${\rm PC}(2\pi)$, and suppose that $f’\in{\rm PC}(2\pi)$. Then the Fourier series of $f$ converges uniformly on $\Bbb R$.

Could anyone help me in doing so?

My attempt: I think that the first term is a constant so it will not make a problem for uniform convergence, and the second series is less than or equal $\sum_{m=1}^{\infty} 1/m^{2}$ which is a convergent series by $p$-test as $p=2>1$ . . . but why would this lead to uniform convergence?

  • $\begingroup$ $f(x)=x^2$ on what interval? That's not a periodic function naturally; we have to cut it off somewhere and repeat for a Fourier series to make sense. $\endgroup$ – jmerry Feb 12 at 5:49
  • $\begingroup$ Try Weierstrass M-test $\endgroup$ – DragunityMAX Feb 12 at 5:52
  • $\begingroup$ The book did not specify, the author said find the Fourier series of $f(x) = x^2$ only .... so what shall I do in this case? @jmerry $\endgroup$ – hopefully Feb 12 at 5:52
  • $\begingroup$ I have edited my question @DragunityMAX so I can apply the Weierstrass M-test by the series I mentioned above and it will be the $\sum_{m=1}^{\infty} M_{m}$ in this test? $\endgroup$ – hopefully Feb 12 at 6:01
  • 1
    $\begingroup$ Your choice of a series $\sum_m a_m\cos mx$ implies that it's an even function on an interval of length $2\pi$ - so you chose the interval $[-\pi,\pi]$. That's good - because if we didn't choose a symmetric interval, the periodic function we got by repeating it wouldn't be continuous, and its Fourier series wouldn't converge uniformly. $\endgroup$ – jmerry Feb 12 at 6:41

$|\frac{(-1)^m}{m^2} \cos (mx)| \le \frac{1}{m^2} $ for all $m \in \mathbb N$ and all $x \in \mathbb R.$

This shows that the series $\sum_{m = 1}^{\infty} \frac{(-1)^m}{m^2} \cos (mx)$ converges uniformly on $ \mathbb R.$


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.