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
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    $\begingroup$ 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.$


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