1

So I looked up that for something to converge in $L^2$ we must have that

$$ \int_{I} |f_n(x) - f(x)|^2 dx \to 0 \text{ as } n \to \infty $$

With

$$ f(x) = \begin{cases} -1, & x\in [-\pi, 0) \\ 1, & x\in [0, \pi]. \end{cases} $$

And $f_n(x) = \sum_{k \ \text{odd}, \ k>0}^{N} \frac{\sin(kx)}{k}$

But it seems difficult to use the definition here. I have proved that $f_n(x)$ does not converge uniformly and neither pointwise to $f(x)$ is that a result I can use here?

Olba12
  • 2,579

1 Answers1

0

First, note that for $f_n(x)=\sum_{k=1}^n \frac{\sin((2k-1)x)}{2k-1}$, we have

$$\lim_{n\to \infty}f_n(x)=\frac\pi4\text{sgn}(x)$$

for $x\in (-\pi,\pi)$, with $f_n(\pm \pi)=0$. Therefore, we have

$$\lim_{n\to \infty}f_n(x)=f(x)=\begin{cases}\frac\pi4&,0<x< \pi\\\\0&,x=0,\pm \pi\\\\-\frac\pi4&,-\pi<x<0\end{cases}$$

Second, Dirichlet's Test for Uniform Convergence guarantees that the convergence is uniform on every closed interval $[a,b]$ for which either $0<a<b<\pi$ or $-\pi <a<b<0$ since there exists a number $M$ such that $\left|\sum_{k=1}^n \sin((2k-1)x)\right|\le M$ and $\lim_{n\to \infty}\frac{1}{2n-1}\to 0$.

Finally, since $f_n(x)$ is the Fourier Series for the function $f(x)$ as given by

$$f(x)=\begin{cases}\frac\pi4&,0<x< \pi\\\\0&,x=0,\pm \pi\\\\-\frac\pi4&,-\pi<x<0\end{cases}$$

and $f(x)\in \mathscr{L}^2([-\pi,\pi])$, then $||f_n(x)- f(x)||_2^2\to 0$.

Mark Viola
  • 179,405
  • $f_n(0) \to 0$ as $n \to \infty \not = f(0)$ for the f I defined, how can it then converge uniformly? Aswell I do not understand why you change $f(x)$ to something new, that is not the $f(x)$ i was given in my exercise. – Olba12 Nov 07 '16 at 03:27
  • The $f$ that is in the OP does not equal the limit of $f_n(x)$ for any $x$. But with $f$ as I've defined it, $f(0)=0$ and $f_n(0) =0$. – Mark Viola Nov 07 '16 at 03:27
  • The $f$ that you have needs to be multiplied by $\pi/4$. And the $L^2$ convergence does not depend on the value of $f$ at discrete points of measure $0$. – Mark Viola Nov 07 '16 at 03:33
  • Yes, but how does that answer my question? I wanted to know if It converge in $L^2$ for the $f(x)$ that I have defined. – Olba12 Nov 07 '16 at 03:34
  • For the $f$ you have, there is no convergence whatsoever - not pointwise, and therefore not uniform, and not $L^2$. – Mark Viola Nov 07 '16 at 03:35
  • So the conclusion is, it does not converge in $L^2$ since $f_n(x) \to \pi/4 sgn(x)$? – Olba12 Nov 07 '16 at 03:36
  • Yes. The series is the Fourier Series for $(\pi/4)\text{sgn}(x)$ and converges in $L^2$ as such. – Mark Viola Nov 07 '16 at 03:39