0

I need to proove or refute these claims.

  1. if $f ∈ L_2[-π,π]$ is integrable then $f$ is Piecewise Continuous
  2. Fourier transform is well defined for every $f ∈ L_2[-π,π]$
  3. if $f:R\rightarrow F$ is Piecewise smooth on $R$ then $f ∈ L_1[R]$

I think 1 is true but I don't know how to show that. about the rest I am just not sure :( Any help will be great

kimchi lover
  • 24,277
Lidor Hacham
  • 21
  • We need to expand to Plancherel transform to do Fourier transform on L2 functions. Usually it is only defined on Schwarz class of infinitely continously differentiable functions of limited support . https://math.stackexchange.com/questions/1840207/extension-of-fourier-transform-to-l2-mathbbr – mathreadler Jan 01 '21 at 17:05

1 Answers1

0
  1. No. Neither Riemann integrable nor Lebesgue integrable functions need to be piecewise continuous. See the Thomae function.
  1. Yes. $L_2[-\pi,\pi] \subset L_1[-\pi,\pi]$, and the Fourier transform is defined for any element of $L_1$. (Note: $L_2(\mathbb R) \not\subset L_1(\mathbb R)$, so this argument does not work there. As mathreader said, the Plancherel transform extends the Fourier transform to $L_2(\mathbb R)$.)

Proof for $L_2[-\pi,\pi] \subset L_1[-\pi,\pi]$. Let $f \in L_2[-\pi,\pi]$. Note $|f(x)| \le |f(x)|^2 + 1$, which we can see by considering the two cases $|f(x)|\le 1$ and $|f(x)| > 1$. So $$ \int_{-\pi}^{\pi} |f(x)|\;dx \le \int_{-\pi}^{\pi} |f(x)|^2\;dx + 2\pi < +\infty $$ and $f \in L_1[-\pi,\pi]$

  1. No. $f(x) = 2+\sin x$ is smooth on $\mathbb R$, but $f$ is not in $L_1(\mathbb R)$.
GEdgar
  • 111,679
  • thank you! about 3 I managed to get that just after I posted the question..XD about 1, do you have specific example? about 2. how do I show L2[−π,π]⊂L1[−π,π]? thank you for your help!!! – Lidor Hacham Jan 01 '21 at 17:10
  • yea got it thank you both I also found another example for 1 f(x) = x^(-0.5) – Lidor Hacham Jan 01 '21 at 17:58