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I am trying to show that for all $f \in L^2([0,2\pi])$, we also have $f \in L^1([0,2\pi])$ by Cauchy-Schwarz.

I really couldn't see how Cauchy-Schwarz could be applied here. If I apply it to $(f,f)$, I still would have an $L^2$ norm. Can you give me some hints?

Thank you!

Mars Plastic
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Alvis Nordkovich
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1 Answers1

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Hint: $\int |f(x)|dx = \int 1\cdot|f(x)|dx$. Can you apply Cauchy-Schwarz to this?

Mars Plastic
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  • Thank you very much! I wish I had thought about this. Could you also explain why this fails for $f\in L^2(\mathbb{R})$? (Is it because of the unboundedness of $\mathbb{R}$ since there is a constant $\frac{1}{2\pi}$ in the inner product of $L^2$?) – Alvis Nordkovich Feb 08 '19 at 03:43
  • As you have figured out yourself, the argument fails for functions defined on $\mathbb R$, because the function $f\equiv 1$ is not integrable there. Even though, in order to show that the statement itself actually fails, one has to give an example of a function $f\colon\mathbb R\to\mathbb R$ with $f\in L^2(\mathbb R)$, but $f\notin L^1(\mathbb R)$. If you play around a little with functions like

    $$ f(x) =\begin{cases} 0, & x\in(-\infty,1), \ x^{-\alpha}, & x\in[1,\infty),\end{cases} \qquad \alpha\in\mathbb R,$$

    you will quickly find such a function.

     – Mars Plastic Feb 08 '19 at 14:12