Today’s problem appears as Problem 5 on the Texas A&M August 2017 Real Analysis Qual:
Prove that for
Solution: The expressions appear to resemble Fourier series, so interpreting them as such, the statements reduce to showing that for 1-periodic functions
and
Now, recall that the Fourier transform
is a bounded operator
and similarly, its inverse
is a bounded operator
with




![Rendered by QuickLaTeX.com \|a\|_{l^p} \leq \|a\|_{l^{p'}}, \|A\|_{L^{p'}([0,1])} \leq \|A\|_{L^p([0,1])}](https://www.stepanmalkov.com/wp-content/ql-cache/quicklatex.com-7bbeff7a2b8502933e4c66d03e48a771_l3.png)

![Rendered by QuickLaTeX.com l^p \hookrightarrow l^{p'}, L^{p'}([0,1]) \hookrightarrow L^{p}([0,1])](https://www.stepanmalkov.com/wp-content/ql-cache/quicklatex.com-edeab5904d9d4980f882c41d9a7e5bf3_l3.png)
![Rendered by QuickLaTeX.com \mathcal{F}: L^{p'}([0,1]) \hookrightarrow L^p([0,1]) \to l^{p'}, 2 \leq p' \leq \infty](https://www.stepanmalkov.com/wp-content/ql-cache/quicklatex.com-f0ed68ce5b338d50a1383da4f3c5d729_l3.png)
![Rendered by QuickLaTeX.com \mathcal{F}^{-1}: l^p \to L^{p'}([0,1]) \hookrightarrow L^{p}([0,1]), 1 \leq p \leq 2](https://www.stepanmalkov.com/wp-content/ql-cache/quicklatex.com-35b991bf203cc91952837e369350bd3e_l3.png)