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Consider the map $ T: L^1 \rightarrow c_0 $ defined by $$ Tf = ( \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t)e^{int} dt)_{n \in \mathbb{N}}.$$ I have shown that this map is continuous and injective, and now I need to show that this map is not surjective. I know that I need to contradict the statement $$ ||f||_{L^1} \leq C || Tf ||_{\infty} $$for all $ f \in L^{1}[-\pi, \pi] $ and $C >0$ by giving a counter example but witl no result. More precisely, I have been trying to show that $$ ||T \frac{D_{m}}{ ||Dm||_{L^1} }||_{\infty} \rightarrow 0$$ as $ m \rightarrow \infty $, where $D_m$ is the mth Dirichlet kernal but have not been successful. In doing so I have tried to show that the Fourier series of the function $ e^{-int} $ converges pointwise at the origin. I thought this would then give me that $ | TD_m | < \infty $ and then the above expression goes to zero as $ m \rightarrow \infty $.

To be honest I'm not even sure my method is correct. Any hint or help would be appreciated! Thanks! $$ $$

Meagain
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Your approach is good: If the map would be surjective, the open map theorem would imply that $\|f\|_1 \leq C \|Tf\|_\infty$ for some $C>0$. In fact, we can use the Dirichlet kernel as a counterexample.

One standard fact about the Dirichlet kernel is that $\|D_m\|_{L^1} \sim \ln(m)$. A proof of this can be found here. Noting that $\|T D_m\|_\infty = 1$ we are already done.

p4sch
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  • Thanks! I in fact was not aware of that the Dirchlet kernel integrate to $ 2\pi $ over $ [-\pi, \pi] $. – Meagain Apr 29 '18 at 11:06