Is $|f(\theta)|$ less than or equal to the sum of the absolute values of $A_n$?

[Dustinsfl]

Supposing $f$ is bounded and $A_n$ is given by 1-8, prove that $\sup_n|A_n|$ is finite.
$$
f(\theta) = \sum_{n = -\infty}^{\infty}A_ne^{in\theta}
$$

Since $f$ is bounded, $|f| < M = |z|\in\mathbb{C}$. Since it could be $\mathbb{C}$, $M$ would be the modulus correct?
We know that the modulus of $e^{in\theta}$ is 1 so $|f| = \sum\limits_{n = \infty}^{\infty}|A_n|$.
How to finish it?

 

[Sudharaka]

Supposing $f$ is bounded and $A_n$ is given by 1-8, prove that $\sup_n|A_n|$ is finite.

Hi dwsmith, :)

I think you haven't written the whole question. What is "1-8" ?

Kind Regards,
Sudharaka.

 

[Dustinsfl]

Hi Jameson, :)

I think you haven't written the whole question. What is "1-8" ?

Kind Regards,
Sudharaka.

Hi Jameson? Have I been promoted?

That is the whole question. 1-8 is the series. I did it like this:
Suppose $\sup_n|A_n|$ is infinite and $f$ is bounded.
Since $f$ is bounded,
$$
|f| = \sum\limits_{n = -\infty}^{\infty}|A_n| < M\in\mathbb{R}
$$
because $|e^{in\theta}| = 1$.
Since $f$ is bounded, $|A_n|\to 0$ but $\sup_n|A_n|$ infinite.
Therefore, we have a contradiction and $\sup_n|A_n|$ must be finite.

 

[Sudharaka]

Hi Jameson? Have I been promoted?

Ha ha, sorry dwsmith, I seem to go nuts here trying to do a couple of things at once. :D

That is the whole question. 1-8 is the series. I did it like this:
Suppose $\sup_n|A_n|$ is infinite and $f$ is bounded.
Since $f$ is bounded,
$$
|f| = \sum\limits_{n = -\infty}^{\infty}|A_n| < M\in\mathbb{R}
$$
because $|e^{in\theta}| = 1$.
Since $f$ is bounded, $|A_n|\to 0$ but $\sup_n|A_n|$ infinite.
Therefore, we have a contradiction and $\sup_n|A_n|$ must be finite.

For the moment I can't think of a way to prove this, but it is not true that,

\[|f(\theta)| = \sum\limits_{n = -\infty}^{\infty}|A_n|\]

It should be,

\[|f(\theta)|=\left|\sum_{n = -\infty}^{\infty}A_ne^{in\theta}\right|=\left|\lim_{m\rightarrow\infty}\sum_{n = -m}^{m}A_ne^{in\theta}\right|\]

Now by >>this<< theorem we can conclude,

\[|f(\theta)|=\lim_{m\rightarrow\infty}\left|\sum_{n = -m}^{m}A_ne^{in\theta}\right|\]

By the triangular inequality,

\[|f(\theta)|\leq\lim_{m\rightarrow\infty}\sum_{n = -m}^{m}\left|A_ne^{in\theta}\right|=\sum_{n = -\infty}^{\infty}\left|A_n\right|\]

Kind Regards,
Sudharaka.

 

[Dustinsfl]

Ha ha, sorry dwsmith, I seem to go nuts here trying to do a couple of things at once. :D
For the moment I can't think of a way to prove this, but it is not true that,

\[|f(\theta)| = \sum\limits_{n = -\infty}^{\infty}|A_n|\]

It should be,

\[|f(\theta)|=\left|\sum_{n = -\infty}^{\infty}A_ne^{in\theta}\right|=\left|\lim_{m\rightarrow\infty}\sum_{n = -m}^{m}A_ne^{in\theta}\right|\]

Now by >>this<< theorem we can conclude,

\[|f(\theta)|=\lim_{m\rightarrow\infty}\left|\sum_{n = -m}^{m}A_ne^{in\theta}\right|\]

By the triangular inequality,

\[|f(\theta)|\leq\lim_{m\rightarrow\infty}\sum_{n = -m}^{m}\left|A_ne^{in\theta}\right|=\sum_{n = -\infty}^{\infty}\left|A_n\right|\]

Kind Regards,
Sudharaka.

$|e^{in\theta}| = 1$ so $\sum|A_ne^{in\theta}| = \sum|A_n|$
since $|z_1z_2| = |z_1||z_2|$.

 

[Sudharaka]

$|e^{in\theta}| = 1$ so $\sum|A_ne^{in\theta}| = \sum|A_n|$
since $|z_1z_2| = |z_1||z_2|$.

Yes that's correct. Hence,

\[|f(\theta)|\leq\sum_{n = -\infty}^{\infty}\left|A_n\right|\]

So to reiterate, you cannot write,

\[|f(\theta)|=\sum_{n = -\infty}^{\infty}\left|A_n\right|\]