# sup of fourier coefficients

#### dwsmith

##### Well-known member
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

##### Well-known member
MHB Math Helper
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.

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#### dwsmith

##### Well-known member
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

##### Well-known member
MHB Math Helper
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.

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.

#### dwsmith

##### Well-known member
Ha ha, sorry dwsmith, I seem to go nuts here trying to do a couple of things at once.

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

##### Well-known member
MHB Math Helper
$|e^{in\theta}| = 1$ 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,

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