Find the norm of the Bounded Linear Functional The whole problem is showing $L$ is a bounded linear functional and to find its norm. I am stuck at the end. I need to find the norm of the bounded linear functional $L$ on $L^2(\mathbb{T})$ where
$$Lf = \sum_{k \in \mathbb{Z}} \frac{e^{ik}}{2^{|k|}} \hat{f}(k)$$
I could find a bound. Let $f : ||f|| = 1$
$$|Lf|^2 = \left| \sum_{k \in \mathbb{Z}} \frac{e^{ik}}{2^{|k|}}\hat{f}(k) 
\right|^2 = \sum_{k \in \mathbb{Z}} \left| \frac{e^{ik}}{2^{|k|}}\hat{f}(k) \right|^2 = \sum_{k \in \mathbb{Z}} \frac{1}{2^{2|k|}}\left|\hat{f}(k) \right|^2$$
by the Pythagorean theorem in inner-product spaces.
Now, by Parseval's identity $||f||_{L^2} = 1 = ||\hat{f}||_{\ell^2} = \sum_{k \in \mathbb{Z}} \left| \hat{f}(k) \right|^2$ since $f \in L^2(\mathbb{T})$
This means that $\forall k \Rightarrow|\hat{f}(k)|^2 \leq 1$
Putting this together
$$|Lf|^2 = \sum_{k \in \mathbb{Z}} \frac{1}{2^{2|k|}} \left| \hat{f}(k) \right|^2 \leq \sum_{k \in \mathbb{Z}} \frac{1}{2^{2|k|}} = \frac{5}{3}$$
Thus
$$||L|| = \sup_{||f|| = 1} |Lf| \leq \sqrt{5/3}$$
Nevertheless, I can't find the exact value.
 A: For any $\ell^{2}(\mathbb Z)$ sequence $(a_k)$ there is a function $f \in L^{2}(T)$ with $\hat {f} (k)=a_k$  for  all $k$. Hence there exists $f \in L^{2}(T)$ with $\hat {f} (k)=e^{-ik}\frac 1 {2^{|k|}}$. By definition of operator norm $\|L\| \geq \frac {|Lf|} {\|f\|}$. Can you now finish the proof of the fact that $\|L\|=\sqrt {\frac  5 3}$?
A: I could achieve an answer thanks to the help from Kabo Murphy in his answer. I will post it for completeness.
$L$ obeys $\frac{|Lf|}{||L||} \leq ||L||$.
Also, since for any $\ell^2(\mathbb{Z})$ sequence $(a_k)$ there is a function $f \in L^2(\mathbb{T})$ with $\hat{f}(k) = a_k$ for all $k$. Therefore, there exists such a function $f$ where $\hat{f}(k)=e^{-ik} \frac{1}{2^{|k|}}$.
In this case,
$$Lf = \sum_{k \in \mathbb{Z}} \frac{1}{2^{2|k|}} = \frac{5}{3} = |Lf|$$
Also, by Parseval's identity:
$$||f(t)||_{L^2} = ||\hat{f}(k)||_{\ell^2} = \left[ \sum_{k \in \mathbb{Z}} \left| \frac{1}{2^{|k|}} \right|^2 \right]^{1/2} = \left[ \sum_{k \in \mathbb{Z}} \frac{1}{2^{2|k|}} \right]^{1/2} = \sqrt{\frac{5}{3}}$$
Thus,
$$\frac{|Lf|}{||f||} = \frac{5/3}{\sqrt{5/3}} = \sqrt{5/3} \leq ||L|| \leq \sqrt{5/3}$$
This last inequality from what I wrote above in the question.
Thus,
$$||L|| = \sqrt{\frac{5}{3}}$$
