Find the norm of linear functional $$f(x) = \int_{-1}^1 sx(s)\,ds,\quad x\in L_1[-1,1]$$
Firstly I try to show boundedness and so I need to find any $M>0$ such that $|f(x)|\le M\|x\|_1$. As shown below, I used Hölder's inequality for norm in $L^1$; I should find right side as $L^1$ but it was $L^2$ norm. How can I obtain $L^1$ norm on the right?
$f$ is linear and bounded functional.
$$f(x) = \int_{-1}^1 s x(s) ds, \quad x \in L_1[-1,1]$$
$$\begin{align} |f(x)| & = \left| \int_{-1}^1 s x(s) ds \right| \\ & \le \int_{-1}^1 |s x(s)| ds \\ & \le \sqrt{\int_{-1}^1 |s|^2 ds} \cdot \sqrt{\int_{-1}^1 |x(s)|^2 ds} \\ & = \left. \frac{|s|^3}{3} \right|_{-1}^1 \sqrt{\int_{-1}^1 |x(s)|^2 ds} \\ & = \underbrace{\left( \frac13 - \frac13 \right)}_{= 0} \cdot \underbrace{\sqrt{\int_{-1}^1 |x(s)|^2 ds}}_{= ?} \end{align}$$
Since $|s| \leq 1$ for $s \in [-1,1]$, one could use the inequalities $$ \big|\,f(x)\,\big| \leq \int_{-1}^{1} \big|\,s x(s)\,\big| ds \leq \int_{-1}^{1} \big|\,x(s)\,\big| ds = \big|\big|\,x\,\big|\big|_{L^1},$$ which show that $f$ is a bounded linear functional on $L^1[-1,1]$ of norm at most $1$.
To show $||f||$ is exactly $1$ we use consider the $L^1[-1,1]$ function
$$x(s) = \begin{cases} 1, & \text{if $0\leq x\leq1$} \\ -1, & \text{if $-1\leq x<0$}, \end{cases}$$ which shows $||f||\geq \left|\int_{-1}^1sx(s)ds\right|=\left|\int_{-1}^1 |s|ds\right|=1$, making $||f||=1$.