$L_{\infty}$ and simple function Suppose $f \in L_{1}[a,b]$ we have a $c >0$ such that for any simple function $\phi$ on $[a,b]$:  
$$\left|\int_{[a,b]}\phi f\right| \le c \cdot \|\phi\|_1$$
Show that $f \in L_\infty[a,b]$ and $\|f\|_\infty \le c$
My attempt:
We can find an increasing sequence of simple fucntions $(\phi_n)_{n=1}^\infty$ such that $\phi_n \to |f|$. Thus $\|\phi_n\|_\infty \to \|f\|_\infty$. After that, I want to find another sequence of simple function $(\psi_n)_{n=1}^\infty$ such that $\|\psi_n\|_1  = 1$ and $\int_{[a,b]}|\psi_n \phi_n| = \|\phi_n\|_{n=1}^\infty$, then using hypothesis I can solve this question. But how can I construct my $(\psi_n)_{n=1}^\infty$? I will appreciate any help.
 A: You already started at a good point! Using sequence of simple functions is powerful. 
Firstly we follow you and get an increasing sequence of non-negative simple functions $(\phi_n)_{n=1}^{\infty}$, $\phi_n\rightarrow |f|$. Thus $\|\phi_n\|_{\infty} \rightarrow \|f\|_{\infty}$. 
Note that for any simple function $\phi_n$, $\|\phi_n\|_{\infty} = M_n<\infty$, thus for any $\varepsilon>0$, there exists $A_n$, such that $\mu(A_n) = a_n>0$ and $|\phi_n(x)|>M_n-\varepsilon,\ \forall x\in A_n$. Let $\psi_n = \frac{1}{a_n}1_{A_n}(x)$, then $\|\psi_n\|_1 = 1$ and $\psi_n$ is a simple function. Furthermore, we have
$$\int_{[a,b]}\phi_n \psi_n = \frac{1}{a_n}\int_{A_n}\phi_n>M_n-\varepsilon.$$
By the monotony of $\phi_n$, for any $m>n$, we have 
$$\int_{[a,b]}\phi_m \psi_n=\frac{1}{a_n}\int_{A_n}\phi_m>M_n-\varepsilon,$$ 
since $\phi_m(x)\geq\phi_n(x)>M_n-\varepsilon,\ \forall x\in A_n$. 
Then by dominant convergence theorem, keep $n$ fixed and let $m\rightarrow \infty$ (note that $f\in L_1[a,b]$), we have
$$c= c\|\psi_n\|\geq|\int_{[a,b]}\psi_n f|\leftarrow |\int_{[a,b]}\psi_n \phi_m| >\|\phi_n\|_{\infty}-\varepsilon$$
Then let $n\rightarrow \infty$ we get $\|f\|_{\infty}<c + \epsilon$. Since $\varepsilon$ is arbitrary, we get $\|f\|_{\infty}\leq c$.
Hopefully this time there is no mistakes haha @_@.
