Let $1\leq p<\infty$, $1/p+1/q=1$. Let $f$ be a measurable function on $[0,1]$ such that for all step functions $g$ on $[0,1]$ $$ \left|\int_0^1 fg d\mu\right|\leq \|g\|_q. $$ Prove that $\|f\|_p\leq 1$.
I have tried following, but failed:
Let $F_f: L^q[0,1]\to\mathbb{R}$, $\displaystyle F_f(g)=\int_0^1 fg $. If for all $g\in L^q[0,1]$ $$ \left|\int_0^1 fg d\mu\right|\leq \|g\|_q, $$ then $$\|f\|_p=\|F_f\|=\sup_{\|g\|_q\neq 0}\frac{|F_f(g)|}{\|g\|_q}\leq 1.$$ Hence we only need to prove for all $g\in L^q[0,1]$ $$ \left|\int_0^1 fg d\mu\right|\leq \|g\|_q. $$ Let $\{\phi_n\}$ be a sequence of step functions such that $\lim\limits_{n\to\infty}\|\phi_n-g\|_q=0$. I want to prove $\displaystyle\lim_{n\to\infty}\int_0^1 f\phi_n d\mu=\int_0^1 fg d\mu$. Then the wanted inequality of $g$ would be obtained.
If $f$ is in $L^p[0,1]$, then by Holder inequality, $$ \left|\int_0^1 f(\phi_n-g) d\mu\right|\leq \|f\|_p\|\phi_n-g\|_q\to \|f\|_p\cdot 0=0. $$ Thus we obtain the proof.
Hence we only need to prove $f\in L^p[0,1]$. How should I continue? Thanks.
One thing you can do is to consider for $N \in \Bbb{N}$ the "truncated" function $\tilde{f} := f \cdot \chi_{|f| \leq N}$.
As you are on a measure space with finite measure, it is then clear that $\tilde{f} \in L^p$. Furthermore, your assumption also holds for $\tilde{f}$, because of
$$ \bigg|\int \tilde{f} \cdot g \,dx \bigg| = \bigg|\int f \cdot (g \cdot \chi_{|f| \leq N}) \, dx \bigg| \leq \Vert g \cdot \chi_{|f| \leq N}\Vert_q \leq \Vert g \Vert_q. $$
Also note that $g \cdot \chi_{|f| \leq N}$ is a step function, if $g$ is.
You can then conclude $\Vert \tilde{f} \Vert_p \leq 1$ and use the monotone convergence theorem to conclude the proof.