$\|f\|\geqslant \|u\|$ functional on ${L_p}_{[a,b]}$ 
Consider the following functional $f(x)=\int_\limits{a}^{b}x(t)u(t)dt$, where $x(t)\in {L_p}_{[a,b]}$ and $u(t)\in {L_q}_{[a,b]}$ The space ${L_q}_{[a,b]}$ is the conjugate of ${L_p}_{[a,b]}$ which implies $\frac{1}{p}+\frac{1}{q}=1$.

It easy to prove $\|f\|\leqslant\|u\|$
However I am not understanding the proof of $$\|f\|\geqslant \|u\|$$

Notes: First it is defined $$u_n(t)=\begin{cases}
|u(t)|, & \mbox{ if }|u(t)|\leqslant n \\
0, & \mbox{ if }|u(t)|> n\end{cases}$$
$$x_n(t)=|u_n(t)|^{q-1}\operatorname{sign}\: u(t)$$
$$\|x_n(t)\|_{L_q}=\left(\int_\limits{a}^{b}|u(t)|^{p(q-1)}dt\right)^{\frac{1}{p}}=\left(\int_\limits{a}^{b}|u(t)|^{q}dt\right)^{\frac{1}{p}}$$
Then,
$$|f(x_n)|\leqslant \|f\|\|x_n\|=\|f\|\left(\int_\limits{a}^{b}|u(t)|^{q}dt\right)^{\frac{1}{p}}$$
$$f(x_n)=\int_\limits{a}^{b}|u(t)|^{q-1}\operatorname{sign}\: u(t)dt=\int_\limits{a}^{b}|u(t)|^{q}dt=|f(x_n)|\leqslant\|f\|\|x_n\|=\|f\|\left(\int_\limits{a}^{b}|u(t)|^{q}dt\right)^{\frac{1}{p}}$$
Therefore:$$\|f\|\geqslant \|u\|$$

Question:
Could someone explain me how it was proven $\|f\|\geqslant \|u\|$? I am having a hard time seeing the inequality in the proof.
Thanks in advance!
 A: I'm assuming that $L^p([a,b])$ and $L^q([a,b])$ are real spaces and that $p, q > 1$ are conjugated exponents.
First notice that $x_n \in L^p([a,b])$ because
$$\int_{[a,b]} |x_n(t)|^p\,dt = \int_{\{|u| \le n\}} |u(t)|^{p(q-1)}\,dt \le \int_{\{|u| \le n\}} n^q\,dt \le n^p(b-a)$$
where $\{|u| \le n\} = \{t \in [a,b] : |u(t)| \le n\}$.
We have
$$\|x_n\|_p = \left(\int_{[a,b]} |x(t)|^p\,dt\right)^\frac1p = \left(\int_{\{|u| \le n\}} |u(t)|^{p(q-1)}\,dt\right)^\frac1p = \left(\int_{\{|u| \le n\}} |u(t)|^{q}\,dt\right)^\frac1p \le \left(\int_{[a,b]} |u(t)|^{q}\,dt\right)^\frac1p = \|u\|_q^{\frac{q}p}$$
On the other hand,
$$f(x_n) = \int_{[a,b]} x_n(t)u(t)\,dt = \int_{\{|u| \le n\}} |u(t)|^{q-1} \operatorname{sgn} u(t) \cdot u(t)\,dt = \int_{\{|u| \le n\}} |u(t)|^q\,dt$$
So we have
$$\|f\| \ge \frac{|f(x_n)|}{\|x_n\|_p} \ge \frac{1}{\|u\|_q^{\frac{q}p}} \int_{\{|u| \le n\}} |u(t)|^q\,dt \xrightarrow{n\to\infty} \frac{1}{\|u\|_q^{\frac{q}p}} \int_{[a,b]} |u(t)|^q\,dt = \frac{\|u\|_q^q}{\|u\|_q^{\frac{q}p}} = \|u\|_q^{q - \frac{q}p} = \|u\|_q$$
Now, I'm not sure why they defined $x_n$ like that. It seems that simply considering $x = |u|^{q-1} \operatorname{sgn} u$ would work just the same. We then have $\|x\|_p = \|u\|^{\frac{q}p}_q$ and $f(x) = \|u\|_q^q$ so $$\|f\| \ge \frac{|f(x)|}{\|x\|_p} = \|u\|_q$$
