Show that the functional $F$ on $L_p(-1,1)$ given by $ F(f) = \int_{-1}^0 f(t)dt - \int_{0}^1 f(t)dt $ is continuous. I have to show that the functional $F$: $$ F(f) = \int_{-1}^0 f(t)dt - \int_{0}^1 f(t)dt  $$
on the space $L_p(-1, 1), p\geqslant 1 $, is continuous.
Solution:
I need to get an inequality $$ \vert F(f) \vert \le M \left(\int_{-1}^1 \big\vert f(t)\big \vert ^p dt\right)^{1/p}.$$
Here $M$ is a constant and $M>0$. 
I started:
$$\big\vert F(f) \big\vert=\left\vert \int_{-1}^0 f(t)dt - \int_{0}^1 f(t)dt \right\vert .$$
And here I stopped. Next I have to evaluate these two integrals by new $\int_{-1}^1 g(t)dt$ function and here I have a problem with this. The next step is to use the Hölder's inequality to get the inequality I need to show (with $M>0$).
Question:
How to find $\int_{-1}^1 g(t)dt$?
 A: Using the Triangle Inequality, we have
$$\big|F(f)\big|\leq \int_{-1}^{+1}\,\big|f(x)\big|\,\text{d}x\text{ for each }f\in L^p\big((-1,+1)\big)\,.$$
By Hölder's Inequality,
$$\begin{align}\int_{-1}^{+1}\,\big|f(x)\big|\,\text{d}x&=\int_{-1}^{+1}\,\big|f(x)\big|\cdot 1\,\text{d}x\\&\leq \left(\int_{-1}^{+1}\,\big|f(x)\big|^p\,\text{d}x\right)^{\frac{1}{p}}\,\left(\int_{-1}^{+1}\,1^q\,\text{d}x\right)^{\frac{1}{q}}\text{ for each }f\in L^p\big((-1,+1)\big)\,,\end{align}$$
where $q\in[1,\infty]$ is such that $\dfrac1p+\dfrac1q=1$.  This shows that
$$\big|F(f)\big|\leq \|f\|_p\,2^{\frac{1}{q}}=2^{1-\frac1p}\,\|f\|_p\text{ for each }f\in L^p\big((-1,+1)\big)\,,$$
or
$$\|F\|_{L^p_\text{op}}\leq 2^{1-\frac1p}\,.$$
By taking $f:(-1,+1)\to\mathbb{C}$ to be the step function
$$f(x)=\begin{cases}-1&\text{if }-1<x<0\,,\\+1&\text{if }0\leq x <+1\,,\end{cases}$$ 
we can see that the operator norm $\|F\|_{L^p_\text{op}}$ of $F$ does indeed equal $2^{1-\frac1p}$ (thus, $F$ is a bounded linear functional, whence continuous).  With a little tweak, this result is also true for $p=\infty$, not just $p\in[1,\infty)$.
