Showing something is increasing I am working with this 
$h(p)=( \int _0 ^1|f(x)|^pdx)^{1/p}$ 
where $1 \leq p < \infty$ and I am trying to show h is increasing.
Attempt 
$\operatorname{sng}(|f|)=1$
so $(\int_0 ^1 \operatorname{sng}(|f|)^q)^{1/q}=1$ thus we can consider
$$((\int_0 ^1 \operatorname{sng}(|f|)^q)^{1/q}= \int _0 ^1|f(x)|^p \, dx)^{1/p} \geq \int_0 ^1 |f(x) \operatorname{sqn}(f)|=\|f\|_1 $$
Basically using the holders inequality does this look good?
 A: Let $1 \leq p < q < \infty$.  The function $a(x):= x^{q/p}$ defined on non-negative reals is convex (it has non-negative second derivative). 
By Jensen's inequality, 
$$ \left( \int_0^1 |f|^p dx \right)^{q/p}= a \left( \int_0^1 |f|^p dx \right)  \leq \int_0^1 a(|f|^p)dx = \int_0^1 |f|^q dx.$$
Take $q$th roots to see that 
$$h(p) \leq h(q)$$
as required.
A: In fact, you can use the Holder Inequality to prove:
$$\int_0^1|f(x)g(x)|dx\le\left(\int_0^1|f(x)|^\alpha dx\right)^{\frac{1}{\alpha}}\left(\int_0^1|g(x)|^\beta dx\right)^{\frac{1}{\beta}}$$
where $\alpha,\beta>1$ satisfy $\frac{1}{\alpha}+\frac{1}{\beta}=1$.
For $p_1<p_2$, choose $\alpha=\frac{p_2}{p_1}$ and then
\begin{eqnarray}
&&\left(\int_0^1|f(x)|^{p_1}dx\right)^{\frac{1}{p_1}}\\
&\le&\left[\left(\int_0^1(|f(x)|^{p_1})^\alpha dx\right)^{\frac{1}{\alpha}}\left(\int_0^11^\beta dx\right)^{\frac{1}{\beta}}\right]^{\frac{1}{p_1}}\\
&\le&\left[\int_0^1(|f(x)|^{p_1\alpha} dx\right]^{\frac{1}{\alpha p_1}}\\
&=&\left(\int_0^1|f(x)|^{p_2}dx\right)^{\frac{1}{p_2}}
\end{eqnarray}
Thus $\left(\int_0^1|f(x)|^{p}dx\right)^{\frac{1}{p}}$ is increasing for $p>1$.
