Apply dominated convergence theorem to show differentiability Let $f,g \in L^p(\mu), 1 < p < \infty$. Show that the function
$$\phi(t)=\int |f+tg|^pd\mu$$ be differentiable at $t=0$ and find $\phi'(0)$.

My try, 
$\psi(t)=|f+tg|^p$ is differentiable at $t=0$, if $p>1$. So
$$\lim_{t\to 0}\frac{|f+tg|^p-|f|^p}{t}=h(x)$$
I also showed 
$$t\in(0,1),\frac{|f+tg|^p-|f|^p}{t}\leq |f+g|^p-|f|^p$$
$$t\in(-1,0),\frac{|f+tg|^p-|f|^p}{-t}\leq |f-g|^p-|f|^p$$
Then I think I need to use dominated convergence theorem to show
$$\lim_{t\to 0}\int\frac{|f+tg|^p-|f|^p}{t}d\mu=\int h(x)d\mu$$
But what's a suitable dominator? I think I only found an integrable upper bound for the function, not the absolute value of the function.
 A: Set
$$F(t) := |f+t \cdot g|^p.$$
Then, by the chain rule,
$$\left| \frac{d}{dt} F(t) \right| = p |f+t \cdot g|^{p-1} \cdot |g| \leq p 2^{p} \big( |f|^{p-1} + |g|^{p-1}\big) \cdot |g| =: h$$
for any $|t| \leq 1$. (Hint for the inequality: Use $|a+b|^{q} \leq 2^q \max\{|a|^q,|b|^q) \leq 2^{q+1} (|a|^q+|b|^q)$ for $q=p-1$.) Consequently, by the mean value theorem,
$$\begin{align*} \left| \frac{|f+tg|^p-|f|^p}{t} \right| &= \left| \frac{F(t)-F(0)}{t} \right|  \leq \sup_{s \in [0,t]} |F'(s)| \leq h \end{align*}$$
for any $ |t|\leq 1$. Finally note that by Hölder's inequality $h$ is integrable. This means that we have found an integrable dominating function.
A: Note that, when $p>1$, the $p$-power function ($x\to x^p$) over positive real numbers is convex. All inequalities below are due to this observation.
Note that:
$$|f|^p-|f-g|^p\leq  |f+g|^p-|f|^p.\tag{1}$$
Moreover, for all $t\in(-1,1)$, $t\not=0$:
$$|f|^p-|f-g|^p\leq\frac{|f+tg|^p-|f|^p}{t}\leq |f+g|^p-|f|^p.\tag{2}$$
Indeed, for $s\in (0,1)$, $p>1$:
$$ |f+sg|^p = |(1-s)f+s(f+g)|^p \leq (1-s)|f|^p +s|f+g|^p,\tag{3}$$
and
$$ |f-sg|^p = |(1-s)f+s(f-g)|^p \leq (1-s)|f|^p +s|f-g|^p.\tag{4}$$
Set $s$ to $t$ when $t$ is positive, and to $-t$ when $t$ is negative, to cover all cases you need. 
Edit1: For (1), we have:
$$ |h+k|^p\leq 2^{p-1}(|h|^p+|k|^p).$$
Indeeed, letting $\theta$ denote the $p$-power function, and using its convexity (and triangle inequality for first step) we have:
$$ |1/2h+1/2k|^p\leq (1/2|h|+1/2|k|)^p=\theta(1/2|h|+1/2|k|)$$ $$\leq 1/2\theta(|h|) +1/2\theta(|k|)=1/2|h|^p+1/2|k|^p.$$
Then we take $h:=f-g$, $k:=f+g$, to obtain (1).
Edit2:
Inequality (3) is obtained using the same technique:
$$ |(1-s)f+s(f-g)|^p\leq ((1-s)|f|+s|f-g|)^p=\theta((1-s)|f|+s|f-g|)$$ $$\leq (1-s)\theta(|f|) +s\theta(|f-g|)=(1-s)|f|^p+s|f-g|^p.$$
Edit3:
Back to (1), replacing $g$ with $tg$ we get:
$$ |f|^p-|f-tg|^p\leq  |f+tg|^p-|f|^p.$$
Now, replacing $g$ by $-g$ in our right inequality (proven for $t\in (0,1)$) we have:
$$ \frac{|f-tg|^p-|f|^p}{t}\leq |f-g|^p-|f|^p$$
equivalent to
$$ |f|^p - |f-g|^p \leq \frac{|f|^p-|f-tg|^p}{t} $$
which is further 
$$ |f|^p - |f-g|^p \leq \frac{|f|^p-|f-tg|^p}{t} \leq \frac{|f+tg|^p-|f|^p}{t}, $$
our left inequality for $t\in (0,1)$.
A: Hint: $\big||f+tg|^p-|f|^p\big| = p|tg|\cdot|\xi|^{p-1} $, where (for any $x$) $\min\{f+tg,f\}\leq \xi \leq \max \{f+tg,f\}$. Hence (wlog assume $t<1$, since you are really interested in $t\to 0$) $|\xi|\leq |f|+|g|$, i.e.
$$\big||f+tg|^p-|f|^p\big| = p|tg|\cdot|\xi|^{p-1}\leq p t |g|(|f|+|g|)^{p-1}\leq pt|g|\cdot \max \{2, 2^{p-2}\}(|f|^{p-1}+|g|^{p-1})$$
and Holder to control $\int |f|^{p-1}|g|$...
