I have a problem with the demonstration of this inequality ('m following Royden) :
If $p$ and $q$ are nonnegative numbers such that $\frac{1}{p}+\frac{1}{q}$ and if $f \in L^p$ and $g \in L^q$, then $f\cdot g \in L^1$ and $$\int |fg| \leqslant ||f||_p \cdot ||g||_q$$ Without loss of generality suppose $f,g \geqslant 0$
Set $h(x)=g(x)^{q-1}=g(x)^{p/q}$,since $q-1=q/p$ also $g(x)=h(x)^{q-1}=h(x)^{p/q}$
Thus for $t \in \bf{R}$: $$ptf(x)g(x)= ptf(x)h(x)^{p-1} \leqslant (h(x)+tf(x))^p-h(x)$$ (because of lemma). Hence $$pt \int fg \leqslant \int |h+tf|^p-\int h^p = ||h+tf||^p-||h||^p$$ and $$pt \int fg \leqslant (||h||+t||f||)^p-||h||^p$$ Now is the passage that I do not understand:
Differentiating both sides with respect to t at t=0, we get $$p \int fg \leqslant p ||f|| \cdot ||h||_p^{p-1}=p||f|| \cdot ||g||$$
I do not see why I can derive from both sides of an INEQUALITY..
If one knows the value at one point and has an inequality, one can actually derive an inequality for the derivative as follows:
Let us assume for simplicity that $f : [0,\infty) \to \Bbb{R}$ fulfils $f(x) \leq 0$ for all $x$ and $f(0) = 0$.
Then (if $f$ is (one-sidedly) differentiable at $0$),
$$ f'(0) = \lim_{h \downarrow 0} \frac{f(h) - f(0)}{h} \leq 0, $$
because the denominator is positive and the numerator is $\leq 0$.
In your case, apply this to the difference
$$ f(t) = pt \int fg \,dx - (\Vert h \Vert +t \Vert f \Vert)^p + \Vert h \Vert^p. $$
As seen in the proof above, it is important to note that $f(0) = 0$ and we have $f(t) \leq 0$ for all $t$ by what you have shown.