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For $2\leq p < \infty$, if we consider $f,g \in L_p(X, \mathcal{M},\mu)$ there is the well-known equality

$$\frac{d}{dt}\Vert f+tg \Vert_p^p = \frac{p}{2} \int_X \vert f(x)+tg(x) \vert^{p-2} \left( 2t \vert g(x) \vert ^2 + f(x)\overline{g(x)} + g(x)\overline{f(x)} \right) d\mu(x),$$

which can be evaluated at $t = 0$ to get the Gateaux derivative of $\Vert \cdot \vert_p^p.$ I'm trying to prove this equality and the trick seems to be to move the differentiation inside. I'm trying to use Lebesgue dominated convergence, and I get stuck trying to bound the integrand of

$$ \frac{\Vert f+(t+h)g \Vert_p^p - \Vert f+tg \Vert_p^p}{h} = \int_X \frac{\vert f(x)+(t+h)g(x) \vert^p - \vert f(x)+tg(x) \vert^p}{h} d\mu(x). $$

The only thing that seems to be available is convexity of $\vert x \vert^p$, but I haven't been able to use it to obtain a useful bound.

Philip Hoskins
  • 2,095

1 Answers1

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You can bound the difference quotient by applying the mean value theorem to the numerator.

But I suggest taking a different approach from the beginning: integrate the right-hand side instead of differentiating the left-hand side. For any finite interval $[a,b]$ the function $$F(x,t)=\frac p2 \vert f(x)+tg(x) \vert^{p-2} \left( 2t \vert g(x) \vert ^2 + f(x)\overline{g(x)} + g(x)\overline{f(x)} \right)$$ is integrable on $X\times [a,b]$ since the triangle inequality implies $$|F(x,t)|\le C(\max(|a|,|b|)+1)^p\max(|f(x)|^p,|g(x)|^p)$$ By Fubini's theorem $$ \int_0^t \int_X F(x,s)\,dx\,ds = \int_X \int_0^t F(x,s)\,ds\,dx = \int_X |f(x)+tg(x)|^p\,dx $$ This simultaneously demonstrates the differentiability of the function $t\mapsto \int_X |f(x)+tg(x)|^p\,dx$ and that its derivative is $\int_X F(x,s)\,dx$.