Proof that dual of $L^p$ is $L^q$

Question

Im working through a proof which shows that for $\frac{1}{p}+\frac{1}{q}$ the map $\Phi:L^q(\Omega;\mathbb{K})\to (L^p(\Omega;\mathbb{K}))^*$, $g\to \Phi(g)[f]:=\int_{\Omega}\overline{g}f\space d\mu$ is an antilinear isometric isomorphism. But I don't understand two important steps when it comes to showing that $\Phi$ is surjective. Since there is a lot of notation and definitions involved I will post a photo of the relevant parts of the proof.

My first question is: Since pointwise we can have $|f_o(x)-Pf_0(x)|=0$, why is $g_T$ as defined below well defined?

My second question is: Why is $F(t)$ differentiable? And assuming it is and assuming we can swap differentiation and integration, how do I get the derivative that is shown below?

Answer 1

For the first part, the map $g$ is well defined because, for all $x$ : $$|g(x)| = \lambda|f_0(x)-Pf_0(x)|^{p-1} $$ And since $p>1$, the above quantity goes to zero when $f_0(x)-Pf_0(x) = 0$ ($p-1>0$ so there is no division by zero occuring).

Edit : Let $y\equiv f_0(x)-Pf_0(x)$, and consider the map $\tilde g:y\mapsto \lambda |y|^{p-2}y\in\mathbb C$. Clearly, $\tilde g$ is continuous and well defined everywhere, except at $y=0$ where we might run into some problems (since $p-2$ might be negative, in which case $|y|^{p-2}$ diverges as $y\to 0$). So let's see what happens when $y$ goes to zero. The modulus of $\tilde g(y)$ is given by : $$|\tilde g(y)| = |\lambda |y|^{p-2}y| = |\lambda|\cdot|y|^{p-2}\cdot|y| =|\lambda|\cdot|y|^{p-1} $$ Now note that by assumption, $p>1$, hence $p-1>0$ from which it follows that $$\lim_{y\to0}|\tilde g(y)| = \lim_{y\to0}\ |\lambda|\cdot|y|^{p-1} = 0 $$ Hence $\tilde g$ is well-defined at $y=0$ and, at that point, is equal to $\tilde g(0) = 0$. You can now deduce from this argument that $g(x) = \tilde g\left(f_0(x)-Pf_0(x)\right)$ is also well defined everywhere, and takes value $0$ when $f_0(x)-Pf_0(x) = 0$.

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Now let's deal with the second part. For notational convenience, I will denote $f_0-Pf_0\equiv f'$ in the following.

A sufficient condition for the differentiability of the map $$F:t\mapsto\int_\Omega |f' +th|^pd\mu $$ Is given by Leibniz integral rule. In the linked article, we are given the three conditions that ensure the existence of $F'$.

• First, $|f' +th|^p$ is integrable by definition of $L^p$ (and the fact that it is a vector space), so the first condition is fulfilled.
• Second, we need to check that $t\mapsto|f' +th|^p$ is continuously differentiable. But it clearly is as a composition of $t\mapsto | f'+ th|$ and $t\mapsto t^p$ which are both continuously differentiable. Furthermore, we can compute the derivative of this function with the chain rule $$\begin{align}\frac{\partial}{\partial t} |f'+ th|^p&=p\cdot |f'+ th|^{p-1}\frac{\partial}{\partial t}|f'+ th|\\ &=p\cdot |f'+ th|^{p-1}\frac{\operatorname{Re}\left[\overline{f'+ th}\cdot h\right]}{|f'+ th|}\tag{*}\\ &=p\cdot |f'+ th|^{p-2} \operatorname{Re}\left[\overline{f'+ th}\right]\cdot h\\ &=p\operatorname{Re}\left[ |f'+ th|^{p-2}\left(\overline{f'+ th}\right)\cdot h\right]\end{align} $$ (In $(*)$, $\partial_t |f'+ th|$ is obtained by applying the chain rule with $\varphi: t\mapsto f'+ th$ and $z\mapsto \sqrt{\mathrm{Re}[z]^2 + \mathrm{Im}[z]^2 }$)
• Lastly, we need to check that $\partial_t |f'+ th|^p$ is dominated by an integrable function (which does not depend on $t$). We have the following inequality : $$\begin{align}\left\vert\frac{\partial}{\partial t} |f'+ th|^p\right\vert &= \left\vert p\operatorname{Re}\left[ |f'+ th|^{p-2}\left(\overline{f'+ th}\right)\cdot h\right]\right\vert\\ &\le\left\vert p |f'+ th|^{p-2}\left(\overline{f'+ th}\right)\cdot h\right\vert\\ &=|f'+ th|^{p-1}\cdot |ph|\\ \end{align} $$ Now it would probably be a bit tricky to get a uniform bound for all $t\in\mathbb R$, but here we are only interested in computing $F'(0)$ so a bound on an interval of the form $[-a,a]$ with some positive real number $a$ is sufficient. But indeed, if you let $a>0$, the map $t\in[-a,a]\mapsto |f'+ th|^{p-1}\cdot |ph| $ is continuous over a compact domain, and thus reaches its maximum value at some point $t_0\in[-a,a]$, we thus have the following bound $$\left\vert\frac{\partial}{\partial t} |f'+ th|^p\right\vert \le |f'+ t_0h|^{p-1}\cdot |ph|\,\,\,\,\forall t\in[-a,a]$$ The third condition is thus fulfilled, and we conclude from Leibniz integral rule that $F$ is differentiable on $[-a,a]$ with $$F'(t) = \int_\Omega\frac{\partial}{\partial t} |f'+ th|^pd\mu =p\operatorname{Re}\left[\int_\Omega |f'+ th|^{p-2}\left(\overline{f'+ th}\right)\cdot hd\mu\right] $$ (Note : Since, $a$ was taken arbitrarily, $F$ is in fact differentiable on all $\mathbb R$. Also, the derivative of $G$ can be computed the exact same way)