Here's the approach that I used on this exercise. I feel like there's probably a simpler way, but I didn't see one without using Fubini's Theorem.
First, use Hölder's inequality to show that $f_n \in L^p$, $f \in L^p$, and $f_n \to f$ in $L^p$ guarantee that $F_n \to F$ pointwise.
In addition, show that $f_1 \le f_2$ pointwise guarantees $F_1 \le F_2$ pointwise.
Verify using Hölder's inequality that $F(x)$ is continuous in $x$ for $x \gt 0$.
The hint which Rudin gives a hint for part a) is to assume the continuous, non-negative compact support case for $f$ to show
$$
\int_0^{\infty}F^p\ dx = -p\int_0^{\infty}F^{p-1}(f-F) \ dx
$$
Rewrite:
$$
\label{a}\tag{*} \int_0^{\infty}F^p\ dx = \frac{p}{p-1}\int_0^{\infty}F^{p-1}f \ dx
$$
Use Lebesgue Monotone Convergence to show ($\ref{a}$) holds when $f$ is the characteristic function of an open set with finite measure.
Show that ($\ref{a}$) holds for non-negative simple functions $s$ which are nonzero only on a set of finite measure as follows: Let $U$ be an open set of finite measure such that $s = 0$ outside of $U$. Use Lusin's theorem to get a sequence $g_n$ of continuous, compactly supported, non-negative functions whose supports are contained in $U$, and such that $g_n \to s$ pointwise a.e., $g_n \to s$ in $L^p$, and with $0 \le g_n \le K\chi_{U}$ for some $K > 0$. Apply Lebesgue Dominated Convergence to show that ($\ref{a}$) holds for $s$.
Show that ($\ref{a}$) holds for any non-negative $f \in L^p$ using Monotone Convergence on a increasing sequence of non-negative simple functions which converges pointwise to $f$.
Now let $f$ be an arbitrary non-negative function in $L^p$.
Suppose (for a contradiction) that $||F||_p =\frac{p}{p-1}||f||_p$ but that it is not the case that $f = 0$ almost everywhere.
Then $||f||_p \gt 0$, and since we are in the equality situation, we also have $||F||_p \gt 0$
Using Hölder's inequality, show that
$$
||F||_p^p = \int_0^{\infty}F^p\ dx = \frac{p}{p-1}\int_0^{\infty}F^{p-1}f \ dx \\
\le \frac{p}{p-1}\left\{ \int_0^{\infty}F^p \ dx \right\}^{1 - \frac{1}{p}} ||f||_p \\
= \frac{p}{p-1} ||F||_{p-1}||f||_p
$$
Thus Hölder's inequality must be an equality, and so there must be a non-negative constant $\alpha$ such that $F = \alpha f$ a.e. or $f = \alpha F$ a.e..
Verify that $\alpha$ must in fact be strictly positive.
The next part is what I wrestled with for some time.
We now have that WLOG $f = \alpha F$ a.e. for some $\alpha \gt 0$.
Verify using ($\ref{a}$) that $\alpha = \frac{p-1}{p}$.
From this we may conclude that
$$
F(x) = \frac{1}{x} \int_0^x \alpha F(t) \ dt
$$
Since $F(t)$ is continuous, it must be the case that $F(x)$ is differentiable for $x \gt 0$. Differentiating gives
$$
xF'(x) = (\alpha - 1)F(x) = -\frac{1}{p}F(x)
$$
Since $||F||_p > 0$, there must be a point $a$ with $F(a) > 0$.
But, the above equation shows that if $F(a) > 0$, then $F(x) > 0$ for $0 < x < a$.
So on $(0, a)$, have $\frac{F'}{F} = -\frac{1}{px}$ and hence $\log F(x) = C -\frac{1}{p} \log(x)$.
So, near zero, we have $F(x) = C x^{-\frac{1}{p}}$ for some $C > 0$. But this is a contradiction since then $F$ is not in $L^p$.
The case where $f$ is an arbitrary (complex-valued) function in $L^p$ follows from the case where $f$ is non-negative.
Let $g = |f|$.
If $||F||_p =\frac{p}{p-1}||f||_p$, then
$$
\frac{p}{p-1}||f||_p = ||F||_p \le ||G||_p \le \frac{p}{p-1}||g||_p = \frac{p}{p-1}||f||_p
$$
So $g = 0$ almost everywhere by the non-negative case, and the same is true for $f$.
Best Answer
The first problem is clear by recalling what is the definition of 1-norm and infinity-norm.
To show the second, apply the generalized AM-GM so we get $$ \frac{1/p|a_i|^p + 1/q|b_i|^q}{1} \geq |a_i||b_i| $$ Now sum the equation through all $i$. Then we get $$ \frac{1}{p}\sum_i |a_i|^p + \frac{1}{q} \sum_j|b_j|^q \geq \sum_i |a_i||b_i| \geq \langle a,b \rangle $$ Now as you assume $||a||_p = ||b||_q = 1$, $\sum_i|a_i|^p = \sum_j|b_j|^q = 1$, so $$ \frac{1}{p}\sum_i |a_i|^p + \frac{1}{q} \sum_j|b_j|^q = \frac{1}{p} + \frac{1}{q} = 1 = ||a||_p||b||_q $$ so the inequality is proved as required.