Interchange max and sup. Here, $\|f \|_{max} = \max_{x \in [a,b]} |f(x)|$. Is the following true?
\begin{align}
\sup_{\|f\|_{max} = 1} \max_{x \in [a,b]} \left|\int_a^b G(x,y) f(y) dy \right| = \max_{x \in [a,b]} \left|\int_a^b G(x,y) dy \right|.
\end{align}
I believe so, since $\|f\|_{max} = 1$ implies $\int G f \leq \int G 1$ and furthermore it seems obvious that $f(y) = 1$ is the best choice for maximizing the integral. Is there any serious justification that needs to be done, or am I overthinking things?
EDIT: Also, the only assumptions on $f$ and $G$ are continuity.
 A: The inequality $Gf \le G1$ only holds when $G$ is non-negative. If $G$ has negative values, we want to define $f$ as
$$
 f(y) = \begin{cases}
          1 & G(x,y) \ge 0 \\
          -1 & G(x,y) < 0
        \end{cases}.
$$
Then $Gf = |G|$ and we should have
\begin{align}
\sup_{\|f\|_{max} = 1} \max_{x \in [a,b]} \left|\int_a^b G(x,y) f(y) dy \right| = \max_{x \in [a,b]} \int_a^b |G(x,y)| dy.
\end{align}

To address your concern, it is true that
$$ \sup_{a \in A} \sup_{b \in B} F(a,b) = \sup_{(a,b) \in A \times B} F(a,b) = \sup_{b \in B} \sup_{a \in A} F(a,b). $$
Let us prove the first equality and the second follows symmetrically.
Let $S_1 = \sup_{a \in A} \sup_{b \in B} F(a,b)$ and $S_2 = \sup_{(a,b) \in A \times B} F(a,b)$. Suppose that $S_1 < S_2$. Then, since $S_2$ is the least upper bound of the set $\{F(a,b) : (a,b) \in A \times B\}$, there exist $(a_0, b_0) \in A \times B$ with $S_1 < F(a,b)$. But then
$$ F(a_0,b_0) \le \sup_{b \in B} F(a_0,b) \le \sup_{a \in A} \sup_{b \in B} F(a,b) = S_1, $$
a contradiction.
Next, suppose that $S_2 < S_1$. Then, because $S_1$ is the least upper bound of $\{\sup_{b \in B} F(a,b) : a \in A\}$, it follows that there is some $a_0 \in A$ such that $S_2 < \sup_{b \in B} F(a_0,b)$. Similarly, there now exists $b_0 \in B$ such that $S_2 < F(a_0,b_0)$, a contradiction.

With that out of the way, let us see why we end up with $|G(x,y)|$ inside our integral since the function $f$ that we defined may not be continuous.
If you want to figure out how to obtain $|G(x,y)|$ by using continuous functions by yourself, stop reading here.
To do this, we want to approximate $f$ by a continuous function. To do this, we will want to partition $[a,b]$ into a collection of intervals such that on each interval, either $G \ge 0$ or $G < 0$. Then, we put $\varepsilon$ small gaps between the intervals and interpolate $f$ linearly.
The set $\{y \in [a,b] : G(x,y) < 0\}$ is relatively open in $[a,b]$. So it is of the form $U \cap [a,b]$ for some open set $U$ in $\mathbf R$. By this question, every open set in $\mathbf R$ is an at most countable union of disjoint intervals. Thus we have some countable collection of disjoint intervals $(I_n)_{n = 1}^\infty$ such that $G(x,y) < 0$ on each $I_n$ and $G(x,y) \ge 0$ if $y \notin \bigcup_{n = 1}^\infty I_n$.
Let $\varepsilon > 0$ be given. Now shrink each interval $I_n$ by some positive $\varepsilon_n$ and define $f = 1$ on the shrinked intervals, $f = -1$ on the complement of $\bigcup_{n = 1}^\infty I_n$ and interpolated linearly in the gaps. Choose $\varepsilon_n$ such that the sum $\sum_{n = 1}^\infty \varepsilon_n = \varepsilon$. For example, $\varepsilon_n = \varepsilon/2^n$.
By construction, the function $G(x,y)f(y)$ agreees with $|G(x,y)|$ except on the gaps, which have a total length of less than $\varepsilon$. On the gaps, the difference $||G(x,y)| - G(x,y)f(y)| \le |G(x,y)| + |G(x,y)||f(y)| \le 2|G(x,y)|$. By the extreme value theorem, $|G(x,y)|$ is bounded by some $M$. Therefore
$$ \left| \int_a^b |G(x,y)|dy - \int_a^b G(x,y)f(y) dy \right| = \left| \int_{\text{gaps}} |G(x,y)| - G(x,y)f(y) dy \right| \le 2M\varepsilon. $$
Since $\varepsilon$ is arbitrary, we get an integral over $|G(x,y)|$ in the supremum.
