Operator norm of integral operator of continuous functions Let $A:C^0[-1,1] \rightarrow \mathbb{R} $ be defined by $$A(f) =\int_{-1}^{1} \gamma (t) f(t) dt $$ for $\gamma \in C^0[-1,1] $.
It's easy to show that $$||A||\leq \int_{-1}^{1}|\gamma (t) | dt$$ but how can the reverse inclusion be shown.
I was thinking something along the lines of constructing a function $f$ based on the roots of $\gamma $, where $\gamma (t) >0 $ let $f(t) =1 $ and where $\gamma (t) <0 $ let $f(t) =-1 $. But to make this $f$ continuous I was thinking of joining - 1 and 1 (near a root of $\gamma $) by a line segment which becomes arbitrarily steep. Its intuitively clear that then we can make $|A(f) |$ arbitrarily close to the value above that I am looking to prove.
Seems a bit difficult to make formal though.
Any help or better ideas? By the way I'm using the supremum norm on the space of continuous functions.
 A: Here is a more formal argument presenting your idea. This is taken from one of my old questions.
Let $\varepsilon > 0$. Since $\gamma$ is continuous, $\gamma^{-1}(\langle 0, +\infty\rangle)$ and $\gamma^{-1}(\langle -\infty,0\rangle)$ are open sets so they can be written as countable (or finite, that case is easier) disjoint unions of intervals:
$$\gamma^{-1}(\langle 0, +\infty\rangle)) = \bigcup_{n=1}^\infty \langle a_n, b_n\rangle, \quad \gamma^{-1}(\langle -\infty,0\rangle) = \bigcup_{n=1}^\infty \langle c_n, d_n\rangle$$
Now define $$S = \bigcup_{n=1}^\infty \left\langle a_n+\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{b_n-a_n}{2}\right\}, b_n-\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{b_n-a_n}{2}\right\}\right\rangle \\
\cup \bigcup_{n=1}^\infty \left\langle c_n+\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{d_n-c_n}{2}\right\}, d_n-\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{d_n-c_n}{2}\right\}\right\rangle\\
\cup \gamma^{-1}(\{0\})$$
and define $f_\varepsilon \in C[-1,1]$ to be equal to $\operatorname{sgn} f$ on $S$, and to an affine function on $[-1,1]\setminus S$ connecting $\pm 1$ with $0$ so that $f_\varepsilon$ is continuous.
Now we have:
$$\begin{align}Af_\varepsilon &= \int_{[-1,1]} \gamma(t)f_\varepsilon(t)\,dt\\
&= \int_{S} \underbrace{\gamma(t)f_\varepsilon(t)}_{=|\gamma(t)|}\,dt + \underbrace{\int_{[-1,1]\setminus S} \gamma(t)f_\varepsilon(t)\,dt}_{\ge 0}\\
&\ge \int_{S} |\gamma(t)|\,dt\\
&= \int_{[-1,1]} |\gamma(t)|\,dt - \int_{[-1,1]\setminus S} |\gamma(t)|\,dt\\
&\ge \int_{[-1,1]} |\gamma(t)|\,dt - \|\gamma\|_\infty \lambda([0,1]\setminus S)\\
&\ge \int_{[-1,1]} |\gamma(t)|\,dt - \varepsilon\\
\end{align}$$
since
$$\lambda([-1,1]\setminus S) = \lambda\left(
\bigcup_{n=1}^\infty
\left[a_n,
a_n+\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{b_n-a_n}{2}\right\}
\right]
\cup
\left[b_n-\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{b_n-a_n}{2}\right\}, b_n
\right]
\right)\\
+\lambda\left(
\bigcup_{n=1}^\infty
\left[a_n,
c_n+\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{d_n-c_n}{2}\right\}
\right]
\cup
\left[d_n-\min\left\{\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, \frac{d_n-c_n}{2}\right\}, d_n
\right]
\right)\\
\le \sum_{n=1}^\infty \lambda\left(\left[a_n, a_n+\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}\right]\right) + \lambda\left(\left[b_n-\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, b_n\right]\right) + \lambda\left(\left[c_n, c_n+\frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}\right]\right) + \lambda\left(\left[d_n- \frac{\varepsilon}{2^{n+3}\|\gamma\|_\infty}, d_n\right]\right)\\
=\frac{\varepsilon}{\|\gamma\|_\infty}\sum_{n=1}^\infty\frac{1}{2^{n+1}} = \frac{\varepsilon}{\|\gamma\|_\infty}$$
Thus:
$$\|A\| \ge \frac{\|Af_\varepsilon\|_\infty}{\|f_\varepsilon\|_\infty} \ge |Af_\varepsilon| = Af_\varepsilon = \int_{[-1,1]} |\gamma(t)|\,dt - \varepsilon \xrightarrow{\varepsilon\to 0} \int_{[-1,1]} |\gamma(t)|\,dt.$$
