A problem in the space $C[a,b]$ Let $E=C[a,b]$ provide with the $\max$ norm. Let $S\neq \emptyset$, let and $D(t,\lambda)$ be a continuous function (for each $\lambda\in S$), from $[a,b]$ to $\mathbb{R}$, such that $\displaystyle D=\sup_{\lambda}\int_{a}^b |D(t,\lambda)|dt<\infty$.  We define for $f\in E$ $$A(f)=y(\lambda)=\int_{a}^b D(t,\lambda)f(t)dt$$
I have to prove that: $\|A\|=D$, where $\sup_{\|x\|=1}{\|A(x)\|}=\|A\|$.
My Approach: I have already showed that $A$ is a linear continuous operator, from $E$ to $F=\{f:S\to \mathbb{R} \mid \text{$f$ is bounded} \}$. 
Furthermore: since 
$$\|A(f)\|=\left\|\int_{a}^bD(t,\lambda)f(t)dt \right\|\leq \int_{a}^b |D(t,\lambda)|\cdot |f(t)|dt\leq \|f\| \int_{a}^b |D(t,\lambda)|dt$$ 
$$\leq \|f\|\cdot \sup_{\lambda}\int_{a}^b |D(t,\lambda)|dt=D\cdot \|f\|$$ 
we have $\|A\|\leq D$.
The difficult part is (I think) $\|A\|\geq D$. I have read some hint:
Consider 
$f_n(t)=-1$ if $t<-1/n$,
$f_n(t)=1$ if $t>1/n$ and 
$f_n(t)$ linear if $t$ is in $[-1/n,1/n]$. 
How this can help me? Please any solution or explanation will be fully appreciated. Thanks!
 A: Since we already know that $\lVert A\rVert \leqslant D$, it remains to see that for every $\varepsilon > 0$ we have $\lVert A\rVert \geqslant D - \varepsilon$. If $D = 0$, that is clear, so let's assume $D > 0$.
Fix an arbitrary $0 < \varepsilon < D$. By definition of $\sup$, there is a $\lambda_0 \in S$ with
$$\int_a^b \lvert D(t,\lambda_0)\rvert\,dt > D-\frac{\varepsilon}{2}.$$
For ease of notation, let $g(t) = D(t,\lambda_0)$. We want to find a continuous $f$ with $\lVert f\rVert \leqslant 1$ so that
$$\int_a^b g(t)\cdot f(t)\,dt > D - \varepsilon.$$
For then we know
$$\lVert A\rVert \geqslant \lVert Af\rVert \geqslant \lvert (Af)(\lambda_0)\rvert = \biggl\lvert \int_a^b g(t)f(t)\,dt\biggr\rvert > D - \varepsilon,$$
and since $\varepsilon$ was arbitrary, it then follows that $\lVert A\rVert \geqslant D$.
This is the place where the $f_n$ come into play. If we did not require $f$ to be continuous, we could just take
$$\sigma(t) = \begin{cases} 1 &, g(t) > 0 \\ 0 &, g(t) = 0 \\ -1 &, g(t) < 0\end{cases}$$
and have $\lVert\sigma\rVert \leqslant 1$ and
$$\int_a^b g(t)\sigma(t)\,dt = \int_a^b \lvert g(t)\rvert\,dt > D - \frac{\varepsilon}{2}.$$
But generally $g$ is positive on some subintervals of $[a,b]$ and negative on others, so $\sigma$ is then not a continuous function. The idea is to approximate $\sigma$ by continuous functions. We set $\sigma_n(t) = 1$ when $g(t) > 1/n$, and $\sigma_n(t) = -1$ when $g(t) < -1/n$, and then we need to continuously extend $\sigma_n$ to the whole interval $[a,b]$ by defining it appropriately on the set $\{ t\in [a,b] : -1/n \leqslant g(t) \leqslant 1/n\}$.
Well, we simply take $\sigma_n(t) = f_n(g(t))$. Since $\sigma_n = f_n \circ g$, it is continuous. Also, by construction $-1 \leqslant \sigma_n(t) \leqslant 1$, so $\lVert \sigma_n\rVert \leqslant 1$. Further we have
$$\lim_{n\to\infty} \sigma_n(t) = \sigma(t)$$
for all $t\in [a,b]$, so we have good reason to expect
$$\lim_{n\to\infty} \int_a^b g(t)\sigma_n(t)\,dt = \int_a^b g(t)\sigma(t)\,dt = \int_a^b \lvert g(t)\rvert\,dt.\tag{1}$$
That is indeed the case. If you have some Lebesgue theory at your service, it follows immediately from the dominated convergence theorem. If not, that requires a little more work: Consider the function
$$\psi_n(t) = \lvert g(t)\rvert - g(t)\sigma_n(t).$$
Since $g$ and $\sigma_n$ are continuous, $\psi_n$ is continuous. If $\lvert g(t)\rvert \geqslant 1/n$, then we have $g(t)\sigma_n(t) = \lvert g(t)\rvert$ and hence $\psi_n(t) = 0$, and if $\lvert g(t)\rvert < 1/n$, then we have
$$\lvert \psi_n(t)\rvert \leqslant \lvert g(t)\rvert + \lvert g(t)\rvert\cdot \lvert \sigma_n(t)\rvert \leqslant 2\lvert g(t)\rvert \leqslant \frac{2}{n},$$
therefore
$$\biggl\lvert \int_a^b \lvert g(t)\rvert - g(t)\sigma_n(t)\,dt\biggr\rvert \leqslant \int_a^b \lvert \psi_n(t)\rvert\,dt \leqslant \frac{2}{n}(b-a),$$
which shows $(1)$.
From $(1)$, it follows that for all large enough $n$ we have
$$\int_a^b g(t)\sigma_n(t)\,dt > \int_a^b \lvert g(t)\rvert\,dt - \frac{\varepsilon}{2},$$
hence
$$\int_a^b g(t)\sigma_n(t)\,dt > D - \varepsilon$$
and $\lVert A\rVert > D-\varepsilon$, as desired.
