Method for finding the norm of an operator by constructing a sequence. I asked a similar question to this one before, and while I am interested to get an answer to the following problems, I am primarily concerned with the actual method for constructing the sequence in $X$ to find the norm of the operator.
I asked a similar question which you can find here: Find the norm of the following operator.
Consider the following problems:

"Let $X=C([0,1])$ and equip $X$ with the uniform metric. If for all $x\in X$,
  $$f(x)=\int_0^1t^2x(t)dt-x(1)$$
  Show that $f\in X'$ and show that its norm is $\frac43$"

and,

"On the space $X=C^1([0,1])$ equipped with the norm,
  $$\|x\|_1=\sup_{t\in[0,1]}|x(t)|+\sup_{t\in[0,1]}|x'(t)|$$
  define a functional $f$ by,
  $$f(x)=\int_0^1tx(t)dt$$
  Show that $f\in X'$ and find its norm."

For the second problem I have that $\|f\|\le\frac12$
I have included two problems because while in both cases I am to find $\|f\|$ both are slightly different, and I want to compare the approach that should be taken for constructing $(x_n)_{n=1}^\infty\subset X:\|x_n\|\le1$, for $X$ as defined in each case, such that $\|f(x_n)\|\to\|f\|,\,n\to\infty.$ I know that the method I am to use is to construct such a sequence, so the method in both cases should be the same.
Could somebody please outline the approach I should take in creating such a sequence? And if you have a go at these examples, could you lead me through and show me why you chose to construct the sequence as you did, and how you knew to do as much with the information you were given?
 A: For the first part, you can write $f$ as a Riemann-Stieltjes integral (or Riemann-Lebesgue),
$$
         f(x) = \int_{0}^{1}x(t)d\mu(t),
$$
where $\mu(t)=\int_{0}^{t}s^2ds$ for $0 \le t < 1$ and $\mu(1)=\int_{0}^{1}s^2ds-1$. Once you have the functional on $C[0,1]$ written in this way, $\|f\|=\|\mu\|$, where $\|\mu\|$ is the variation of $\mu$. In this case, the variation of $\mu$ is easy:
$$
    \|\mu\|=|\mu(1-0)-\mu(0)|+|\mu(1)-\mu(1-0)|= \int_{0}^{1}s^2ds+1=\frac{1}{3}+1=\frac{4}{3}.
$$
The idea in finding $x$ to maximize $|f(x)|$ is to construct $f$ so that $|f|\le 1$ and $f(x) \approx \int_{0}^{1}d|\mu|(t)=\|\mu\|$. The variation is approximated by
$$
                \sum_{k=1}^{K}|\mu(t_k)-\mu(t_{k-1})|,
$$
where $\{ 0=t_{0} < t_1 < \cdots < t_k = 1\}$ is a partition of $[0,1]$ that is sufficiently refined.
To see how this works, let $\epsilon > 0$ be given. You can choose a partition $\mathcal{P}=\{ 0=t_0 < t_1 < \cdots < t_K =1 \}$ such that every interior $t_k$ is a point of continuity of $\mu$, and such that
$$
                 \|\mu\|-\frac{\epsilon}{2} < \sum_{k=1}^{K}|\mu(t_k)-\mu(t_{k-1})| \le \|\mu\|
$$
Let $s_k$ be the sign of $\mu(t_k)-\mu(t_{k-1})$. Define a function $\chi_{\delta}(x)$ to be $s_1$ on $[t_0,t_1-\delta]$, to be $s_k$ on $[t_{k-1}+\delta,t_{k}-\delta]$, and to be $s_K$ on $[t_{K-1}+\delta,t_{K}]$; then extend to be linear and continuous on $[t_{j}-\delta,t_{j}+\delta]$ for $j=1,2,\cdots,K-1$. Then $|\chi_{\delta}|\le 1$ and
$$
         f(\chi_{\delta})=\int_{0}^{1}\chi_{\delta}d\mu
            \approx \sum_{k=1}^{K}|\mu(t_k)-\mu(t_{k-1})|
$$
with the difference being bounded by the sum of variations
$$
               2\sum_{k=1}^{K-1}V_{t_{k}-\delta}^{t_{k}+\delta}(\mu),
$$
which can be made arbitrarily small for small enough $\delta$ because of choosing $t_k$ to be points of continuity of $\mu$. Hence, for small enough $\delta$,
$$
               \left|\int_{0}^{1}\chi_{\delta}(t)d\mu(t)-\|\mu\|\right| < \epsilon,\;\;\;\; \|\chi_{\delta}\|\le 1.
$$
If you apply this construction to your particular case, there is a partition $\{ 0 = t_0 < t_1 < t_2 = 1 \}$ such that
$$
       \|\mu\|+\frac{\epsilon}{2} < |\mu(t_1)-\mu(0)|+ |\mu(1)-\mu(t_1)| \le \|\mu\| = \frac{4}{3}
$$
The partition can be so minimal because $V_{0}^{t}(\mu)=\mu(t)-\mu(0)$ for $0 \le t < 1$. So you can use a function $\chi_{\delta}$ that is $1$ on $[0,1-\delta]$, is $-1$ at $t=1$, and is linear in between.
I'd have to think more about your second problem.
A: For the first problem, continuity of $f$ follows from continuity of $x$ and the fundamental theorem of calculus, and linearity $f$ from linearity of the integral, so $f\in X'$. For any $x\in C([0,1])$ we have
\begin{align}
|fx| &= \left|\int_0^1 t^2 x(t)\ \mathsf d t - x(1) \right|\\
&\leqslant \left|\int_0^1 t^2 x(t)\ \mathsf dt\right| + |x(1)|\\
&\leqslant \int_0^1 t^2 \|x\|\ \mathsf dt + \|x\|\\
&= \frac43\|x\|,
\end{align}
and hence $$\|f\|\leqslant\frac43. $$
Define the sequence $x_n$ by
$$x_n(t) = \chi_{\left[0,\frac n{n+1}\right]} + (-2(n+1)t+2n+1)\chi_{\left(\frac n{n+1},1\right]}. $$
Then
$$\|fx_n\|=\left|\int_0^{\frac n{n+1}} t^2\ \mathsf dt +1 \right| = \frac13\left(\frac n{n+1}\right)^3 + 1\stackrel{n\to\infty}\longrightarrow\frac43, $$
so that $\|f\|\geqslant\frac43$. 
Here we can find an upper bound for $\|f\|$ by simple inequalities, and from
$$\int_0^1 t^2\ \mathsf dt = \frac13 $$
I found a sequence of continuous functions that converged pointwise to the constant function $1$, but decreased linearly to the value $-1$ at the point $1$ so that the norm would be as large as possible. I couldn't tell you a systematic way to find a sequence like that; I just had to think for a bit.
For the second problem, continuity and linearity of $f$ are again clear from basic properties of the integral. For any $x\in C^1([0,1])$ we have
\begin{align}
|fx| & = \left|\int_0^1 tx(t)\ \mathsf dt\right|\\
&\leqslant \int_0^1|tx(t)|\ \mathsf dt\\
&\leqslant \sup_{t\in[0,1]}|x(t)|\int_0^1 t\ \mathsf dt\\
&= \frac12\sup_{t\in[0,1]}|x(t)|\\
&\leqslant\frac12\|x\|_1,
\end{align}
and hence $\|f\|\leqslant\frac12$. The constant function $x(t)=1$ satisfies $\|x\|_1=1$ and 
$$|fx| = \left|\int_0^1 t\ \mathsf dt \right| = \frac12, $$
so indeed $\|f\|=\frac12$.
