Show that $L$ is not bounded Given $X=C([0,1])$ the norm $\|f\|=\int{|f(t)|dt}$ and define $L:X\to F$, (where $F=\mathbb{R}$, or $\mathbb{C}$) by $L(f)=f\left(\dfrac{1}{2}\right)$. Show that L is not bounded.
My approach: I thought, take a sequence $f_n=x^n\in C([0,1])$, then $\|f_n\|=\int_{0}^{1}{|f_n(t)|dt}=\frac{1}{n+1}$. But, $L(f_n)=\dfrac{1}{2^n}$, then $\lim_{n\to\infty}{\dfrac{\|L(x^n)\|}{\|x^n\|}}=\dfrac{n+1}{2^n}$, but this approach to zero, so I need find a sequences that works in this problem ( if this is correct), Thanks!
 A: Consider functions of the form:

A: Fix $\beta>2$. Define
\begin{align*}
f(x)=\begin{cases}0&\text{if $0\leq x\leq \dfrac{1}{2}-\dfrac{1}{\beta}$,}\\\dfrac{2\beta^2x-\beta(\beta-2)}{2}&\text{if $\dfrac{1}{2}-\dfrac{1}{\beta}<x\leq\dfrac{1}{2}$,}\\
\dfrac{\beta(\beta+2)-2\beta^2x}{2}&\text{if $\dfrac{1}{2}<x\leq\dfrac{1}{2}+\dfrac{1}{\beta}$,}\\
0&\text{if $\dfrac{1}{2}+\dfrac{1}{\beta}<x\leq 1$.}
\end{cases}
\end{align*}
The algebra is messy, but with a little imagination one can see that the graph of this function essentially draws out an isosceles triangle between the three points $$\left(\frac{1}{2}-\frac{1}{\beta},0\right)\quad\left(\frac{1}{2},\beta\right)\quad\left(\frac{1}{2}+\frac{1}{\beta},0\right)$$ in the plane. The area of this triangle is $$\frac{1}{2}\left[\left(\frac{1}{2}+\frac{1}{\beta}\right)-\left(\frac{1}{2}-\frac{1}{\beta}\right)\right]\beta=1.$$ Therefore, $$\|f\|=\int_0^1|f(t)|\,\mathrm dt=1.$$ The function $f$ is continuous because it is piecewise linear. Finally, $$f\left(\frac{1}{2}\right)=\beta,$$ which can be taken to be arbitrarily large.
