Is the set $\{|f(0)|: \int_{0}^{1}|f(t)|dt\le1\}$ bounded? Let $x_0 \in [0,1]$ and define $T:C[0,1] \rightarrow \mathbb{R}$ by $T_{x_0}(f)=f(x_0)$. Let $||\cdot||_1$ be a norm on $C[0,1]$. Is $T_0$ bounded or not? That is, is the set 
$$
\left\{|T_{0}(f)|:||f||_1 \leq 1\right\}=\{|f(0)|:||f||_1 \leq 1,f \in C[0,1]\}
$$  bounded? Since $||f||_1:=\int_{0}^{1}|f(t)|dt$, the question may be equivalent to the following:
Let $f:[0,1] \rightarrow \mathbb{R}$ be continuous. Is the set $$\left\{|f(0)|: \int_{0}^{1}|f(t)|dt \leq 1\right\}$$ bounded?
I guess the answer is no. Because, for example, we can have a function whose graph is a narrow spike at the origin but with infinite height. The area enclosed by the graph may be 1 but the value at the origin $f(0)$ which is its height is infinite.
But how can I prove this formally?
 A: No, it is not bounded. Define, for each $n\in\mathbb N$,$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\mathbb R\\&t&\mapsto&\begin{cases}n-n^2t&\text{ if }t\leqslant\frac1n\\0&\text{ otherwise.}\end{cases}\end{array}$$Then$$\int_0^1\bigl\lvert f_n(t)\bigr\rvert\,\mathrm dt=\frac12,$$but $f_n(0)=n$.
A: We can take a function $f_n$ such that $f_n$ is affine on $(0,1/n)$, $f_n(0)=2n$ and $f_n(1/n)=0$ and $0$ for the other values of the interval. Then $\left\lVert f_n\right\rVert_1=1$.
For a formal example: define for each positive integer $n$ the function $f_n$ in the following way: $f_n(t)=-2n^2t+2n$ for $0\leqslant t\leqslant 1/n$ and $f_n(t)=0$ for $1/n\lt t\leqslant 1$. 
A: Consider the function $f_n(x)=2n(1/n-x)1_{[0,1/n]}(x).$ Then $\int_0^1 |f_n(x)|dx=1$ but $f_n(0)=2n.$
A: For every $a>0$, the function
$$f_a(x)=\frac{2 a e^{-a^2 x^2}}{\sqrt{\pi} \textrm{erf}(a)}$$ with the error function $\textrm{erf}(a)=\frac{2}{\pi}\int_0^a e^{-t^2}dt$ is in the set $\{|f(0)|: \int_{0}^{1}|f(t)|dt=1\}$ and evaluates to $f_a(0)=\frac{2 a }{\sqrt{\pi} \textrm{erf}(a)}$. Because $\textrm{erf}(a) \rightarrow 1$ as $a\rightarrow \infty$, we obtain $f_a(0)$ arbitrarily large as we increase $a$. As a consequence, $f_a(0)$ and thus your set are unbounded.
For every $a>0$, the function $f_a(x)$ is in $C[0,1]$ and even infinitely differentiable on $[0,1]$.
