neighborhood of $0$ according to the integral metric ex 
If $a,b\in\mathbb{R}$, in which $a< b$.Let $\mathscr{C}([a,b])$ be the space of continuous functions of $[a,b]$ in $\mathbb{C}$ with the supremum metric $d_{\infty}$.$(d_\infty(f,g)=\max|f(x)-g(x)|)$. Let $0$ be the null function and let $V=\{f\in\mathscr{C}([a,b]) : (\forall x\in[0,1])\,|f(x)|\leqslant \frac{1}{2}\}$.
So $V$ is a neighbourhood of $0$, because it contains the ball $B(0,\frac{1}{2})$.
Using the same notation, $V$ is not the neighbourhood of $0$ if it the metric is the integral metric.

I tried to solve this $\int_\limits{0}^{1}|f(x)-0|=\frac{\frac{1}{2}\times 1}{2}=\frac{1}{4}$ so it contains a ball $B(0,\epsilon)$ for $\epsilon>0$.
Question:
What am I doing wrong? The ball according to my computation should be inside $V$ for an arbitrary small epsilon.
 A: You want to prove that $V$ is not a neighborhood of $0$. So, take $\varepsilon>0$; you want to prove that $B(0,\varepsilon)\varsubsetneq V$. Take a continuous function $f\colon[a,b]\longrightarrow\mathbb R$ such that its graph is


*

*the line segment from $(a,1)$ to $(c,0)$ for a point $c\in(a,b)$;

*the line segment from $(c,0)$ to $(b,0)$.


The distance from $f$ to $0$ is the area of the triangle whse vertices are $(a,0)$, $(a,1)$, and $(c,0)$, which is equal to $\frac{c-a}2$. So, pick $c$ such that this number is smaller thatn $\varepsilon$. Then $f\in B(0,\varepsilon)$, but $f\notin V$, since $f(a)=1$. So, $B(0,\varepsilon)\varsubsetneq V$
A: You showed that if $f\in V$, then it is in the integral metric ball $B(0,1/2)$.
This does not show that $V$ contains an integral metric ball.
The set $V$ is not a neighborhood of $0$ in the integral metric because it has empty interior (in integral metric). Indeed, for small $\varepsilon>0$, there is a function $f\in B(0,\varepsilon)$ (this is a ball in the integral metric) such that $f\not\in V$. To see this, fix $0<\varepsilon<1$. Define $f:[0,1]\to\mathbb{R}$ by
$$
f(x) = \begin{cases}
1-x/\varepsilon & \text{if $x\in[0,\varepsilon)$}, \\
0 & \text{if $x\in[\varepsilon,1]$}.
\end{cases}
$$
(Note that $f$ is just the piecewise linear function whose graph connects the points $(0,1)$, $(\varepsilon,0)$, and $(1,0)$.)
Then $f$ is continuous, $f\not\in V$, and
$$
\int_0^1 |f(x)-0|\,dx = \varepsilon/2 < \varepsilon.
$$
Therefore every ball $B(0,\varepsilon)$ satisfies $B(0,\varepsilon)\cap V^c$, which shows that no ball is contained by $V$.
