Prove that $f_0$ is not an inner point of $A$ relative to $\lVert\cdot\rVert_1$ Let $V=\big\{f:[0,1]\longrightarrow\mathbb{R}:f\text{ is continuous}\big\}$ and define the following norms:
$$\lVert f\rVert_1=\int_0^1|f(x)|dx$$
$$\lVert f\rVert_\infty=\max\big\{|f(x)|:x\in[0,1]\big\}$$
Now define $A=\big\{f\in V:\lVert f\rVert_\infty<1\big\}$ and $f_0(x)=0$, for all $x\in [0,1]$.
Prove that $f_0$ is not an inner point of $A$ relative to $\lVert\cdot\rVert_1$
I thought that the problem was in the functions that are below the $x$-axis whose integral is less than $1$, but the norm clearly does not allow negative values, any suggestions?
 A: Hint
Can you construct a sequence $f_n$ of continuous functions such that

*

*$f_n \notin A$ for all $n$

*$\|f_n-f_0\|_1 \to 0$ as $n \to \infty$
If such a sequence exists, why does it prove the statement ?
Further help
$f_0$ is not an inner point of $A$ if for every open set containing $f_0$, there is at least one element that is not in $A$. If a sequence $f_n$ that does not belong to $A$ but still converges to $f_0$, you have constructed such elements.
Indeed, assume $U$ is an open set containing $f_0$, then we can find a ball $B_{f_0,\varepsilon}$ centered at $f_0$ and of radius $\varepsilon$ such that $B_{f_0,\varepsilon} \subset U$. It is then enough to consider $f_n$ with $n$ large enough so $||f_n-f_0||_1<\varepsilon$.
A: Let $\,\varepsilon\!\in\left]0,1\right[\,$ and consider the following function :
$f_\varepsilon(x)=\begin{cases}-\dfrac 1\varepsilon x+1\qquad\text{if }x\in\big[0,\varepsilon\big]\\0\qquad\qquad\quad\text{if }x\in\big]\varepsilon,1\big]\quad.\end{cases}$
The function $\,f_\varepsilon\,$ belongs to $\,V\,$ because it is continuous on $[0,1]\,,\,$ but it does not belong to $\,A\,$ since $\,\lVert f_\varepsilon\rVert_\infty=1\,.$
Moreover ,
$\displaystyle\lVert f_\varepsilon-f_0\rVert_1=\lVert f_\varepsilon\rVert_1=\int_0^1\left|f_\varepsilon(x)\right|\mathrm dx=\dfrac\varepsilon2<\varepsilon\,.$
Therefore, for any ball $\,B\left(f_0,\delta\right),\,$ where $\,\delta>0\,,\,$ there exists $\varepsilon=\min\left\{\delta,\dfrac12\right\}\in\big]0,1\big[\,$ such that $\,f_\varepsilon\in B\left(f_0,\delta\right)\setminus A\,.$
Consequently, there does not exist any ball $\,B\left(f_0,\delta\right)\subseteq A\,,$ hence $f_0$ is not an inner point of $A$ relative to $\lVert\cdot\rVert_1\,.$
