Interior and closure of a set in $(C^0[0,1], \|\cdot\|_p)$

Question

Let $$A=\{f\in C^0[0,1] : |f(x)|\le 1 \forall x\in [0,1]\}$$

How can I find the interior and the closure of $A$ for each $p\in [1,\infty]$ in $(C^0[0,1], \|\cdot\|_p)$

Any hints, comments or suggestions would be highly appreciated

Answer 1

The interior of $A$ consists of all points $f\in A$ so that $B_f(r) \subset A$ for some $r >0$. One can actually show that $\text{int}A = \emptyset$ when $p<\infty$ and $\text{int}A = \{ f: |f|<1\}$ when $p = \infty$. (The latter part is easier and I will leave that to you.

For $p<\infty$. Let $f\in A$ and $r>0$. We will show that there is $g\notin A$ so that $g\in B_f(r)$ (Thus $\text{int}A = \emptyset$). To see this, let $g \in C^0([0,1])$ be defined so that

$$ |g(x)|\le 2,\ \ g(x) = f(x)\ \ \text{on } \ [0,1-\epsilon], \ \ g(1) = 2.$$

(Do you know how to construct this $g$?) Then

$$\begin{split} \|f-g\|_p &= \left(\int_0^1 |f(x) - g(x)|^p dx\right)^{\frac 1p} \\ &= \left(\int_{1-\epsilon}^1 |f(x) - g(x)|^p dx\right)^{\frac 1p} \\ &\le \left(\int_{1-\epsilon}^1 3^p dx\right)^{\frac 1p} \\ &= 3 \epsilon^{\frac 1p}. \end{split}$$

Then if we choose $\epsilon$ small so that $3\epsilon^{\frac 1p} <r$, $g\in B_f(r)$. But as $g(1) = 2$, $g\notin A$. As $r$ is arbitrary, $f\notin \text{Int}A$.

For the second part, one can actually show that the closure of $A$ is $A$ itself for all $p\in [1, \infty]$. I will leave the $p = \infty$ to you again.

We will show if $g\notin A$, then $g\notin \overline A$ (in particular $\overline A \subset A$).

Now assume $g\notin A$. Then by definition of $A$, there is $x\in [0,1]$ so that $|g(x)| >1$. As $g$ is continuous, there is $\epsilon, \delta$ so that $|g(y)| > 1+ \delta$ for all $y\in [0,1]$ so that $|y-x|\le \epsilon$. Let $f\in A$ be arbitrary, then

$$\begin{split} \|f-g\|_p = &= \left(\int_0^1 |f(x) - g(x)|^p dx\right)^{\frac 1p} \\ &\ge \left(\int_{x-\epsilon}^{x+\epsilon} |f(x) - g(x)|^p dx\right)^{\frac 1p} \\ &\ge \left(\int_{x-\epsilon}^{x+\epsilon} \delta^p dx\right)^{\frac 1p} \\ &= (2\epsilon)^{\frac 1p} \delta. \end{split}$$

Note that $\epsilon, \delta$ can chosen independent of $f$, thus the ball $B_g(c)$ with $c <(2\epsilon)^{\frac 1p} \delta$ does not intercept $A$. Thus $g\notin \overline A$.

Answer 2

To show the interior of $A$ is empty for $1\le p < \infty,$ let $f\in A.$ Define $f_n(x) = f(x) + nx^{n!}.$ Check that $\|f_n\|_\infty \to \infty,$ while $\|f_n - f\|_p \to 0$ for all $p\in [1,\infty).$ This gives the result.