$B_\infty(0,1)$ is not closed w.r.t. $\|\cdot\|_1$

Question

Let $B_\infty(0,1)$ be the unit ball of $C([0,1],\mathbb{R})$ endowed with the $\|\cdot\|_\infty$ norm.

How to show that it's not closed with respect to the $\|\cdot\|_1$ norm ?

That is, I need to find a sequence $f_n$ of continuous functions such that $\|f_n\|_\infty\leq 1$ but $f_n \xrightarrow[n\to\infty]{L^1}f$ and $\|f\|_\infty > 1$, or $f$ is not even continuous.

Answer 1

The ball $B_\infty(0,1)$ is actually closed in $C([0,1], \mathbb{R})$ equipped with the $L^1$ norm (which, I assume, is the space of continuous function $[0,1] \to \mathbb{R})$. It seems you've misunderstood what is means for $B \subset X$ to be closed: it's closed if for all convergent sequences $(b_n)$ with $b_n \in B \, \forall n$ (with a limit in $X$, because where else would it be?), then $\lim_{n \to \infty} b_n \in B$. Thus, if you find a sequence with a discontinuous limit, then the sequence doesn't converge in $X = C([0,1], \mathbb{R})$ and so this isn't a counterexample for the closedness of $B = B_\infty(0,1)$.

Indeed, suppose that $(f_n)$ is a sequence in $B_\infty(0,1)$ (i.e. $\|f_n\|_\infty \le 1$ for all $n$) and $f_n \xrightarrow{L_1} f$ with $f \in C([0,1], \mathbb{R})$, i.e. $\lim_{n \to \infty} \int_0^1 |f(x) - f_n(x)| dx = 0$. Then I claim that $\|f\|_\infty \le 1$.

Suppose otherwise. Then $|f(a)| > 1$ for some $a$, WLOG we can assume $f(a) > 1$. Since $f$ is continuous, this implies that there is some $\epsilon > 0$ and $\delta > 0$ such that $f(x) \ge 1+\epsilon$ for all $x \in (a-\delta, a+\delta)$.^*

Since $\|f_n\|_\infty \le 1$ for all $n$, this implies that $|f(x) - f_n(x)| \ge \epsilon$ for all $x \in (a-\delta, a+\delta)$, and hence $$\int_0^1 |f(x) - f_n(x)| dx \ge \int_{a-\delta}^{a+\delta} |f(x) - f_n(x)| \ge \int_{a-\delta}^{a+\delta} \epsilon \ge 2 \epsilon \delta$$ for all $n$, so it's bounded below by a positive number and we cannot have $\lim_{n \to \infty} \int_0^1 |f(x) - f_n(x)| dx = 0$, a contradiction.

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^* If $a$ is $0$ or $1$, you can take instead $[0, \epsilon)$ or $(1-\epsilon, 1]$ for the interval where $f(x) \ge 1+\epsilon$, and then bound below the integral by $\epsilon \delta$ instead of $2 \epsilon \delta$. This doesn't change anything.

Answer 2

Take a sequence $$f_n (t) =\begin{cases}1 \hspace{1.1cm}\mbox{ for } -1 \leq x \leq 0\\ 1-nx \mbox{ for } 1\leq x\leq n^{-1} \\ 0 \hspace{1.1cm}\mbox{ for } n^{-1} \leq x \leq 1\end{cases}$$

and define $k_n (t) =f_n (2t-1).$ Then $k_n $ are continuous and $||k_n||_{\infty} =1 $ but $\lim_n k_n (t) $ is not continuous (lim with respect to $L_1$ norm ).