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.

  • $\begingroup$ @Siminore And if the limit is a discontinuous function then it's not in $C([0,1], \mathbb{R})$, so you're not talking about closedness of $B_\infty(0,1)$ in $C([0,1], \mathbb{R})$ anymore. $\endgroup$ – Najib Idrissi Oct 25 '16 at 11:39

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.

* 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.

  • $\begingroup$ Thank you for your nice answer :) $\endgroup$ – anonymus Oct 25 '16 at 12:01

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 ).

  • $\begingroup$ The sequence $k_n$ does not converge in $C([0,1], \mathbb{R})$ at all. The question doesn't ask if the ball is compact, it's asking if it's closed, i.e. if all sequence in $B_\infty(0,1)$ that converge in $C([0,1], \mathbb{R})$ have their limit in $B_\infty(0,1)$.Since $k_n$ does not converge in $C([0,1], \mathbb{R})$, the condition is void. $\endgroup$ – Najib Idrissi Oct 25 '16 at 11:17
  • $\begingroup$ Define a function $$k(t)=\begin{cases} 1 \mbox{ for } 0\leqslant t \leqslant 2^{-1} \\ 0 \mbox{ for } 2^{-1}\leqslant t \leqslant 1\end{cases}$$ then $||k_n -k ||_{L_1} =\int_{[0,1]} |k_n (t) - k(t) | dt \to 0$ as $n\to \infty.$ But $k\notin C([0,1] , \mathbb{R} ).$ So if you consider a $C([0,1] , \mathbb{R} )$ as subspace of $L_1 ([0,1] , \mathbb{R} ) $ then the unit ball is not closed. I think thats was the oryginal meaning of this question. $\endgroup$ – MotylaNogaTomkaMazura Oct 25 '16 at 11:55
  • $\begingroup$ The question said nothing about $L^1([0,1], \mathbb{R})$. If this is what the question meant, then OP should have been clearer. $\endgroup$ – Najib Idrissi Oct 25 '16 at 12:00
  • $\begingroup$ Thank you guys. I thought i was clear enough though. My original question was about closedness in $C^0(0,1)$. Anyway, MotylaNogaTomkaMazura examples are interesting. $\endgroup$ – anonymus Oct 25 '16 at 12:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.