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Define $D:=\{ f \in C[0,1] \mid f$ is differentiable $\}$. Find $\overline{D}$ . Is $D$ open or closed ?

My attempt: I think $D$ is neither open nor closed in $(C[0,1],||.||_{\infty})$. consider the function $f(x)=0$ $ \,\forall\,$ $x \in [0,1]$ , clearly $f \in D$. Let $r$ be any arbitrary positive real number. Then $g \in B(f,r)$ , where $g(x) = r\mid x-\frac{1}{2}\mid$. $g$ is a continuous function but it is not differentiable at $x=\frac{1}{2}$ . Hence $B(f,r) \not\subseteq D$ . This proves that $D$ is not an open set.

To show that $D$ is not closed, we want to construct a sequence of differentiable function which uniformly converge to a non-differentiable function. Consider the function $f_n(x)$ which is defined as below.

$$f_n(x)=\left \{ \begin{array}{ll} -x+\frac{1}{2}\left( 1-\frac{1}{n}\right) & \mbox{, if } 0\le x <-\frac{1}{n}+\frac{1}{2} \\ -\frac{n}{8}(2x-1)^2 & \mbox{, if } -\frac{1}{n}+\frac{1}{2}\le x <\frac{1}{n}+\frac{1}{2} \\ x-\frac{1}{2}\left( 1+\frac{1}{n}\right) & \mbox{, if } \frac{1}{n}+\frac{1}{2} \le x \le 1 \\ \end{array} \right. $$

The corresponding function to which it will converge is $f(x)=\mid x-\frac{1}{2}\mid$ . But $f \not\in D$, hence $D$ is not closed. This finishes the proof. Kindly verify if this proof is correct or not. I believe that $\overline{D}$ will be the whole metric space. But I am not able to prove this.


Definition 1: A point $x \in X$ is called a limit point of $E\subseteq X$ if $B(x,r)\cap E \ne \emptyset $ $\,\forall\, r>0$.

Definition 2(Interior of a set): Let $S\subset X$ be a subset of a metric space. We say that $x \in S$ is an interior point of $S$ if $\,\exists\,$ $r > 0$ such that $B(x, r) \subset S$ . The set of interior points of $S$ is denoted by $S^o$ and is called the interior of the set $S$

What will be $D^o$? is it the empty set?

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    $\begingroup$ Indeed, $\overline D$ is all of $C([0, 1])$. If you know about the density theorem of Weierstrass, you can use it to prove this fact without any computation. Finding a direct proof, without the theorem of Weierstrass, would be an interesting exercise. Also, $D°=\varnothing$. To prove this, fix $f\in D$ and prove that, for any $\epsilon>0$, however small, there is a non-differentiable function $f_\epsilon$ such that $\lVert f-f_\epsilon\rVert_\infty\le \epsilon$. $\endgroup$ – Giuseppe Negro Jan 24 at 10:54
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    $\begingroup$ I am sorry @GiuseppeNegro, I am not aware of the density theorem of Weierstrass. Can we give a prove without using this thm? $\endgroup$ – Sabhrant Jan 24 at 10:56
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    $\begingroup$ Of course. Actually you have already done almost all. Here you approximated the function $x\mapsto |x-\tfrac12|$. Ok. Now you have to approximate a generic function $x\mapsto f(x)$. Try to do so by splitting $[0, 1]$ in sub-intervals of length $\frac1n$. $\endgroup$ – Giuseppe Negro Jan 24 at 11:09
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I will give a couple of hints.

To prove that the interior $D°$ is empty, you need to prove that for each $f\in D$ and for each $\epsilon>0$ there is $f_\epsilon\notin D$ such that $\lVert f-f_\epsilon\rVert_\infty <\epsilon$. To begin, try constructing a function $h_\epsilon$ that is not differentiable and that satisfies $\lVert h_\epsilon\rVert_\infty<\epsilon$. (Think of a tiny saw-tooth). Once you have this, you are finished: just let $f_\epsilon:=f+h_\epsilon$.

To prove that $D$ is dense, there are many ways, of course. I would do the following. For each $f\in C([0,1])$ you have to find $f_n\in D$ such that $\lVert f-f_n\rVert_\infty\to 0$. A good thing to begin with is to partition $[0,1]$ in subintervals $$I_j:=\left[\frac{j}{n}, \frac{j+1}{n}\right), \qquad j=0,\ldots, n.$$ Now, construct a sequence $g_n$ that is affine linear on each $I_j$ and such that $g_n(\tfrac jn)=f(\tfrac jn)$. You can easily prove that $\lVert g_n-f\rVert_\infty\to 0$. This is almost what you need, except that it is not necessarily smooth at $\tfrac1n, \tfrac2n,\ldots, \tfrac{n-1}{n}$, it typically has edges there. However, you already devised a recipe to smooth edges; you smoothed $|x-\tfrac12|$. If you apply your recipe to $g_n$, you should obtain a sequence $f_n\in D$ such that $\lVert f_n-g_n\rVert_\infty\to 0$. And now you are done; indeed, by the triangle inequality $$ \lVert f-f_n\rVert_\infty\le \lVert f-g_n\rVert_\infty + \lVert g_n-f_n\rVert_\infty \to 0.$$

Hope this helps, but if not, feel free to completely ignore these hints and do as you prefer, of course.

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Yes, it is correct. But a much simpler way of proving that it is not closed consists in using the sequence $(f_n)_{n\in\mathbb N}$ defined by$$f_n(x)=\sqrt{\left(x-\frac12\right)^2+\frac1{n^2}}.$$Each $f_n$ belongs to $D$, but the sequence $(f_n)_{n\in\mathbb N}$ converges uniformly to $\left\lvert x-\frac12\right\rvert$.

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