# Closure of set of all differentiable functions in $(C[0,1],||.||_{\infty})$

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?

• 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 $\lVert f-f_\epsilon\rVert_\infty\le \epsilon$. – Giuseppe Negro Jan 24 at 10:54
• I am sorry @GiuseppeNegro, I am not aware of the density theorem of Weierstrass. Can we give a prove without using this thm? – Sabhrant Jan 24 at 10:56
• 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 $\frac1n$. – Giuseppe Negro Jan 24 at 11:09

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 $$\lVert f-f_n\rVert_\infty\le \lVert f-g_n\rVert_\infty + \lVert g_n-f_n\rVert_\infty \to 0.$$
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 $$\left\lvert x-\frac12\right\rvert$$.