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Let $X=C([0,1])$ with the $||\cdot||_\infty$-norm. Show that $\overline{C^1([0,1])}=X$.

Every differentiable function is continuous. So $C^1([0,1])\subset X$. Now I have to show $D=\overline{C^1([0,1])}\subset X$. I also have to show: $X\subset D$. So every continuous function which is not differentiable is in $\overline{D}/\dot{D}$.

Every continuous function f in a compact intervall is bounded, so $||f||_\infty<\infty$. $C^1([0,1])\subset X$, so $||f||_\infty<\infty \forall f\in C^1([0,1])$. So for the functions g in the closure of $C^1([0,1])$ we have $||g||_\infty<\infty$. How can I show that these functions g are also in X? And how can I show that every continuous non-differentiable function is in $\overline{D}/\dot{D}$? Can someone help me?

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Here is a direct way of showing density.

Pick $f \in C[0,1]$. Without loss of generality we can extend $f$ to $[-1,2]$ by defining $f(t) = f(0)$ for $t < 0$ and $f(t) = f(1)$ for $t >$1.

Let $f_n(t) = n \int_{t-{1 \over n}}^{t+{1 \over n}} f(\tau) d \tau$. Then $f_n$ is differentiable with derivative $f_n'(t) = {f(t+{1 \over n})-f(t-{1 \over n}) \over {1 \over n}}$ and it is clear that $f_n'$ is continuous.

Suppose $\epsilon>0$, then since $f$ is uniformly continuous there is some $\delta>0$ such that if $|x-y| < \delta$ then $|f(x)-f(y)| < \epsilon$. Now suppose ${2 \over n} < \delta$, then $|f(t)-f_n(t)| \le n \int_{t-{1 \over n}}^{t+{1 \over n}} |f(t)-f(\tau)| d \tau \le n \int_{t-{1 \over n}}^{t+{1 \over n}} \epsilon d \tau = \epsilon$, and so $\|f-f_n\|_\infty \le \epsilon$.

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  • $\begingroup$ Just to clarify things: This is a proof to show that the differentiable functions in the intervall $[0,1]$ are dense in the set of continuous functions in the intervall $[0,1]$. Why do you have to suppose $\frac{2}{n}<\delta$ ? $\endgroup$ – Tobi92sr Nov 23 '17 at 17:44
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    $\begingroup$ Yes, continuously differentiable functions. The interval of integration is ${2 \over n} $ long. $\endgroup$ – copper.hat Nov 23 '17 at 17:57
  • $\begingroup$ @Tobi92sr: Actually, I should have chosen $|f(x)-f(y)| < {1 \over 2} \epsilon$ (also ${1 \over n} < \delta$ would do) so that we end up with $\|f-f_n\|_\infty \le \epsilon$. Sloppy on my part! $\endgroup$ – copper.hat Nov 23 '17 at 19:38
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Stone–Weierstrass theorem says that any continuous function $f$ on $[0,1]$ can be approximated by a polynomial in the supremum norm. In particular, the polynomial is $C^{1}$.

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