# Diameters, distances and contraction mappings on a subset of $C_{\mathbb{R}}[0,1]$

Let $$M=\{f\in C_{\mathbb{R}}([0,1]): f(0)=0\le f(t)\le f(1)=1,\text{ for }t\in [0,1]\}$$ where $$C_{\mathbb{R}}([0,1])=\{f:[0,1]\to \mathbb{R}:f\text{ is continuous on }[0,1]\}$$ is Banach space with norm $$\|f\|_\infty=\sup \{|f(t):t\in [0,1]\}$$ . Prove

(a) $$M$$ is closed subset of $$C_{\mathbb{R}}([0,1]).$$

(b) $$\delta(f, M)=\delta(M),$$ where $$f(t)=t$$.

(c) $$\delta(f_n, M)=\delta(M),$$ where $$f_n(t)=t^n, n=2,3,...$$.

(d) Fix $$f_0\in M$$ . Define $$T_n:M\to M$$ by $$T_n(f)=\frac{(n-1)T(f)}{n}+\frac{f_0}{n}, n\in \mathbb{N}$$.Then $$T_n$$ is a contraction mapping

(e) if $$g_n\in M$$ is a fixed point of $$T_n$$ then $$\lim_{n\to \infty}\|g_n-T(g_n)\|=0$$

Here $$\delta(M)= \dim M=\sup\{\|x-y\|:x,y\in M\}$$ and

$$\delta (x,M)=\sup \{\|x-y\|:y\in M\}$$

i am trying to prove (a)

for proving (a)

let $$\{x_n\}$$ be a sequence in $$M$$ such that $$x_n\to x$$

we have to prove that $$x\in M$$

so consider $$\|x_n-x\|_\infty=\sup \{|x_n(t)-x(t)|:t\in [0,1]\}$$

since $$x)n\to x$$ so $$\|x_n-x\|<\epsilon$$ this implies $$|x_n(t)-x(t)|<\epsilon$$

from i here how to prove $$x\in M$$

and for proving (e)

since $$g_n\in M$$ is a fixed of $$T_n$$ so $$T_n(g_n)=g_n$$

so $$\lim_{n\to \infty}\|g_n-T(g_n)\|=\lim_{n\to \infty}\|T_n(g_n)-T(g_n)\|=\lim_{n\to \infty}\|(T_n-T)(g_n)\|$$ from this step can we say ?

$$\lim_{n\to \infty}\|g_n-T(g_n)\|=0?$$

and remaining problem i dont know how to prove can some one help thank you

(a): Assume $$(f_n)_n$$ is a sequence in $$M$$ such that $$f_n \to f \in C[0,1]$$ uniformly. We claim that $$f \in M$$.

Since uniform convergence implies pointwise convergence, we have $$f(0) = \lim_{n\to\infty} f_n(0) = \lim_{n\to\infty} 0 = 0$$ $$f(1) = \lim_{n\to\infty} f_n(1) = \lim_{n\to\infty} 1 = 1$$ $$f(x) = \lim_{n\to\infty} \underbrace{f_n(x)}_{\in[0,1]} \in [0,1], \quad\forall x \in [0,1]$$ since $$[0,1]$$ is a closed set in $$\mathbb{R}$$. Hence $$f \in M$$ so $$M$$ is a closed set in $$C[0,1]$$.

(b) and (c): For any $$g,h \in M$$ we have $$-1 = 0 - 1\le g(x) - h(x) \le 1 - 0 = 1$$ so $$\|g-h\|_\infty = \sup_{x \in [0,1]}|g(x) - h(x)| \le 1$$

It follows $$\delta(M) \le 1$$. On the other hand, we have $$f, f_n \in M$$ so $$\delta(f,M), \delta(f_n, M) \le \delta(M) \le 1$$

Also plugging in $$t = \frac1{\sqrt[n-1]{n}}$$ gives $$\delta(M) \ge \delta(f,M), \delta(f_n, M) \ge \|f_n-f\|_\infty = \sup_{t \in [0,1]}|t^n - t| = \sup_{t \in [0,1]}|t||t^{n-1} - 1| \ge \frac1{\sqrt[n-1]{n}}\left(1 - \frac1n\right) \xrightarrow{n\to\infty} 1$$ so we conclude $$\delta(f,M) = \delta(f_n, M) = \delta(M) = 1$$.

For (e):

\begin{align} \|g_n - Tg_n\|_\infty &= \|T_ng_n - Tg_n\|_\infty \\ &= \left\|\left(\frac{n-1}n - 1\right)Tg_n + \frac{f_0}{n}\right\|_\infty \\ &= \left\|-\frac1n Tg_n + \frac{f_0}{n}\right\|_\infty \\ &= \frac1n\|f_0 - Tg_n\|_\infty \end{align} To conclude that this converges to $$0$$ we have to know what is $$T$$.

• i tried something on (e) is it right and how to prove (d)...thank you so much – Inverse Problem Mar 24 '19 at 16:03
• @InverseProblem Can you tell me what exactly is $T$? How is $T_n$ defined? – mechanodroid Mar 24 '19 at 16:05
• @mechanodroid....you correct beacuse i also got same doubt let me check again any way thanks for your fast response... – Inverse Problem Mar 24 '19 at 16:07
• is my attempt is correct for (e)? – Inverse Problem Mar 24 '19 at 16:08
• @InverseProblem I'm still not sure what is $T$, I have added my attempt for $(e)$ in the answer. – mechanodroid Mar 24 '19 at 16:18

Your subspace $$M$$ is the intersection of three closed sets $$M = \{ f \in C_{\mathbb{R}}[0,1] : \|f\| \le 1\} \cap \{ f : f(0)=0 \} \cap \{ f : f(1)=1 \}.$$ The first set is closed because it is the closed unit ball of radius $$1$$ in $$C_{\mathbb{R}}[0,1]$$. The second set is closed because it is the inverse image of $$\{0\}$$ under the continuous function $$f\in C_{\mathbb{R}}[0,1] \mapsto f(0)$$. Similarly the third set is closed.

• Sir how to prove remaining problems – Inverse Problem Mar 24 '19 at 1:09
• @InverseProblem : What is $\delta$? – DisintegratingByParts Mar 24 '19 at 1:30
• I defined in my question what is delta – Inverse Problem Mar 24 '19 at 1:32
• @InverseProblem : $M$ is not a subspace. o $\delta(M)$ is not defined, and I don't see how distance relates to either. – DisintegratingByParts Mar 24 '19 at 4:22
• @InverseProblem : I'm just asking you to clarify your notation. I don't like the chat on this forum; it's poorly done. – DisintegratingByParts Mar 24 '19 at 4:42

I'll try to give a more detailed proof of (a) because you'll probably see similar exercises and problems involving function spaces.

Let $$f_n$$ be a sequence in $$M$$ such that $$f_n \to f$$. So let us prove that $$f \in M$$:

• $$f \in C([0,1],\mathbb{R})$$:

Note that with this norm $$f_n \to f$$ means uniform convergence of continuous functions, since by definition: $$\begin{equation*} \|f_n - f\|_{C([0,1],\mathbb{R})} \to 0 \iff\sup_{t\in[0,1]} |f_n(t)-f(t)| \to 0 \end{equation*}$$

It is a well-known fact that uniform limit of continuous functions is also continuous.

• $$f(0) = 0$$ and $$f(1) = 1$$: It should be clear from the sequence.

• $$0 \leq f(t) \leq 1$$ for all $$t\in[0,1]$$:

Suppose that $$f(t_0) < 0$$ for some $$t_0\in[0,1]$$; then there is $$\varepsilon > 0$$ such that $$f(t_0) +\varepsilon < 0$$.

Since $$f_n$$ converges to $$f$$ uniformly, then it also converges pointwise. Hence for $$n > n_0$$ (for $$n_0$$ big enough), we have: $$\begin{gather} |f_n(t_0)-f(t_0)| < \varepsilon \\ -\varepsilon < f_n(t_0)-f(t_0) < \varepsilon \\ f(t_0)-\varepsilon < f_n(t_0) < f(t_0)+\varepsilon < 0\\ \end{gather}$$

This contradicts the fact that $$f_n \in M$$ (that is, $$f_n(t) \geq 0$$ for all $$t\in[0,1]$$). The same reasoning can be applied to prove that $$f(t) \leq 1$$.

Hence $$f \in M$$, and $$M$$ is closed.

Remark 1: Note that we had to check that $$f \in C([0,1],\mathbb{R})$$. It is an important step (albeit most of the time it'll be satisfied).

Remark 2: For convergence of the sequence, generally you need to apply a convergence theorem (uniform limit, Arzelà-Áscoli, Lebesgue's Dominated Convergence, among others). That's one of the reasons you learn them!

• can you know how to solve remaining questions – Inverse Problem Mar 24 '19 at 3:32