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

Question

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

Answer 1

(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 $T$.

Answer 2

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.

Answer 3

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!