2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

(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 $T$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ i tried something on (e) is it right and how to prove (d)...thank you so much $\endgroup$ – Inverse Problem Mar 24 '19 at 16:03
  • $\begingroup$ @InverseProblem Can you tell me what exactly is $T$? How is $T_n$ defined? $\endgroup$ – mechanodroid Mar 24 '19 at 16:05
  • $\begingroup$ @mechanodroid....you correct beacuse i also got same doubt let me check again any way thanks for your fast response... $\endgroup$ – Inverse Problem Mar 24 '19 at 16:07
  • $\begingroup$ is my attempt is correct for (e)? $\endgroup$ – Inverse Problem Mar 24 '19 at 16:08
  • $\begingroup$ @InverseProblem I'm still not sure what is $T$, I have added my attempt for $(e)$ in the answer. $\endgroup$ – mechanodroid Mar 24 '19 at 16:18
1
$\begingroup$

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.

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

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!

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ can you know how to solve remaining questions $\endgroup$ – Inverse Problem Mar 24 '19 at 3:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.