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Let $$M=fin C_mathbbR([0,1]): f(0)=0le f(t)le f(1)=1,text for tin [0,1]$$ where $C_mathbbR([0,1])=f:[0,1]to mathbbR:ftext is continuous on [0,1]$ is Banach space with norm $|f|_infty=sup $ . Prove

(a) $M$ is closed subset of $C_mathbbR([0,1]).$

(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 $f_0in M$ . Define $T_n:Mto M$ by $T_n(f)=frac(n-1)T(f)n+fracf_0n, nin mathbbN$.Then $T_n$ is a contraction mapping

(e) if $g_nin M$ is a fixed point of $T_n$ then $lim_nto infty|g_n-T(g_n)|=0$

Here $delta(M)= dim M=sup:x,yin M$ and

$delta (x,M)=sup x-y$

i am trying to prove (a)

for proving (a)

let $x_n$ be a sequence in $M$ such that $x_nto x$

we have to prove that $xin M$

so consider $|x_n-x|_infty=sup $

since $x)nto x$ so $|x_n-x|<epsilon $ this implies $|x_n(t)-x(t)|<epsilon$

from i here how to prove $xin M$

and for proving (e)

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

so $lim_nto infty|g_n-T(g_n)|=lim_nto infty|T_n(g_n)-T(g_n)|=lim_nto infty|(T_n-T)(g_n)|$ from this step can we say ?

$lim_nto infty|g_n-T(g_n)|=0?$

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
$$f(0) = lim_ntoinfty f_n(0) = lim_ntoinfty 0 = 0$$
$$f(1) = lim_ntoinfty f_n(1) = lim_ntoinfty 1 = 1$$
$$f(x) = lim_ntoinfty underbracef_n(x)_in[0,1] in [0,1], quadforall x in [0,1]$$
since $[0,1]$ is a closed set in $mathbbR$. 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
$$-1 = 0 - 1le g(x) - h(x) le 1 - 0 = 1$$
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 $t = frac1sqrt[n-1]n$ gives
$$delta(M) ge delta(f,M), delta(f_n, M) ge |f_n-f|_infty = sup_t in [0,1]|t^n - t| = sup_t in [0,1]|t||t^n-1 - 1| ge frac1sqrt[n-1]nleft(1 - frac1nright) xrightarrowntoinfty 1$$
so we conclude $delta(f,M) = delta(f_n, M) = delta(M) = 1$.

For (e):

beginalign
|g_n - Tg_n|_infty &= |T_ng_n - Tg_n|_infty \
&= left|left(fracn-1n - 1right)Tg_n + fracf_0nright|_infty \
&= left|-frac1n Tg_n + fracf_0nright|_infty \
&= frac1n|f_0 - Tg_n|_infty
endalign
To conclude that this converges to $0$ we have to know what is $T$.

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

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 $f in M$:

• 
  $f in C([0,1],mathbbR)$:

Note that with this norm $f_n to f$ means uniform convergence of continuous functions, since by definition:
beginequation*
|f_n - f|_C([0,1],mathbbR) to 0 iffsup_tin[0,1] |f_n(t)-f(t)| to 0
endequation*

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 $tin[0,1]$:

Suppose that $f(t_0) < 0$ for some $t_0in[0,1]$; 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:
begingather
|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\
endgather

This contradicts the fact that $f_n in M$ (that is, $f_n(t) geq 0$ for all $tin[0,1]$). 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 $f in C([0,1],mathbbR)$. 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!