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$\textbf{Ex. 1.2.16.}$ Let $f,g: [0,1] \longrightarrow \mathbb{R}$ be continuous and $f(t) < g(t)$ for all $t \in [0,1]$. Consider the set

$U := \{ h \in \mathcal{C}[0,1] \ ; \ f(t) < h(t) < g(t), t \in [0,1] \}$

in the space $X = \left( \mathcal{C}[0,1], || \ ||_{\infty} \right)$. Is $U$ a ball in $X$? If not, can you think of a condition of $f$ and $g$ will ensure that the set $U$ is an open ball?

My attempt:

I thought if even I define $\epsilon := \inf_{h \in \mathcal{C}[0,1]} \ \{ \sup_{t \in [0,1]} \{ |(g-h)(t)| \}, \sup_{t \in [0,1]} \{ |(f - h)(t)| \} \}$, I won't conclude that have that $U$ is an open ball, but I don't know how to show this.

Anyone can help me? Thanks in advance!

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George
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2 Answers2

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$U$ is a ball if and only if $f-g=\text{constant function}$!

Pr00f: By translating $U$ toward the direction $-\frac{g+f}{2}$, we may assume $g >0$ and $f= -g.$ Therefore in this case $U = \{ h \in C[0,1] ~|\quad -g < h < g \}. $ Since $U$ is symmetric about origin, then $U$ is a ball if and only if $U$ is a ball centered origin. This means there is a positive real number, say $r >0$ such that $$U = \{ h \in C[0,1] ~|\quad -r < h < r \}$$

A simple comparing between later set and $U$, we get $g(x) =r$ for all $x \in [0,1],$ (note that in this argument the compactness of $[0.1]$ is the key ingredient)

Red shoes
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  • Nothing wrong but I don't understand the last sentence,in brackets. I think it has typos. – DanielWainfleet May 28 '17 at 03:10
  • Yes it is a typos,, also there is a typos in first sentence I 'll correct it. thanks – Red shoes May 28 '17 at 04:15
  • I don't think the compactness of [0,1] is involved. – DanielWainfleet May 28 '17 at 13:50
  • The compactness of $[0,1]$ implies that $U$ is open. By contrast, on $C[\mathbb R]$ with $ |f-g|=\min (1, sup_{x\in \mathbb R}|f(x)-g(x)|),$ and $f(x)=e^{-x}$ and $g(x)=0,$ the set ${h:g<h<f}$ is not open. – DanielWainfleet May 28 '17 at 14:26
  • @nonlinearthought, I have two questions about this proof:

    1. When you translated $U$, you obtained $\frac{f(t) - g(t)}{2} < h(t) - \frac{g(t)+f(t)}{2} < \frac{g(t) - f(t)}{2}$, but in your proof you put that $U = { h \in \mathcal{C}[0,1] \ | \ -g < h < g }$. Is your $g$ my $\frac{g(t) - f(t)}{2}$ and your $h$ my $h(t) - \frac{g(t)+f(t)}{2}$?

    2. Is the $r := \sup_{t \in [0,1]} g(t)$? I think you need the compactness of [0,1] and the continuity of $g$ to ensure the existence of $r$.

     – George May 30 '17 at 23:50
  • @George First Yes. Second, Actually $$U = { h \in C[0,1] ~|\quad -r< h < r } = { h \in C[0,1] ~|\quad -g < h <g } \quad \text{iff} ~ r=g(x) $$

    Third, In fact, we need compactness and continuity in order to $C[0,1]$ be well-defined!

     – Red shoes May 31 '17 at 00:01
  • The definition of a ball states that there is a positive real number fixed $r > 0$, but, by the way you defined $r$, you didn't define a number fixed, did you, did you? I think you defined a number fixed $r_x > 0$ for each $x \in [0,1]$ – George May 31 '17 at 01:05
  • @George of course I did! look at third line, I said $r$ is a positive real number, not a function. – Red shoes May 31 '17 at 01:13
  • Oh, sorry, you put a condition in $f$ and $g$ in order to $U$ be an open ball, I understood now, thanks! – George May 31 '17 at 01:57
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Let $B(j,r)=\{h\in C[0,1]: \|h-j\|<r\}$ for $r>0$ and $j\in C[0,1].$

Suppose there exist $x,x'$ with $A=f(x)-g(x)\ne f(x')-g(x')=B.$

(I). If $r> B/2$ and $j(x')\geq (f(x')+g(x'))/2$ there exists $h\in B(j,r)$ with $h(x')=j(x')+B/2\geq f(x')$. So $h\not\in U.$

(II). If $r>B/2$ and $j(x')\leq (f(x')+g(x'))/2$ there exists $h\in B(j,r)$ with $h(x')=j(x')-B/2\leq g(x')$. So $h\not \in U.$

(III). If $r\leq B/2$ and $j(x)\geq (f(x)+g(x))/2$ there exists $h\in U$ with $h(x)=j(x)-B/2$. So $h\not \in B(j,r).$

(IV). If $r\leq B/2$ and $j(x)\leq (f(x)+g(x))/2$ there exists $h\in U$ with $h(x)=j(x)+B/2$. So $h\not \in B(j,r).$

.......................In all cases we have $U\ne B(j,r).$......................

We can also see it this way: For all $y\in [0,1]$ we have $\sup \{h_1(y)-h_2(y): h_1,h_2\in B(j,r)\}=2r.$ Then:

(I'): If $r>B/2$ there exist $h_1,h_2 \in B(j,r)$ with $h_1(x')-h_2(x')>B,$ so $h_1,h_2$ can't both belong to $U.$

(II'): If $r\leq B/2$ there exist $h_1,h_2\in U$ with $h_1(x)-h_2(x)>B\geq 2r$ so $h_1,h_2$ can't both belong to $B(j,r).$

Remark. The set $U$ is open: For $h\in U,$ let $a=\min_{x\in [0,1]}h(x)-g(x)$ and $b=\min_{x\in [0,1]}f(x)-g(x).$ Then $c=\min (a,b)$ is positive, and $B(h,c/2)\subset U.$

DanielWainfleet
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  • I think there is a typo in (I) and (II), please confirm : In (I), $h(x') \geq g(x')$ and so $h \not \in U$. In (II), $h(x') \leq f(x')$ and so $h \not \in U$. – P-addict Mar 14 '21 at 12:01