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Show that the subspace $S=\{f : f\text{ is continuous and }f(0)=f(1)\} $ equipped with the infinity norm is Banach.

(It's enough to show that it's closed)

My try to solve it

Take $f_n$ sequence of functions such that $f_n$ converges to $f$ in the norm Then$ |f_n(x)-f(x)| \rightarrow 0 $ for all $ x$

Letting $x=0$

$|f_n(0)-f(0)| \rightarrow0$

$|f_n(1)-f(0)| \rightarrow0$

But I think this will not lead me to the answer.

Can we find a finite set of bases that spans this set so we can say it's finite dimensional?

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  • $\begingroup$ You certainly can't say the set is finite-dimensional - for any $n$ you can have $f()$ take on arbitrary values at $i\cdot 2^{-n}$ for $1\leq i\leq 2^n-1$ and (continuously) linearly interpolate between them, so the dimension is at least $2^n-1$. Since $n$ was arbitrary... $\endgroup$ – Steven Stadnicki Nov 30 '16 at 23:22
  • $\begingroup$ No, that will lead to the answer: $f(0)= \lim f_n(0) = \lim 0 = 0.$ $\endgroup$ – zhw. Nov 30 '16 at 23:23
  • $\begingroup$ It is sufficient to show the complement is open. A ball in our space is {g| max{|(f-g)(x)|} is less than epsilon}. Assume f(0) is not f(1). $\endgroup$ – Jacob Wakem Dec 1 '16 at 0:08
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You said it's enough to show it's closed; that's true. Note now that $f(0)=f(1)$ is equivalent to $f(0)-f(1)=0$. So your set is the preimage of $\{ 0 \}$ under $G(f)=f(0)-f(1)$. Can you conclude?

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  • $\begingroup$ Since it's continuous linear operator so kerlG is closed ? $\endgroup$ – Sara Suradi Nov 30 '16 at 23:26
  • $\begingroup$ @user283366 The use of linearity is a bit overkill, but yeah. $\endgroup$ – Ian Nov 30 '16 at 23:37
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@Ian's answer already shows a nice way to finish your proof. Another approach to see $S$ is a Banach space is to realize that $S$ is isometric to $C(T)$, where $C(T)$ denotes the Banach space of continuous functions on the torus $T:=\mathbb{R}\big/\mathbb{Z}$. To prove it just define $\varphi:S\to C(T)$ by $$ \varphi(f):=\tilde{f},\hspace{.5cm}\tilde{f}(\bar{x}):=f(\{x\}), $$ where $\{x\}$ denotes the fractional part of $x$. The condition $f(0)=f(1)$ in the definition of $S$ guarantees that this is well-defined. As $\varphi$ is trivially a isometry we have that $S$ is isometric to $C(T)$, and therefore a Banach space.

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