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Let’s start with connectedness. One can attempt to proceed via the open set/clopen set characterisation of connected spaces, but it’s very difficult to determine which subsets of \(X\) are open and which are closed. Rather, it’s far easier to consider the fact that path-connected metric spaces are connected!

Thus, given two functions \(f, g\in X\), we wish to show there is a continuous path \(F:\left[ 0, 1 \right]\to X\) such that \(F(0)=f\) and \(F(1)=g\). There is a rather naïve thing to try, and it ends up working. For \(t\in \left[ 0, 1 \right]\), define the function \(F(t)\) via \[F(t)(x) = (1-t)\cdot f(x) + t \cdot g(x). \] We first check that \(F(t)\) is in \(X\): clearly \(F(t)(0)=0\) since \(f(0)=g(0)=0\). If \(x,y\in \left[ 0,1 \right]\), we have \[\left\lvert F(x)-F(y) \right\rvert \leq \left( 1-t \right)\left\lvert f(x)-f(y) \right\rvert + t \left\lvert g(x) - g(y) \right\rvert\leq \left\lvert x-y \right\rvert\] by expanding and spamming the triangle inequality. Finally, to show continuity, fix \(\epsilon>0\). If \(t, s\in \left[ 0, 1 \right]\), we have \[\begin{align*}d\left( F(t), F(s) \right) &= \sup _{x\in [0, 1]} \left\lvert (s-t) f(x) + (t-s) g(x) \right\rvert \\&= \sup _{x\in \left[ 0, 1 \right]}\left\lvert s-t \right\rvert \cdot \left\lvert f(x) - g(x) \right\rvert \\&= \left\lvert s-t \right\rvert d\left( f, g \right). \end{align*}\] Thus, if \(\delta=\frac{\epsilon}{d\left( f, g \right)}\), then \(\left\lvert t-s \right\rvert< \delta\) implies \(d\left( F(t), F(s) \right)< \epsilon\). Of course, if \(d\left( f, g \right)=0\), then \(F\) is a constant function, which is continuous anyways. Thus \(X\) is path-connected, and it follows that \(X\) is connected.

For completeness, we note that \(C\left( \left[ 0,1 \right] \right)\) is complete with respect to \(d\). Thus, it suffices to show that \(X\) is closed in this larger space, as closed subsets of complete metric spaces are complete.

Let \(\left\lbrace f_n \right\rbrace\) be a sequence of functions in \(X\) converging uniformly to a limit \(f\) in \(C\left( \left[ 0,1 \right] \right)\). Since \(f_n(0)\to f(0)\) and \(f_n(0)=0\) for all \(n\), \(f(0)=0\). Likewise, if \(x, y\in \left[ 0, 1 \right]\), we have \[\left\lvert f(x)-f(y) \right\rvert = \lim _{n\to\infty}\left\lvert f_n(x)-f_n(y) \right\rvert \leq \lim _{n\to\infty} \left\lvert x-y \right\rvert=\left\lvert x-y \right\rvert.\] Here, we used continuity of the absolute value. We conclude that \(f\in X\). Since \(X\) contains all of its limit points, it is closed, and thus \(X\) is complete. \(\square\)

The big idea behind this problem is knowing the different ways to characterise continuous functions. First, the “only if” direction is trivial — the restriction of any continuous function will remain continuous.

We have that \(F\) is continuous if and only if for all sequences \(\left\lbrace f_n \right\rbrace\subseteq \ell^\infty(\mathbb{Z})\) converging to some limit \(f\), then \(F\left( f_n \right)\) also converges to \(F(f)\). So, we need to find a compact subset of \(\ell^\infty\left( \mathbb{Z} \right)\) containing \(\left\lbrace f_n \right\rbrace\) and containing \(f\).

Let \(S=\left\lbrace f_n : n\in \mathbb{N} \right\rbrace\cup \left\lbrace f \right\rbrace\). We claim \(S\) is compact. This is because if \(\left\lbrace g_m \right\rbrace\) is any sequence in \(S\), then either \(g_m\) repeats an element of \(S\) infinitely often or \(g_m\) contains a subsequence of \(\left\lbrace f_n \right\rbrace\). In either case, \(g_m\) has a converging subsequence whose limit lies in \(S\).

Since \(\left. F\right\rvert_{S}\) (the restriction to \(S\)) is continuous by assumption, \(F\left( f_n \right)\to F(f)\), as desired. \(\square\)

Consider the function \(F:\left[ 0,1 \right]\to \mathbb{R}\) given by \(F(t) = f\left( (1-t)x+ty \right)\), which is well-defined since \(X\) is convex. Then \(F(1) = f(y)\) and \(F(0)=f(x)\). We will compute the derivative, then appeal to the mean value theorem.

By the definition of the derivative, \[\lim _{\left\lVert \vec h \right\rVert\to 0} \frac{f\left(\vec a+\vec h\right)-f\left(\vec a\right)-\nabla f\left( \vec a \right)\cdot \vec h}{\left\lVert \vec h \right\rVert}=0.\] Now replace \(\vec a\) with \((1-t)x+ty\) (dropping the vector hats for wrist health) and \(\vec h\) with \(h\left( y-x \right)\), where \(h\) is a real number instead of a vector: \[\lim _{h\to 0} \frac{f((1-t-h)x+(t+h)y-f((1-t)x+ty) - \nabla f\left( (1-t)x+ty\right)\cdot\left( h(y-x) \right)}{ \left\lvert h \right\rvert\cdot \left\lVert y-x \right\rVert} = 0.\] We may replace \(\left\lvert h \right\rvert\) with \(h\) here since the sign doesn’t matter — the resulting limit is \(0\) either way. Since \(\left\lVert y-x \right\rVert\) is just a constant, we can multiply through by \(\left\lVert y-x \right\rVert\). Thus, our result after rearranging is \[\lim _{h\to 0}\frac{f\left( (1-t-h)x+(t+h)y \right)-f\left( (1-t)x+ty \right)}{h} = \nabla f\left( (1-t)x+ty \right)\cdot (y-x).\] But the left-hand-side is just \(F’(t)\)! By the mean value theorem, there exists a time \(t\) such that \[f(y)-f(x) = F(1)-F(0)= F’(t) = \nabla f\left( (1-t)x+ty \right)\cdot (y-x).\] Taking \(z=(1-t)x+ty\) concludes the proof. \(\square\)