Exercise 1

(1)

Suppose \(f\) is continuous and let \(x_n \to x\) be a sequence. For any \(\epsilon>0\), \(B_Y(f(x), \epsilon)\) the \(\epsilon\)-ball around \(f(x_n)\) in \(Y\) is a neighbourhood and as \(f\) is continuous there is a \(\delta\) such that \(f(B_X(x,\delta)) \subseteq B_Y(f(x), \epsilon)\). As \(x_n\to x\) there is some \(N\) such that \(n\geq N \implies d(x_n, x) < \delta\), therefore \(f(x_n) \to f(x)\).

Conversely suppose that \(f\) fails to be continuous at \(x\in X\), then there exists a neighbourhood \(M\) of \(f(x)\) such that for all neighbourhoods of \(x\) \(N\) we have \(f(N) \nsubseteq M\). Fix \(x, M\) as above and for each \(n\), we choose a point \(x_n \in B(x, \frac1n)\) such that \(f(x_n) \notin M\). Then this sequence \(x_n \to x\) but each \(f(x_n) \notin M\) so \(f(x_n) \to f(x)\) is impossible.

(2)

For any \(U\subseteq Z\) we have \((g\circ f)^{-1}(U) = g^{-1}(f^{-1}(U))\). So in particular, when \(U\) is open \((g\circ f)^{-1}(U)\) is also open by continuity of \(f\) and \(g\).

Exercise 2

(1)

Positivity, definiteness and symmetry easily carry from the fact that \(X_1, X_2\) are metric spaces. Triangle inequality also carries over, consider any \(x_1,y_1,z_1\in X_1\) and \(x_2,y_2,z_2\in X_2\), \[\begin{align*} d((x_1, x_2), (z_1, z_2)) &= d_1(x_1, z_1) + d_2(x_2, z_2) \\ &\leq d_1(x_1, y_1) + d_1(y_1,z_1) + d_2(x_2, y_2) + d_2(x_2,z_2) \\ &= d((x_1, x_2), (y_1,y_2)) + d((y_1,y_2), (z_1,z_2)) \end{align*}\]

(2)

Let \((X,\tau)\) denote the product topology and \((X,\sigma)\) denote the one induced by the metric \((X,d)\). We know that \[ \mathcal{B} = \left\{ B(x_1,r_1)\times B(x_2,r_2): x_1\in X_1, x_2\in X_2, r_1,r_2>0 \right\} \] is a base for \((X,\tau)\). Fix \(B(x_1,r_1)\times B(x_2,r_2)\in\mathcal{B}\) and it suffices to show that this set is open in \((X,\sigma)\). Let \((y_1,y_2) \in B(x_1,r_1)\times B(x_2,r_2)\) choose \(0 < \delta < \min\left\{r_1 - d(x_1,y_1), r_2 - d(x_2, y_2)\right\}\) and we can see that \(B\left((y_1,y_2), \delta\right) \subseteq B(x_1,r_1)\times B(x_2,r_2)\) which shows that \(B(x_1,r_1)\times B(x_2,r_2)\) is open in \((X,\sigma)\).

Conversely fix \((x_1,x_2)\in X, r>0\) and we want to show that \(B((x_1,x_2), r)\) is open in the product topology. Let \((y_1,y_2)\in B((x_1,x_2), r)\), and let \[m = r - d_1(x_1,y_1) - d_2(x_2,y_2) > 0\] choose \(\delta < \frac{m}2\) and we can see that \(B(y_1,\delta)\times B(y_2,\delta) \subseteq B((x_1,x_2),r)\), so \(B((x_1,x_2),r)\) is also open in \(\tau\).

Exercise 3

(1)

To show that \(\phi\) is Lipschitz we want to find an upper bound for \[ \frac{\left\lVert\phi\lambda - \phi\mu\right\rVert}{\left\lvert\lambda -\mu\right\rvert} = \frac{\max_{x\in[0,1]}\left\lvert\lambda - \mu\right\rvert x^2}{\left\lvert\lambda -\mu\right\rvert} = 1 \]

(2)

The closed unit ball is not compact. We just need to see that \(\mathscr{C}^0\) is an infinite-dimensional vector space. We consider the set \(\left\{x\mapsto 1,x\mapsto x,x\mapsto x^2,\dots\right\}\) which is an infinite set that is linearly independent.

(3)

Let \(g\in\overline{\mathscr{F}}\setminus\mathscr{F}\), then every neighbourhood of \(g\) contains some \(f\in\mathscr{F}\). In particular for all strictly positive integer \(n\) choose \(f_n \in \mathscr{F}\) such that \(\left\lVert g-f_n\right\rVert < \frac1n\). Now for any \(x,y\in[0,1]\), by triangle inequality for all \(n\) \[\begin{align*} \left\lvert g(y)-g(x)\right\rvert &\leq \left\lvert g(y)-f_n(y)\right\rvert + \left\lvert f_n(y)-f_n(x)\right\rvert + \left\lvert f_n(x)-g(x)\right\rvert \\ &< \frac2n + \left\lvert f_n(y)-f_n(x)\right\rvert \\ &\leq \frac2n + \left\lvert y-x\right\rvert \end{align*}\] so it turns out that \(g\in\mathscr{F}\). Hence \(\mathscr{F}\) is closed.

(4)

Choose any sequence \(f_n\) in \(\mathscr{F}_1\), we see that the sequence \((f_n)\) is uniformly bounded as \(\left\lVert f_n\right\rVert_0 \leq 1\). The sequence is also uniformly equicontinuous as each \(f_n\) is 1-Lipschitz. By Arzelà–Ascoli theorem \((f_n)\) has a convergent (wrt \(\left\lVert\cdot\right\rVert\)) subsequence, hence \(\mathscr{F}_1\) is sequentially compact, and therefore compact (in the metric space \(\mathscr{C}^0\)).

FIXUP: also state that the convergent subsequence converges to some function in \(\mathscr{F}_1\).

(5)

Choose any \(f\in\mathscr{F}_2\), if \(f\not\equiv 0\) since \(f\) is continuous, intermediate value property gives us \(c\in[0,1]\) with \(f(c) = 0\). Then as \(f\) is 1-Lipschitz we still have a uniform bound \(\left\lVert f\right\rVert_0\leq 1\). Now repeat the argument in (4).

(6)

For any \(f\in\mathscr{F}_3\) by mean value theorem for any \(x,y\in[0,1]\), we have \[ \left\lvert f(y) - f(x)\right\rvert\leq\left\lvert y-x\right\rvert \] due to the uniform bound on \(f'\). Repeat the argument in (4).

FIXUP: this is wrong

Exercise 4

Let \(U^c\) denote \(X\setminus U\), note that \(U^c\) is closed and disjoint from \(K\) as assumed. Choose \(\delta>0\) such that \(d(x,U^c) \geq\delta\) for all \(x\in K\). Define \(f\) as \[ f(x) = \frac{\max\left\{d(x,U^c),\delta\right\}}\delta \] and it follows from our choice of \(\delta\) that \(f=0\) on \(X\setminus U\) and \(f=1\) on \(K\).

We first show that \(d(-,U^c)\) is Lipschitz. Let \(x,y\in X\) and for all \(z\in U^c\), \[\begin{align*} d(x,U^c) &\leq d(x,z) \\ &\leq d(x,y) + d(y,z) \\ d(x,U^c) - d(x,y) &\leq d(y,z) \end{align*}\] therefore \(d(x,U^c) - d(y,U^c) \leq d(x,y)\). Symmetric argument shows \(d(y,U^c) - d(x,U^c) \leq d(x,y)\).

To see that \(f\) is Lipschitz fix \(x,y\in X\) and consider the following cases,

Both \(d(x,U^c)\) and \(d(y,U^c)\) bounded below by \(\delta\)

then \(\left\lvert f(y) - f(x)\right\rvert = 0 \leq d(x,y)\).

Both \(d(x,U^c)\) and \(d(y,U^c)\) less than \(\delta\)

by earlier result \[\left\lvert f(y) - f(x)\right\rvert = \frac1\delta\left\lvert d(y,U^c) - d(x,U^c)\right\rvert \leq \frac{d(x,y)}{\delta}.\]

WLOG \(d(x,U^c)\geq\delta\) and \(d(y,U^c)<\delta\)

\[ \left\lvert f(y)-f(x)\right\rvert = \frac1\delta\left\lvert d(y,U^c)-\delta\right\rvert \leq \frac1\delta\left\lvert d(y,U^c)-d(x,U^c)\right\rvert\leq \frac{d(x,y)}\delta. \]

Hence \(f\) is \(\frac1\delta\)-Lipschitz as desired.