Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 2

A norm on a real vector space $X$ is a map $\Vert \cdot \Vert :X\to [0,\infty )$ such that, for $x,y\in X$ and $\lambda \in \mathbb{R}$,The last property is the triangle inequality; the middle property is absolute homogeneity of a norm. The induced metric isPositivity and separation follow from the norm axioms, symmetry from absolute homogeneity of a norm with $\lambda =-1$, and the triangle inequality from $x-z=(x-y)+(y-z)$.

First verify that $V_0$ is a vector space. Sums and scalar multiples preserve continuity and the condition at zero. For every partition of an interval $D$, the triangle inequality and absolute homogeneity of a norm of the absolute value implyTaking suprema yieldsFor $\lambda =0$ the second equality follows directly from $V(0)=0$. In particular, finite variation is preserved, so $V_0$ is closed under these operations.

Nonnegativity is immediate. For any $t\in [0,1]$, use the partition of an interval $(0,t,1)$, omitting repeated endpoints if necessary, to getThus $V(f)=0$ forces $f=0$, and conversely $f=0$ has zero total variation of a function. These observations prove that $\Vert f\Vert =V(f)$ is a norm on $V_0$.

Applying the same bound to $f_n-f$ givesTherefore convergence in the anchored variation norm implies uniform convergence on $[0,1]$. The anchoring condition $f(0)=0$ removes the nonzero constant functions which otherwise have zero total variation of a function.

Both properties are preserved by a continuous map.

For sequential compactness, take any sequence $(y_n)$ in $f(X)$ and choose $x_n\in X$ with $f(x_n)=y_n$. If $X$ is sequentially compact, some subsequence satisfies $x_{n_k}\to x\in X$. By continuity, $y_{n_k}=f(x_{n_k})\to f(x)\in f(X)$. Hence $f(X)$ is a sequentially compact space.

For connectedness, suppose instead that $f(X)=U\cup V$ is a separation into disjoint nonempty relatively open sets. Then $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint nonempty open sets of $X$ whose union is $X$, contradicting connectedness. Therefore $f(X)$ is a connected space.

We prove uniform continuity by contradiction. If it fails, there is ${\epsilon }_0>0$ such that, for every $n\ge 1$, we can choose $x_n,y_n\in X$ withBy sequential compactness, pass to a subsequence with $x_{n_k}\to x\in X$. The triangle inequality givesBy continuity, both $f(x_{n_k})$ and $f(y_{n_k})$ converge to $f(x)$. A second triangle inequality therefore givescontradicting the lower bound. Thus the requested $\delta$, independent of $x,y$, exists for every $\epsilon >0$.

We first show that a continuous injection $j:S^1\to S^1$ must be surjective. Otherwise its image misses a point $q$. Stereographic projection identifies $S^1\setminus \{q\}$ with $\mathbb{R}$, giving a continuous injection $u:S^1\to \mathbb{R}$. The extreme value theorem supplies distinct points $p_{-},p_{+}$ where $u$ attains its minimum $m$ and maximum $M$, with $m<M$ because $u$ is injective.

The two circular arcs from $p_{-}$ to $p_{+}$ each contain a point at which $u=(m+M)/2$, by the intermediate value theorem. Those points are distinct because they lie in the disjoint interiors of the two arcs. This contradicts injectivity, proving the claim about $j$.

Now suppose $F:S^2\to S^1$ were a continuous injection. Its restriction to the equatorial circle $E=\{(x,y,0):x^2+y^2=1\}$ would be a continuous injection from $S^1$ to itself, hence $F(E)=S^1$. The north pole $N=(0,0,1)$ is outside $E$, but $F(N)$ must equal $F(e)$ for some $e\in E$. This contradicts injectivity. Therefore

Assume condition (1). For fixed $a$, the supremumis nondecreasing as $r$ increases. Therefore its limit as $r\downarrow 0$ exists in the extended real numbers and equals ${\inf }_{r>0}{M}_a(r)$. Because $a\in B(a,r)$, it is at least $f(a)$. Condition (1) says precisely that this limit is $f(a)$.

It follows that for every $\epsilon >0$ there is $r>0$ with $M_a(r)<f(a)+\epsilon$. Fix $C\in \mathbb{R}$ and $a$ with $f(a)<C$. Taking $\epsilon =C-f(a)$ produces an open ball on which $f(x)<C$. Thus $\{x:f(x)<C\}$ is an open set, and its complementThis proves (1) implies (2). The property being characterized is upper semicontinuity.

Assume condition (2). Fix $a\in X$ and $\epsilon >0$. The superlevel setis a closed set and does not contain $a$. Its complement is therefore an open neighbourhood of $a$, so some open ball $B(a,r)$ avoids $F$. Consequently $f(x)<f(a)+\epsilon$ throughout that ball, givingThe same bound holds for every smaller radius. Taking $r\downarrow 0$ and then $\epsilon \downarrow 0$ yieldsTogether with the previous implication this proves the equivalence. A supremum may equal $f(a)+\epsilon$ even when every individual value is strictly smaller; the non-strict estimate here handles that possibility.

If $f$ is continuous at $a$, then for every $\epsilon >0$ there is $r_0>0$ such that $d(x,a)<r_0$ implies $|f(x)-f(a)|<\epsilon$. In particular, for $0<r\le r_0$,The squeeze theorem giveswhich implies condition (1). Thus continuity implies upper semicontinuity.

Apply the local bound from condition (1) separately to $f$ and $-f$. For any $a\in X$ and $\epsilon >0$, there are $r_{+},r_{-}>0$ such thatWith $\delta =\min \{r_{+},r_{-}\}$, both inequalities hold when $d(x,a)<\delta$, soThis is exactly continuity at $a$. Since $a$ was arbitrary, $f$ is continuous on $X$: upper semicontinuity of $f$ and of $-f$ is equivalent to simultaneous upper semicontinuity and lower semicontinuity of $f$.