Flashback to Undergrad Analysis (2012)

As I leisurely browsed through my old emails, I ran into a short sequence of them from a professor of a class I took in the early days of 2012. It was titled "Analysis 2", and I recall it being, without a shadow of a doubt, the most demanding one I took as an undergrad, by far. He would habitually send us polished proofs of claims we discussed in class, and as I browsed through them, although simple from my current standpoint, it escapes me how he expected an undergrad to produce them during an exam!

I picked two claims which I liked from his list and want to share them and their abridged proofs here.

 

Functions discontinuous at countably many points can be Riemann integrable.

 

Let $f$ be such that $f\left(x\right)=1$ if $x=n^{-1}$ for some $n\in\mathbb N$ and $f\left(x\right)=0$ otherwise.

Claim: $f$ is Riemann integrable on $\left[0,1\right]$.

Proof: Clearly, $L\left(f,P\right)=0$. Let $P_N=\left\{0,1/N,x_1,x_2,\dots,x_{N^2-N}\right\}$, where $x_k=1/N+k/N^2$. Then $U\left(f,P_N\right)=2/N$. Let $N>2/\varepsilon$. Then 

$U\left(f,P_N\right)-L\left(f,P_N\right)=2/N<\varepsilon$. Moreover, as a bonus fact, it follows that $\int_0^1f=0$. $\blacksquare$

 

Moving limits into integrals is a delicate matter.

 

Let $g_n,g:\left[0,1\right]\to\mathbb R$ be integrable and bounded by $M$, and let $g_n\to g$ pointwise.

Claim: If $g_n\to g$ uniformly on any $\left[\delta,1\right]$, where $\delta\in\left(0,1\right)$, then $\lim\int_0^1g_n=\int_0^1g.$

Proof: Uniform convergence implies that $\left|g_n\left(x\right)-g\left(x\right)\right|<\varepsilon/2$ whenever $x\in\left[\delta,1\right]$ and $n\geq N$. Therefore,
$$\begin{align}\left|\int_0^1g_n-\int_0^1g\right|&=\left|\int_0^1\left(g_n-g\right)\right|\\&=\left|\int_0^\delta\left(g_n-g\right)+\int_\delta^1\left(g_n-g\right)\right|\\&\leq\int_0^\delta\left|g_n-g\right|+\int_\delta^1\left|g_n-g\right|\\&<2M\delta+\frac\varepsilon2.\end{align}$$Letting $\delta=\varepsilon/4M$ yields the desired result. $\blacksquare$

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The Regular Representation

I am taking a class this semester in which we are studying representation theory on infinite groups. We are currently focusing on compact groups. This class has so far been manageable yet incredibly abstract.

Let $G$ be a compact, metrizable group and, for $g\in G$, define $\lambda_g:L^2(G, m)\to L^2(G, m)$ by $\xi\mapsto\xi(g^{-1}\cdot)$, where $m$ is the Haar measure such that $m(G)=1$.

Claim: $\lambda_g$ is a unitary operator.

Proof: Let $\xi\in L^2(G,m)$. Then

$$\int_G|(\lambda_g(\xi))(h)|^2\,dm(h)=\int_G|\xi(g^{-1}h)|^2\,dm(h)=\int_G|\xi(h)|^2\,dm(h)<\infty.$$
Thus $\lambda_g$ is well defined.

Observe that

$$\lambda_{g^{-1}}(\lambda_g(\xi))=\lambda_{g^{-1}}(\xi(g^{-1}\cdot))=\xi(g^{-1}g\cdot)=\xi=\xi(gg^{-1}\cdot)=\lambda_g(\xi(g\cdot))=\lambda_g(\lambda_{g^{-1}}(\xi)).$$
Thus $\lambda_g$ is bijective.

Let $\eta\in L^2(G,m)$ and let $c\in\mathbb C$. Then

$$\lambda_g(c\xi+\eta)=c\xi(g^{-1}\cdot)+\eta(g^{-1}\cdot)=c\lambda_g(\xi)+\lambda_g(\eta).$$
Thus $\lambda_g$ is linear.

We showed above that $\|\lambda_g\xi\|=\|\xi\|$. Thus $\lambda_g$ is bounded.

Observe that

$$\begin{align*}\langle\lambda_g(\xi),\lambda_g(\eta)\rangle&=\int_G(\lambda_g(\xi))(h)\overline{(\lambda_g(\eta))(h)}\,dm(h)\\&=\int_G\xi(g^{-1}h)\overline{\eta(g^{-1}h)}\,dm(h)=\int_G\xi(h)\overline{\eta(h)}\,dm(h)=\langle\xi,\eta\rangle.\end{align*}$$
Thus $\lambda_g$ preserves the inner product.

Hence, $\lambda_g$ is a unitary operator. $\blacksquare$

Claim: $\lambda:G\to\mathcal U(L^2(G,m))$ defined by $g\mapsto\lambda_g$ is a continuous, unitary representation.

Proof: Let $g,h\in G$. Then

$$(\lambda(gh))(\xi)=\lambda_{gh}(\xi)=\xi(h^{-1}g^{-1}\cdot)=\lambda_g(\xi(h^{-1}\cdot))=\lambda_g(\lambda_h(\xi))=(\lambda(g)\circ\lambda(h))(\xi).$$
Thus $\lambda$ is a homomorphism.

Let $\xi,\eta\in L^2(G,m)$, define $\varphi:G\to\mathbb C$ by $g\mapsto\langle\lambda(g)\xi,\eta\rangle$, and let $\{g_n\}$ be a sequence in $G$ converging to $g$. Then

$$\begin{align*}\lim|\varphi(g_n)-\varphi(g)|&=\lim|\langle\lambda(g_n)\xi,\eta\rangle-\langle\lambda(g)\xi,\eta\rangle|\\&=\lim|\langle\lambda(g_n)\xi-\lambda(g)\xi,\eta\rangle|\\&\leq\lim\|\lambda(g_n)\xi-\lambda(g)\xi\|\|\eta\|\\&=\lim\|\xi(g_n^{-1}\cdot)-\xi(g^{-1}\cdot)\|\|\eta\|\\&=0.\end{align*}$$
Thus $\varphi$ is continuous.

Hence, $\lambda$ is a continuous, unitary representation called the regular representation. $\blacksquare$

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Example of a Unital Banach Algebra

While studying spectral theory to prepare for a rigurous treatment of $C^*$-algebras, I ran into the following example of a unital Banach algebra:

Let $S$ be a set. Then $\ell^\infty\left(S\right)$, the space of bounded functions on $S$ taking values in $\mathbb C$, is a unital Banach algebra with respect to the supremum norm. The operations are defined pointwise.

It is clear that $\ell^\infty\left(S\right)$ is an algebra, that $1:S\to\mathbb C$ defined by $s\mapsto1$ is the unit, and that $\left\|1\right\|=1$. Moreover, let $f,g\in\ell^\infty\left(S\right)$. Then

$$\left\|fg\right\|=\sup\left|f(x)g(x)\right|\leq\sup\left|f(x)\right|\sup\left|g(x)\right|=\left\|f\right\|\left\|g\right\|.$$

Therefore, $\ell^\infty\left(S\right)$ is a unital normed algebra.

To show completeness, let $\left\{f_n\right\}$ be Cauchy. Then $\left\{f_n\left(x\right)\right\}$ is Cauchy for every $x\in S$ and thus convergent to some $f\left(x\right)$. Define $f:S\to\mathbb C$ by $x\mapsto f\left(x\right)$ and let $\varepsilon>0$. Note that $\left\{f_n\right\}$ is uniformly Cauchy, which implies that it converges uniformly to $f$. Therefore, there is an $N$ such that $\left\|f_n-f\right\|\leq\varepsilon/2<\varepsilon$ whenever $n\geq N$. Moreover, $\left\|f\right\|\leq\left\|f_n-f\right\|+\left\|f_n\right\|<\varepsilon+\left\|f_n\right\|$.

It is a highly instructive exercise to show that a sequence of functions into a Banach space is uniformly convergent if and only if it is uniformly Cauchy.

Let $\Omega$ be a topological space. Then $C_b\left(\Omega\right)$, the set of continuous, bounded functions on $\Omega$ taking values in $\mathbb C$, is a closed subalgebra of $\ell^\infty\left(\Omega\right)$. To see this, let $\left\{f_n\right\}$ be a sequence in $C_b\left(\Omega\right)$ converging to $f\in\ell^\infty\left(\Omega\right)$. Then $\left\{f_n\right\}$ converges uniformly to $f$, which implies that $f$ is continuous. If $\Omega$ is compact, then $C(\Omega)=C_b(\Omega)$.

It is also an instructive exercise to show that if a sequence of continuous functions from a topological space to a normed space is uniformly convergent, then its limit is continuous. The stronger hypothesis that the functions take values in a metric space is not much more difficult to show. In this case, recall that $f:X\to\left(Y,\rho\right)$ is bounded if there are $y\in Y$ and $M\geq0$ such that $\rho\left(f\left(x\right),y\right)\leq M$ for all $x\in X$.

Let $\Omega$ be a locally compact, Hausdorff topological space. Then $C_0\left(\Omega\right)$, the set of continuous functions taking values in $\mathbb C$ that vanish at infinity, is a closed subalgebra of $C_b\left(\Omega\right)$. Recall that $f$ vanishes at infinity if for every $\varepsilon>0$, $\left\{\omega\in\Omega:\left|f\left(\omega\right)\right|\geq\varepsilon\right\}$ is compact. To see that it is closed under addition, let $f,g\in C_0\left(\Omega\right)$, let $\varepsilon>0$, let $A=\left\{\omega\in\Omega:\left|f\left(\omega\right)\right|<\varepsilon/2\right\}$, let $B=\left\{\omega\in\Omega:\left|g\left(\omega\right)\right|<\varepsilon/2\right\}$, and let $\omega\in A\cap B$. Then $\left|f\left(\omega\right)+g\left(\omega\right)\right|\leq\left|f\left(\omega\right)\right|+\left|g\left(\omega\right)\right|<\varepsilon$. Therefore, $A\cap B\subseteq C=\left\{\omega\in\Omega:\left|f\left(\omega\right)+g\left(\omega\right)\right|<\varepsilon\right\}$, which implies that $C^c\subseteq A^c\cup B^c$. Since $C^c$ is closed and $A^c\cup B^c$ is compact, $C^c$ is compact.

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Convergence in $L^1$ Implies Convergence in Measure

I have not posted in a gargantuan amount of time, so here is a quick result:

Claim: Let $(X,\mathcal M,\mu)$ be a measure space, let $f_n,f\in L^1(\mu)$ for all $n\in\mathbb N$, and let $f_n\to f$ in $L^1(\mu)$. Then $f_n\to f$ in measure.

Proof: Let $E_{\varepsilon,n}=\{x\in X:|f_n(x)-f(x)|\geq\varepsilon\}$, where $\varepsilon>0$ and $n\in\mathbb N$. Then $$\int|f_n-f|d\mu\geq\int_{E_{\varepsilon,n}}|f_n-f|d\mu\geq\varepsilon\mu(E_{\varepsilon,n}).$$ Therefore, $$\mu(E_{\varepsilon,n})\leq\varepsilon^{-1}\int|f_n-f|\xrightarrow{n\to\infty}0.\tag*{$\blacksquare$}$$
The converse of this claim does not hold:

Let $f_n=n\chi_{[0,n^{-1}]}$, let $\varepsilon>0$, and let $n\in\mathbb N$. Then
$$\mu(\{x\in X:n\chi_{[0,n^{-1}]}(x)\geq\varepsilon\})\xrightarrow{n\to\infty}0$$
since $\mu(\{x\in X:n\chi_{[0,n^{-1}]}(x)\geq\varepsilon\})\leq n^{-1}$. Therefore, $f_n$ converges to $0$ in measure. However,
$$\int n\chi_{[0,n^{-1}]}d\mu=n\mu([0,n^{-1}])=1.$$

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Interesting Result (Question from One of My Finals)

Claim: There is no Kähler-Einstein metric on $\mathbb P^2$  blown up at two distinct points.
Proof: Without loss of generality, let $\widetilde{\widetilde{\mathbb{P}^2}}$ be the blow up of $\mathbb{P}^2$ at the points $\left[1,0,0\right]$ and $\left[0,1,0\right]$. Then$$\text{Aut}_{\mathcal{O}}\left(\widetilde{\widetilde{\mathbb{P}^2}}\right)=\left\{\begin{pmatrix}1&0&*\\0&*&*\\0&0&*\end{pmatrix}\in\text{GL}_3\left(\mathbb{C}\right)\right\}\ni\begin{pmatrix}1&0&*\\0&a&*\\0&0&b\end{pmatrix}.$$Observe that$$\begin{pmatrix}1&0&*\\0&a&*\\0&0&b\end{pmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\0\\0\end{pmatrix}\qquad\text{and}\qquad\begin{pmatrix}1&0&*\\0&a&*\\0&0&b\end{pmatrix}\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\a\\0\end{pmatrix}=\begin{pmatrix}0\\1\\0\end{pmatrix}.$$Let$$H=\left\{\begin{pmatrix}1&0&x\\0&1&0\\0&0&1\end{pmatrix}\in\text{GL}_3\left(\mathbb{C}\right)\right\}.$$Then$$\begin{pmatrix}1&0&a\\0&b&c\\0&0&d\end{pmatrix}\begin{pmatrix}1&0&*\\0&1&0\\0&0&1\end{pmatrix}=\begin{pmatrix}1&0&*+a\\0&b&c\\0&0&d\end{pmatrix}=\begin{pmatrix}1&0&a+*\\0&b&c\\0&0&d\end{pmatrix}=\begin{pmatrix}1&0&*\\0&1&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&a\\0&b&c\\0&0&d\end{pmatrix},$$which implies $H$ is an Abelian and thus normal subgroup of $\text{Aut}_{\mathcal{O}}\left(\widetilde{\widetilde{\mathbb{P}^2}}\right)$. Let $h\in H$. Then$$\left(h-I_3\right)^2=\begin{pmatrix}0&0&x\\0&0&0\\0&0&0\end{pmatrix}^2=\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix},$$which implies $\left\{I_3\right\}$ is not the only unipotent radical. Therefore, $\text{Aut}_{\mathcal{O}}\left(\widetilde{\widetilde{\mathbb{P}^2}}\right)$ is not reductive, which, by Matsushima, implies that there is no Kähler-Einstein metric on $\widetilde{\widetilde{\mathbb{P}^2}}$. $\blacksquare$

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The Hahn-Banach Theorem

The Hahn-Banach theorem is considered one of the most important results of functional analysis.

Theorem

 

Let $\mathcal X$ be a vector space over $\mathbb R$, let $p$ be a sublinear functional on $\mathcal X$, let $\mathcal M$ be a subspace of $\mathcal X$, and let $f$ be a linear functional on $\mathcal M$ such that $f\leq p$. Then there exists a linear functional $F$ on $\mathcal X$ such that $F\leq p$ and $F|_{\mathcal M}=f$.

Proof

 

We will first show that if $x\in\mathcal X\setminus\mathcal M$, then there exists a linear functional $g$ on $\mathcal M+\mathbb Rx$ such that $g\leq p$ and $g|_{\mathcal M}=f$.

Let $y_1,y_2\in\mathcal M$. Then$$f\left(y_1\right)+f\left(y_2\right)=f\left(y_1+y_2\right)\leq p\left(y_1+y_2\right)=p\left(y_1-x+x+y_2\right)\leq p\left(y_1-x\right)+p\left(x+y_2\right),$$which implies that$$f\left(y_1\right)-p\left(y_1-x\right)\leq p\left(x+y_2\right)-f\left(y_2\right).$$Let $\alpha$ be such that$$\sup\left\{f\left(y\right)-p\left(y-x\right):y\in\mathcal M\right\}\leq\alpha\leq\inf\left\{p\left(x+y\right)-f\left(y\right):y\in\mathcal M\right\}$$and define $g:\mathcal M+\mathbb Rx\to\mathbb R$ by $y+\lambda x\mapsto f\left(y\right)+\lambda\alpha$. Then $g$ is clearly linear and $g|_{\mathcal M}=f\leq p$.

Let $\lambda>0$. Then$$g\left(y+\lambda x\right)=f\left(y\right)+\lambda\alpha=\lambda\left[f\left(\frac{y}{\lambda}\right)+\alpha\right]\leq\lambda\left[f\left(\frac{y}{\lambda}\right)+p\left(x+\frac{y}{\lambda}\right)-f\left(\frac{y}{\lambda}\right)\right]=p\left(y+\lambda x\right).$$Let $\lambda=-\mu<0$. Then$$g\left(y+\lambda x\right)=g\left(y-\mu x\right)=f\left(y\right)-\mu\alpha=\mu\left[f\left(\frac{y}{\mu}\right)-\alpha\right]\leq\mu\left[f\left(\frac{y}{\mu}\right)-f\left(\frac{y}{\mu}\right)+p\left(\frac{y}{\mu}-x\right)\right]=p\left(y-\mu x\right)=p\left(y+\lambda x\right).$$Therefore, $g\leq p$.

Let $\mathcal F$ be the set of linear extensions $F$ of $f$ such that $F\leq p$. Note that elements of $\mathcal F$ are subsets of $\mathcal X\times\mathbb R$. Then $\mathcal F$ is partially ordered by inclusion. Since increasing unions of subspaces of $\mathcal X$ are subspaces of $\mathcal X$, unions of linearly ordered elements of $\mathcal F$ lie in $\mathcal F$. Thus, Zorn's lemma completes the proof.

$\blacksquare$

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