Monthly Archives: June 2015

Equivalent condition for bounded linear operator and continuity

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오늘 실해석학 시험을 쳤는데, 다른 건 답변을 했지만(다른 것도 마이너한 실수들이 있긴 하다) 이 한 문제만 답변을 제대로 못했다.
쉬운건 알고 있었는데, 증명이 생각이 나질 않았다. 바로 증명을 보고 내 큰 실수를 또 발견했는데, ‘정의’에 충실하지 않았다는 것이다.
언제나 블로그에는 내 부끄러움도 올려야 한다고 생각하기 때문에, 못 풀었던 문제의 증명도 실어 놓는다.

Theorem. Let T be a linear operator from \mathfrak{X} to \mathfrak{Y}. Then the following are equivalent:

  1. T is continuous on \mathfrak{X}
  2. T is continuous at 0
  3. T is bounded

Proof. (1) implies (2): clear.
(2) implies (3): Assume T is continuous at 0. Then there is \delta>0 such that if \norm{h}\delta, then we have

    \[ \left\Vert T(x)-T(0) \right\Vert <1\]

Then for all nonzero x\in \mathfrak{X}, we have

    \[\left\Vert T(x)\right\Vert \leq \left\Vert T\left(\delta\frac{x}{\left\Vert x\right\Vert}\right) \delta^{-1}\left\Vert x\right\Vert\right\Vert < 1\cdot \delta^{-1} \left\Vert x \right\Vert \]

Since x was arbitrary chosen nonzero vector, T is bounded.
(3) implies (1): Assume T is bounded. Choose x,y in \mathfrak{X}. Then we have

    \[ \left\Vert T(x)-T(y) \right\Vert \leq \left\Vert T \right\Vert \left\Vert x-y\right\Vert\]

So this is Lipschitz continuous on \mathfrak{X}. So it is continuous on \mathfrak{X}

Calculation of Integral with log term and some lessons

Problem. Evaluate

    \[ \int_{0}^{\infty}\frac{\left(\ln x\right)^{2}}{1+x^{2}}dx. \]

이 값을 구하는 과정에서 다음과 같은 결과가 필요하다.

Proposition. We have

    \[ \int_{0}^{\infty}\frac{\ln x}{1+x^{2}}dx=0. \]

Proof. Define f\left(z\right)=\frac{\log z}{1+z^{2}} where \left|z\right|>0 and -\frac{\pi}{2}<\arg z<\frac{3\pi}{2}. Consider the following contour:

Untitled-2

Here 0<\rho<1<R. Then inside the contour, f has simple pole at i. So

    \[ \mathrm{Res} (f;i)=\frac{\log i}{2i}=\frac{i\left(\frac{\pi}{2}\right)}{2i}=\frac{\pi}{4}. \]

So by Residue theorem, we get

    \[ \int_{\rho}^{R}f\left(x\right)dx+\int_{C_{R}}f\left(z\right)dz+\int_{-R}^{-\rho}\frac{\ln\left(-x\right)+i\pi}{1+x^{2}}dx-\int_{C_{\rho}}f\left(z\right)dz=2\pi i\left(\frac{\pi}{4}\right). \]

First, note that

    \[ \int_{-R}^{-\rho}\frac{\ln\left(-x\right)+i\pi}{1+x^{2}}dx=\int_{\rho}^{R}\frac{\ln x+i\pi}{1+x^{2}}dx=\int_{\rho}^{R}f\left(x\right)dx+i\pi\int_{\rho}^{R}\frac{1}{1+x^{2}}dx. \]

Also, we have

    \[ \left|\int_{C_{R}}f\left(z\right)dz\right|\le\pi R\cdot\frac{\sqrt{\left(\ln R\right)^{2}+\pi^{2}}}{R^{2}-1}. \]

Note that R\cdot\frac{\sqrt{\left(\ln R\right)^{2}+\pi^{2}}}{R^{2}-1}=O\left(\frac{\ln R}{R}\right) as R\rightarrow\infty and R\ln R\rightarrow0 as R\rightarrow0^{+}. Hence, as \rho\rightarrow0 and R\rightarrow\infty, we have

    \[ 2\int_{0}^{\infty}f\left(x\right)dx+i\pi\int_{0}^{\infty}\frac{1}{1+x^{2}}dx=\pi i\left(\frac{\pi}{2}\right). \]

Since i\pi\int_{0}^{\infty}\frac{1}{1+x^{2}}dx=\pi i\left(\frac{\pi}{2}\right), we get \int_{0}^{\infty}f\left(x\right)dx=0.


이제 본격적으로 문제의 적분을 계산해보도록 하자.

Solution to Problem. Define f\left(z\right)=\frac{\left(\log z\right)^{2}}{z^{2}+1} where \left|z\right|>0 and -\frac{\pi}{2}<\arg z<\frac{3\pi}{2}. Consider the following contour:

Untitled-2

Here 0<\rho<1<R. Then inside the contour, f has simple pole at i. So

    \[ \mathrm{Res} (f,i)=\frac{\left(i\frac{\pi}{2}\right)^{2}}{2i}=\frac{\pi^{2}i}{8}. \]

Hence, by Residue theorem, we get

(1)   \begin{equation*} \int_{\rho}^{R}f\left(x\right)dx+\int_{C_{R}}f\left(z\right)dz+\int_{-R}^{-\rho}\frac{\left(\ln\left(-x\right)+i\pi\right)^{2}}{1+x^{2}}dx-\int_{C_{\rho}}f\left(z\right)dz=2\pi i\left(\frac{\pi^{2}i}{8}\right)=-\frac{\pi^{3}}{4}. \end{equation*}

Note that

(2)   \begin{align*} \int_{-R}^{-\rho}\frac{\left(\ln\left(-x\right)+i\pi\right)^{2}}{1+x^{2}}dx & =\int_{\rho}^{R}\frac{\left(\ln x\right)^{2}+2\pi i\left(\ln x\right)-\pi^{2}}{1+x^{2}}dx\\ & =\int_{\rho}^{R}f\left(x\right)dx+2\pi i\int_{\rho}^{R}\frac{\ln x}{1+x^{2}}dx-\pi^{2}\int_{\rho}^{R}\frac{1}{1+x^{2}}dx.\nonumber \end{align*}

Also, we have

    \[ \left|\int_{C_{R}}f\left(z\right)dz\right|\le\frac{\pi R\left(\left(\ln R\right)^{2}+\pi^{2}\right)}{R^{2}-1}. \]

So the essential part of our calculation is calculating

    \[ \lim_{R\rightarrow\infty}\frac{R\left(\ln R\right)^{2}}{R^{2}-1}=\lim_{R\rightarrow\infty}\frac{\frac{\left(\ln R\right)^{2}}{R}}{1-\frac{1}{R^{2}}}=0 \]

since \lim_{R\rightarrow\infty}\frac{\left(\ln R\right)^{2}}{R}=\lim_{R\rightarrow\infty}\frac{2\left(\ln R\right)/R}{1}=0.
Similarly,

    \[ \lim_{\varepsilon\rightarrow0}\frac{\varepsilon\left(\ln\varepsilon\right)^{2}}{\varepsilon^{2}-1}=0. \]

So as \rho\rightarrow0 and R\rightarrow\infty, from (1) and (2) and by Proposition, we get

    \[ 2\int_{0}^{\infty}f\left(x\right)dx-\pi^{2}\int_{0}^{\infty}\frac{1}{1+x^{2}}dx=-\frac{\pi^{3}}{4}. \]

Since \int_{0}^{\infty}\frac{1}{1+x^{2}}dx=\frac{\pi}{2}, we get

    \[ \int_{0}^{\infty}f\left(x\right)dx=\frac{\pi^{3}}{8}. \]

이 문제가 오늘 시험문제에 나왔었는데, 나는 간단한 극한 계산 마무리를 그때 당시에 잘못 했었다. Estimate를 정교하지 못하게 했나라는 생각이 들었었는데, 쓸데없는 생각이 나를 막았던 것 같다.

Application of Urysohn’s metrization theorem

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Theorem. Let X be a compact metric space, Y a Hausdorff space, and f:X\rightarrow Y a continuous function from X onto Y. Then Y is metrizable.

Proof. Note that since f\left(X\right)=Y and f is continuous and X is compact, Y is compact. Since Y is Hausdorff space, Y is T_{4}. So Y is T_{3}. It suffices to show Y is second
countable. Since X is compact, it is second countable. Let \mathcal{B} be a countable basis for X. Define

    \[ \mathcal{A}=\left\{ \bigcup_{n\in F}B_{n}:F\subset\mathbb{N},\text{ finite subset}\right\} . \]

Clearly, \mathcal{A} is countable. For A\in\mathcal{A}, define A^{*}=Y\setminus f\left(X\setminus A\right). Since A is open, X\setminus A is closed. Since f is closed map, Y\setminus f\left(X\setminus A\right)=A^{*} is open. So if we collect these A^{*}, say \mathcal{A}^{*}, we claim that \mathcal{A}^{*} is basis for Y.

Let U be an open set in Y and let y\in U. Then f^{-1}\left(U\right) is open and \left\{ y\right\} is closed. So f^{-1}\left(\left\{ y\right\} \right) is closed since f is continuous. Since X is compact space, f^{-1}\left(\left\{ y\right\} \right) is compact. So there is a finite collection \left\{ B_{i}\right\} _{i=1}^{N} of \mathcal{B} such that

    \[ f^{-1}\left(\left\{ y\right\} \right)\subset\bigcup_{i=1}^{N}B_{i}\subset f^{-1}\left(U\right) \]

since \mathcal{B} is a basis for X and f^{-1}\left(U\right) is open in X. Let A=\bigcup_{i=1}^{N}B_{i}. Then A\in\mathcal{A} by definition and A^{*}\in\mathcal{A}^{*}. Moreover, y\in\left(Y\setminus f\left(X\setminus A\right)\right)=A^{*}. For any point f\left(x\right)\in A^{*}, f\left(x\right)\notin f\left(X\setminus A\right). So x\notin X\setminus A. This shows x\in A. Since A\subset f^{-1}\left(U\right), then f\left(x\right)\in U. This implies for each open set U in Y and y\in U, there is a member A^{*} of \mathcal{A}^{*} such that

    \[ y\in A^{*}\subset U. \]

The second condition of basis easily comes from \mathcal{B}. This shows \mathcal{A}^{*} is a countable basis for Y. So by Urysohn’s metrization theorem, Y is metrizable.