Equivalent condition for bounded linear operator and continuity

By | June 18, 2015

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}

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