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Let $V$ be a Hilbert space and let $A:V \to V^*$ be a bounded linear operator such that $$\langle Av, v \rangle \geq C|v|_V$$ for all $v \in V$.

Why does this mean that

  1. $A$ is an isomorphism

  2. $A^{-1}:V^* \to V$ is continuous?? For 1), one can show that $|Av_1 - Av_2| \geq C|v_1 - v_2|$ so $A$ maps distinct elements to distinct elements, so it must have an inverse.

For 2), I tried $$|A^{-1}| = \sup_{f \in V^*} \frac{|A^{-1}f|}{|f|}$$ but can't get anywhere..

matt.x
  • 605
  • For 1), you proved that $A$ is injective but surjectivity is missing. For 2), take $f\in V^*$ and replace $v=A^{-1}f$ in your inequality. – TerranDrop Jan 28 '14 at 16:48

1 Answers1

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Apply the theorem of Lax-Milgram...

mkk030572
  • 65