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Hey I could use a lot of help with this problem please!

Let $(X, \langle\,\cdot,\cdot\,\rangle)$ be a Hilbert space over $\mathbb{R}$. Then, let $A\colon X \to X$ be a linear operator. Suppose that $A$ is symmetric and positive definite. Let $(\,\cdot,\cdot\,)$ be the inner product defined as $(\,\cdot,\cdot\,) \colon X \times X \to \mathbb{R}$ such that $(x,y) = \langle x, Ay\rangle$. Then, suppose that $A$ is a bounded linear operator when endowing X with $\lVert\,\cdot\rVert_{1}$, which is the norm induced from the Hilbert space, and let $\lVert\,\cdot\,\rVert_2$ be the norm induced from $(\,\cdot,\cdot\,)$. Then, denote the operator norm as usual,

$$\lVert A\rVert_{\text{op}1} = \inf \bigl\{ C > 0 : \lVert Ax\rVert_1 \leqslant C\lVert x\rVert_1 \text{ for all } x \in X\bigr\}.$$

Let $C_1 = \sqrt{\lVert A\rVert_{\text{op}1}}$. Show that for all $x\in X$

$$\lVert x\rVert_2 \leqslant C_1 \lVert x\rVert.$$

I know that I need to show that

$$(\lVert x\rVert_2)^2 = (x,x) \leqslant \lVert A\rVert_{\text{op}1} (\lVert x\rVert_1)^2$$

and that the proof revolves around Cauchy-Schwarzing everything, I just cannot figure out how. Any help would be greatly appreciated.

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You're right that Cauchy-Schwarz is important. It comes in here: $$\langle x, Ax \rangle \leq ||x||_1 ||Ax||_1$$ Can you fill in the rest?

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  • $\begingroup$ oh god I'm an idiot thank you so much haha sorry complete brain fart $\endgroup$ – user143731 Apr 17 '14 at 5:01

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