Let $X$ be an inner product space and for each $x \in X$ define $ \phi_x (y) : X \to \mathbb{K} $ by $ \phi_x(y) = \langle x,y \rangle.$ Show that $\phi_x$ is a continuous linear map and that $ \|\phi_x\| = \|x\|$.

I have managed to show that $ \phi_x $ is a linear map and I think that the following proves continuity:

Take $(y_n) \in X$ such that $ y_n \to y. $ Then:

$|\langle x,y_n \rangle - \langle x,y\rangle| = |\langle x-x,y_n-y \rangle + \langle x, y_n - y \rangle + \langle x-x, y \rangle | \leq \|x\|\|y_n-y\| \to 0 $ as $ n \to \infty$.

However, I have no idea how to compute the norm of $\phi_x$ and so I can't do the last part of the question which asks me to show that $\|\phi_x\| = \|x\|.$ Any help with this last step would be greatly appreciated.

  • $\begingroup$ However, you already used Cauchy-Schwartz to prove it is continuous... Do the same with the definition of norm of an operator $\endgroup$ – Exodd Apr 25 '17 at 13:33
  • $\begingroup$ What's the definition of the norm? $\endgroup$ – Nosrati Apr 25 '17 at 13:46

You have by Cauchy-Schwarz that

$$|\phi_x(y)|^2 = |\langle x,y\rangle|^2 \le |x|^2 |y|^2$$

That is $|\phi_x|\le |x|$. Also we have:

$$\phi_x(x) = \langle x, x\rangle = |x|^2 $$

so you have $|\phi_x|\ge|x|$.


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