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Let $H$ be a Hilbert space and $f: H → \mathbb{C}$ be a linear functional. Denote $(H, \| \cdot \|)$ to be the topological space induced by the inner product over $H$, $(H, w)$ to be the topological space under the weak topology.

Is it true that $f: (H, \| \cdot \|) → (\mathbb{C} , | \cdot |)$ is continuous if and only if $f: (H, w) → (\mathbb{C} , | \cdot |)$ is continuous?

I have learned that a basic open set under the weak topology can be written as a finite intersection $\cap_{x^* \in \phi} \{ x \in X | |x^*(x) - x^*(x_0)| < \epsilon\}$ , but I don't know how to go on.

Any hints would be most welcomed.

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As every $(H, \| \cdot \|)$-open set is a $(H, w)$-open set, we trivially have that $f_w$ is continuous implies $f_{\| \cdot \|}$ is continuous (of course using the inverse-image-of-open-set-is-open definition).

For the other implication, use that $\bigcap_i f^{-1}(U_i)=f^{-1}(\bigcap_i U_i)$. Your second-last line basically means that any $(H, w)$-open set can be written as a finite intersection of $(H, \| \cdot \|)$-open sets, so combining these two facts you can show that $f_w$ is continuous whenever $f_{\| \cdot \|}$ is continuous.

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