How to read $f(K)=\bigcup_{k\in K}\langle f,k\rangle$ and what it means, mathematically

Question

Let $E$ be a real Banach space and let $K$ be a subset of $E$. Suppose for each $f\in E^{*}$, the set $$f(K)=\bigcup_{k\in K}\langle f,k\rangle$$ is bounded in $\Bbb{R}$. Then, $K$ is a bounded subset of $E.$

Question:

How to read $f(K)=\bigcup_{k\in K}\langle f,k\rangle$ and what does it mean, mathematically?

Answer 1

I can tell you what it means mathematically, but I prefer to leave your question about how you should read the symbols out loud to a native English speaker.

A functional $f$ is a function $f: E \to \mathbb{R}$ (sometimes $\mathbb{C})$ that takes a vector in $E$ and outputs a real(complex) number. When you have a Hilbert space, every functional arises as $\varphi: E \to \mathbb{R}$ where $\varphi(\cdot)=\langle v,\cdot\rangle$ for some $v \in E$. This follows from the Riesz representation theorem for Hilbert spaces.

Because of this, you can denote a functional $f$ using the bra-ket notation which is popular in physics and real analysis. Note that a Banach space does not necessarily have to be an inner product space. So, when we're dealing with a Banach space, $\langle f, k \rangle$ is just a notation that means the evaluation of $f$ at $k$. Your equality is now this:

$$f(K) = \bigcup_{k\in K} \{f(k)\} =\bigcup_{k \in K} \{\langle f,k\rangle\}$$

Answer 2

Since $K$ is a Banach space $\langle f,k \rangle$ is the evaluation of $f$ at $k$ where $f\in E^*$ is a functional on $E$ (i.e. $f(k)$). So $f(K)$ here is defined completely literally: it's the union of the evaluation of the functional $f$ at each point $k\in K$.

The point of writing it this way though, is that the norm of $f$ is $\|f\| = \sup_{\|k\|=1} |\langle f,k \rangle |$ so this is suggestive of how to go about proving boundedness.