# Continuity on a topological vector space

This is related to a question I posted Continuity of a composition of functions on complex field, Parrallelogram law, inner product, which has not gained too much attention. It has a long context but my essential question is topological. So I extract the essential question here. Hope it is not considered as duplicate.

Suppose we have a topological vector space $V$ over field $\mathbb{C}$ and we assume this space is induced by a norm $\|\cdot\|$. For fixed $x, y \in V$, $\alpha \in \mathbb{C}$, we define a function $f \colon \mathbb C \to \mathbb C$ which is given by \begin{align*} \lambda \to \|{\lambda x + y}\|^2 + \| \lambda x - y\|^2 + \lambda \alpha \|x+ \alpha y\|^2 - \lambda \alpha \|x- \alpha y\|^2. \end{align*} Now I think $f$ is continuous. My thinking is to consider $f$ as following composition \begin{align*} \lambda \xrightarrow{g} (\lambda, \lambda x + y, \lambda x - y) \xrightarrow{h} f(x) \end{align*} I have convinced myself the map $g : \mathbb C \to \mathbb{C} \times V \times V$ is actually continuous. Let $g=(g^1, g^2, g^3)$. $g^1$ is identity and $g^2, g^3$ are continuous by topological vector space properties. Then the sum of the first two terms of $f$ should be continuous operation. The sum of last two terms of $f$ is also continuous because $\lambda \alpha$ (as multiplication in $\mathbb{C}$) is continuous and then follows $\lambda \alpha \|x+ \alpha y\|^2 - \lambda \alpha \|x- \alpha y\|^2$ is continuous.

My question is :(1)is my argument correct? (2) If it is, do I need to go through all this to prove $f$ is continuous? Since $x,y,\alpha$ are fixed, I feel there should be some very easy way to assert the continuity.

• You could also use the equivalent condition to continuity in terms of nets: so if $\lambda_i \to \lambda_0$, then by continuity of scalar multiplication, $\lambda_i x \to \lambda_0 x$; by continuity of addition, $\lambda_i x + y \to \lambda_0 x + y$; etc. etc. – Daniel Schepler Oct 6 '17 at 21:17
Seems about right. This is continuous even if you let $x,y, \alpha$ vary. You have several operations, all of which are continuous (taking norm, addition and multiplication in real numbers or in a topological vector space, scalar multiplication), all of which are continuous, so the end result is continuous. You also use elementary properties of product topology, like that if $f_1\colon X\to Y_1$ and $f_2\colon X\to Y_2$ are continuous, then $(f_1,f_2)\colon X\to Y_1\times Y_2$ is continuous, but all of this easily follows from the definitions.