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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.

Thanks in advance.

user1101010
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  • 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 06 '17 at 21:17
  • @DanielSchepler: I am under the impression only when the topology is not 'nice' enough, we consider nets. If we go this way, are sequences sufficient in this setting? – user1101010 Oct 06 '17 at 23:17
  • I guess you're right, since you're assuming the vector space is normed (therefore metrizable, therefore first-countable). – Daniel Schepler Oct 06 '17 at 23:21

1 Answers1

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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.

If you still have doubts, you can just check continuity directly step by step by looking at sub(basic) open sets, or you can write your function explicitly as a big composition of continuous functions (there will be a whole bunch of them).

About (2), yes, technically, you need to go through all this (and actually quite a bit more if you go into all the details). In practice, once you do such things two or three times, it is quite obvious how you can do it in general, and you will probably never find such proof in a research, or even advanced graduate level texts -- it is understood implicitly that such a thing is continuous.

tomasz
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