Skip to main content

Quick Analysis Trick 1

05 Sep 2020 -

This is the start of a series of short posts which will come sporadically. To say that I struggle with analysis would be akin to saying that Stravinsky’s “Rite of Spring” wasn’t well received during its premiere. While technically true, the reality is much more dramatic than you might expect. I’m obviously being a bit hyperbolic here, but it really is something that I find quite difficult (this is probably due in part to lack of exposure).

That said, I recognize the importance of analysis, and I wouldn’t make a very good mathematician if I didn’t do my best to improve. Oftentimes I’ll come across an argument and think “Wow! That’s really clever!”, but I always struggle to remember it. I had two of these moments today, and so I’ve decided to write up a short post whenever I see one. In the best case, this will help me internalize the argument by putting me in a position where I have to teach it. In the worst case, at least I’ll have a searchable list of all the arguments I thought were clever enough to save. Either way, future readers will hopefully find these useful for their own studies.

So what’s the inaugural post, then? It’s probably something that many analytically minded readers will find obvious: a technique for proving lower bounds. I came across this while reading Helemskii’s “Lectures and Exercises in Functional Analysis”, which a friend recommended me because it uses the word “category” throughout.

In proposition 4 of chapter 1, Helemskii proves

Let $E$ be a vector space equipped with a seminorm, and let $F$ be a closed subspace. For any $y \in E \setminus F$ and $x \in F$, we have the following estimate (which is uniform in $x$!):

\(|| \lambda y + x || \geq C_y |\lambda|\)

The idea that I want to take from this proof is simple: If $y$ were able to get arbitrarily close to $F$, then since $F$ is closed we would have $y \in F$. Thus $y$ has to remain a bounded distance away from $F$. In other words, we can find a $C$ so that $||y-x|| > C$ for every $x \in F$.

Here we’ve turned merely the closedness of $F$ into a uniform lower bound, whereas my brain would expect a compactness assumption to be necessary. In the interest of completeness, we then see

\[|| \lambda y + x || = |\lambda| \; ||y - (-\lambda^{-1}x)|| \geq C |\lambda|\]

where the constant $C$ can depend on $y$ but not $x$.