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I don’t remember where I first saw the following inequality (and operator-theoretic generalizations):

Lemma 1. If $r$ and $s$ are non-negative real numbers, then $(r-s)^2 \leq |r^2-s^2|$.

The lemma is most easily/intuitively proved by noticing that $\vert r-s\vert \leq \max(r,s) \leq r+s$ (draw a picture!) and then multiplying both sides of this inequality by $|r-s|$. The lemma also shows, without need for any calculus, that the square-root function on $[0,\infty)$ is Hölder continuous with exponent 1/2.

I was recently reminded of the inequality in Lemma 1 by some discussions with Matt Daws, who had pointed out to me that it gives an easy proof that the square-root function $\Phi_+: L^1[0,1]_+ \to L^2[0,1]_+$ is continuous. (See also comments by Matt and others on this old MO question.) In some ongoing discussions with Matt and Jon Bannon, I found myself wanting a version of this for not-necessarily real-valued functions in $L^1[0,1]$.

Specifically, consider the function $\phi: \mathbb C \to \mathbb C$ defined by

$\displaystyle \phi(0)=0\quad\hbox{and}\quad \phi(z) = |z|^{-1/2}z$

and then define $\Phi: L^1[0,1] \to L^2[0,1]$ by

$\Phi(f) = \phi\circ f$

Note that the restriction of $\Phi$ to the positive cone $L^1[0,1]_+$ agrees with the square root map $\Phi_+$ already mentioned above.

A direct calculation shows that $\Phi$ is norm-preserving, but since it is nonlinear this does not automatically ensure continuity; and it is continuity of $\Phi$ which we wanted to know/check. Continuity of $\Phi$, or very similar maps, is presumably folklore, but having failed in some half-hearted attempts to find a reference for this result, I decided it would be easiest to try and come up with a proof with bare hands. Guided by the easy proof that $\Phi_+$ is continuous, it is natural to try and prove some 2-dimensional version of the inequality at the start of this post, perhaps at the expense of worse constants. Specifically, we would like to know that the following result holds.

Claim 2. There exists a constant $c\geq 1$ such that $|\phi(w)-\phi(z)|^2 \leq c|w-z|$ for all $w,z\in \mathbb C$.

By the same argument used to deduce continuity of $\Phi_+$ from Lemma 1, one sees that Claim 2 implies continuity of $\Phi$; in fact, we get Hölder continuity with exponent $1/2$.

In what follows, I will sketch one possible way to prove this claim, not necessarily with the best constant $c$. It is meant as a tidied version of a train of thought, rather than an attempt to present the cleanest and most polished approach. (I should acknowledge the influence, in spirit if not the fine detail, of these old blogposts-before-weblogs-existed writings by Timothy Gowers, which were quite influential on me during my years as a PhD student and postdoctoral researcher.)

The first thing to notice is that $\phi$ preserves arguments of complex numbers; its effect is merely radial (hence the title of this blog post). This means that in trying to prove Claim 2, we are always free to rotate $w$ and $z$ by a fixed angle, so that either can be assumed real if this is convenient. Introducing the change of variables $r=|w|^{1/2}$, $s=|z|^{1/2}$ and rotating to make $w$ real, we see that the claim is equivalent to

$\displaystyle |r-se^{i\theta} |^2 \overset{\hbox{\bf?}}{\leq} c |r^2- s^2e^{i\theta} | \quad\hbox{for all }r,s\geq 0\hbox{ and }\theta\in{\mathbb R}\quad\quad(*)$

Next: what happens if we expand out both sides of the desired inequality (*)? The left hand side is $r^2+s^2 -2rs \cos\theta$ — which you can also see by drawing a picture and using the cosine rule from school trigonometry — and the right hand side is then $\sqrt{r^4+s^4-2r^2s^2\cos\theta}$. At this point it looks unappealing to square both sides again and attempt to compare terms; this might work, but it looked messy when I tried it, so let’s take a step back and think again.

Recall that the case θ=0 is covered by Lemma 1. How or why does the proof break down for other values of θ? Well, for θ=0 the left hand side of (*) is $|r-s|^2$ and the right hand side is $c|r^2-s^2| = c|r-s| |r+s|$, and we won in this case because $|r-s|\leq |r+s|$, so we can take $c=1$. But for general θ we don’t have the same convenient factorization of the right-hand side.

What would be nice is if we had $c|r^2-s^2e^{2i\theta}|$ on the right hand side of (*). For this does factor as $c|r-se^{i\theta}||r+se^{i\theta}|$, and then we would be hoping to dominate $|r-se^{i\theta}|$ by a multiple of $|r+se^{i\theta}|$. Of course $e^{2i\theta}$ is not the same as $e^{i\theta}$ for general θ, but perhaps for small values of θ we can control the discrepancy with some crude bound, using calculus and the Mean-Value theorem if necessary?

Putting this thought on one side for the moment, let us think what to do when $e^{i\theta}$ is far from 1. In fact, as a “stress-test”, what happens when $e^{i\theta}=-1$? (With hindsight we should have thought of this case sooner, since it corresponds to looking at Lemma 1 and wondering whether it applies to all $r,s\in\mathbb R$, not just positive values.) In this case the left hand side of (*) is $|r+s|^2$ and the right hand side is $c|r^2+s^2|$, so clearly we cannot take $c=1$ any more; nevertheless, the AM-GM inequality shows that we could take $c=2$ in this case.

At this point we have two separate working arguments for θ=0 and θ=π, so how can we handle intermediate cases? We already had a brief look at what happens for small values of θ, so let’s look at what happens when θ is close to π. Drawing a picture of the appropriate obtuse-angled triangle, we realise that for the right-hand side of (*) to be small, both $r^2$ and $s^2$ must be small. In fact, as we let θ vary between π/2 and 3π/2, $|r^2-s^2e^{i\theta}|$ is minimized at the endpoints (i.e. when we have a right-angled triangle), and so

$\displaystyle \min_{\pi/2\leq \theta\leq\pi/2} |r^2-s^2e^{i\theta}| = \sqrt{r^4+s^4}$

(The geometric intuition can be backed up by an appeal to the cosine rule; recall that on this interval, $\cos\theta \leq 0$.)
What about the left hand side of (*)? Well, we may as well replace $|r-se^{i\theta}|^2$ with its “worst-case scenario”, namely $(r+s)^2$, and we already saw this is bounded above by $2r^2+2s^2$. Applying Cauchy-Schwarz, we see that this in turn is bounded above by $2\sqrt{2}(r^4+s^4)^{1/2}$. So putting things together, we have established

Lemma 3. For $\pi/2 \leq\theta\leq 3\pi/2$, and any $r,s\geq 0$, we have $|r-se^{i\theta}|^2 \leq 2\sqrt{2} |r^2-s^2e^{i\theta} |$.

Let’s turn back to the case of acute-angled triangles, i.e. $-\pi/2 \leq \theta\leq \pi/2$, or equivalently the region where $\cos\theta\geq 0$. Can we make good on the earlier hopes that

1. $|r-se^{i\theta}|$ is dominated by (a multiple of) $|r+se^{i\theta}|$ ?
2. $|r^2-s^2e^{2i\theta}|$ is dominated by (a multiple of) $|r^2-s^2e^{i\theta}|$ ?

Recall that both of these statements do hold when θ=0. In fact, drawing a picture, we see that $|r-se^{i\theta}|\leq |r+se^{i\theta}|$ for any θ in this range; once again, the geometric intuition from drawing triangles can be backed up with explicit expansion of both sides using the cosine formula and the fact that $\cos\theta\geq 0$. So we do indeed have $|r-se^{i\theta}|^2 \leq |r^2-s^2e^{2i\theta}|$, and it only remains to prove that the second of our two statements holds.

We seem to be doing well drawing pictures, so let’s do this. While we’re at it, let us rescale the second statement to reduce notational clutter, so that we are aiming to prove

$\displaystyle |t-e^{2i\theta}| \overset{\hbox{\bf?}}{\lesssim} |t-e^{i\theta}| \quad\hbox{ for all } \theta\in [-\pi/2,\pi/2] \hbox{ and all } t\geq 0\quad\quad(**)$

(Clearly, if we can prove this then the 2nd statement above will follow.)

Now, in drawing pictures, it seems that we should distinguish between the cases $0\leq t\leq 1$ and $t\geq 1$. Let’s look at the second case, and consider the triangle formed by the three points $t,e^{i\theta},e^{2i\theta}$. Then $|t-e^{2i\theta}|$ is bounded above by the sum of the other two side lengths of the triangle, but it is clear from the picture that $|e^{2i\theta}-e^{i\theta}| \leq \hbox{arclength}(e^{i\theta},e^{2i\theta})$ while $|t-e^{i\theta}|\geq \hbox{arclength}(e^{i\theta},e^{2i\theta})$. Hence we have proved that

$\displaystyle |t-e^{2i\theta}| \leq 2 |t-e^{i\theta}| \quad\hbox{ for all } \theta\in [-\pi/2,\pi/2] \hbox{ and all } t\geq 1\quad\quad(**)$

(Actually, now that we have written this down, we can simplify the proof of (**) slightly to get rid of the arguments using arc length. Observe that $|t-e^{2i\theta}| \leq |t-e^{i\theta} | + |e^{i\theta}-e^{2i\theta}| = |t-e^{i\theta} | + |1-e^{i\theta}|$ and then observe that since $t\geq 1$, we have $|1-e^{i\theta}|\leq |t-e^{i\theta}|$.)

But what if $0\leq t <1$? Well, in this case we can apply (**) with $t$ replaced by $t^{-1}$ and θ replaced by -θ, to get

$\displaystyle |t^{-1}-e^{-2i\theta}| \leq 2 |t^{-1}-e^{-i\theta}|$

and then we have

$\displaystyle |e^{2i\theta}-t| = |t| |t^{-1}-e^{-2i\theta}| \leq 2|t| |t^{-1}-e^{-i\theta}| = 2 |e^{i\theta}-t |$

as required. Hence we have established the inequality (**) with a constant 2 on the right-hand side, and therefore by rescaling back up again, we have proved the following lemma.

Lemma 4. For $-\pi/2 \leq\theta\leq \pi/2$, and any $r,s\geq 0$, we have $|r-se^{i\theta}|^2 \leq 2 |r^2-s^2e^{i\theta} |$.

Combining Lemma 3 and Lemma 4, we have proved Claim 2, with a value of $c=2\sqrt{2}$.

Looking back on this, it is clear that the arguments above could be written up in a more concise and more formal way, but this will be left for a shorter follow-up blog post, which might say more about the motivation for the original problem. However, at present I don’t see how to avoid the division into two cases (Lemma 3 and Lemma 4). It should be possible to improve the constant in Lemma 3, since on one side of the inequality we were using worst-case behaviour of the left hand side at θ=π while using worst-case behaviour of the right hand side at θ=π/2.

Update: after writing the bulk of this post, I learned from Matt Daws that he can get $c=2$. His argument is more calculus-based and less explicitly geometric. Both Matt and I would be happy to hear of any explicit references for this inequality, either in the sharp form $c=2$ or even just for some explicit value of $c$.

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