Is the matrix square root uniformly continuous?

Question

Let $\operatorname{Psym}_n$ be the cone of symmetric positive-definite matrices of size $n \times n$.

How to prove the positive square root function $\sqrt{\cdot}:\operatorname{Psym}_n \to \operatorname{Psym}_n$ is uniformly continuous?

I am quite sure this is true, since on any compact ball this clearly holds, and far enough from the origin, I think the rate of change should decrease (analogous to the one-dimensional case where $(\sqrt{x})'=\frac{1}{2\sqrt{x}}$ tends to zero when $x \to \infty$).

A naive approach is to try to use the mean value inequality:

For that we need to show the norm $\|d(\sqrt{\cdot})_A\|$ is bounded for $\|A\|$ large enough. We know the derivative satisfies:

$$d(\sqrt{\cdot})_A(B) \cdot \sqrt{A} + \sqrt{A} \cdot (\sqrt{\cdot})_A(B)=B$$ for every $B \in \operatorname{sym}_n$.

Thus,

$$ \|B\| \le \| d(\sqrt{\cdot})_A(B) \cdot \sqrt{A}\| +\| \sqrt{A} \cdot (\sqrt{\cdot})_A(B) \|\le 2 \| \sqrt{A}\| \| d(\sqrt{\cdot})_A(B)\|,$$

so we only get a bound from below: $$\|d(\sqrt{\cdot})_A\|_{op} \ge \frac{1}{2\|\sqrt{A}\|}$$

Answer 1

The cheapest way is to write some representation of the square root in terms of functions whose continuity is obvious. Note that the square root is homogeneous of degree $1/2$, so it suffices to show that if $\|A-B\|\le 1$, then $\|A^{1/2}-B^{1/2}\|\le C$. Now consider the function $$ f(x)=\int_0^1\left[1-\frac1{1+tx}\right]t^{-3/2}\,dt $$ Making the obvious change of variable $tx=s$, we get $$ f(x)=x^{1/2}\int_0^x\frac{s}{1+s}s^{-3/2}\,ds=Kx^{1/2}-x^{1/2}\int_x^\infty\frac{s}{1+s}s^{-3/2}\,ds=Kx^{1/2}+g(x)\,. $$ Note that $|g(x)|\le 2$ for all $x>0$. Thus, $$ \|KA^{1/2}-f(A)\|\le 2 $$ for an arbitrary positive definite self-adjoint $A$. Now it will suffice to show that $f$ is "operator Lipschitz", but that is obvious since $$ f(A)-f(B)=\int_0^1(1+tA)^{-1}(B-A)(1+tB)^{-1}t^{-1/2}\,dt $$ and $\|(1+tX)^{-1}\|\le 1$ for any positive definite self-adjoint $X$ and $t\ge 0$. (The resolvent identity $X^{-1}-Y^{-1}=X^{-1}(Y-X)Y^{-1}$ has been used here, of course).

In fact, we can say much more: every $\alpha$-Holder continuous function $F$ is operator Holder continuous ($0<\alpha<1$) on the space of self-adjoint matrices. The Lipschitz case is more subtle, however, and is not fully resolved yet. I know neither how curious you are about all this stuff, nor how much you know already, so I'm stopping here.

Answer 2

$\newcommand{\id}{\operatorname{Id}}$

This is merely a more detailed version of fedja's answer:

Lemma 1:

Let $f$ be a real function defined on the positive reals. Assume $|f(x)| \le C$ for every $x >0$. Then $\|f(A)\|_{op} \le C$ for every $A \in \operatorname{Psym}_n$.

Proof:

First, we note that $f$ can be extended to the cone of symmetric positive definite matrices, since their eigenvalues are strictly positive. It is enough to prove the statement for diagonal positive definite matrices:

Let $A=\operatorname{diag}(\sigma_1,...,\sigma_n)$. Then:

$$ f(A)=\operatorname{diag}(f(\sigma_1),...,f(\sigma_n)),$$ thus

$$ \|f(A)\|_{op} = \max(|f(\sigma_i)|) \le C.$$

Lemma 2

It is enough to prove that $\|A-B\|_{op} \le 1 \Rightarrow \|A^{1/2}-B^{1/2}\|_{op}\le C$.

Proof:

Indeed, let $A,B \in \operatorname{Psym}_n$. Define $\lambda=\|A-B\|$, and let $\tilde A = \frac{1}{\lambda}A, \tilde B= \frac{1}{\lambda}B$. Note that $\sqrt{\tilde A}=\frac{1}{\sqrt \lambda} \sqrt{A},\sqrt{\tilde B}=\frac{1}{\sqrt \lambda} \sqrt{B}$, and that $\|\tilde A- \tilde B\|=1$. Thus, by our assumption,

$$ \frac{1}{\sqrt{\lambda}}\|A^{1/2}-B^{1/2}\| = \|\tilde A^{1/2}-\tilde B^{1/2}\|\le C$$

Thus, $$ \|A^{1/2}-B^{1/2}\| \le C \|A-B\|^{\frac{1}{2}}$$

So, the matrix square root is $\frac{1}{2}$-Holder on $\operatorname{Psym}_n$ and in particular uniformly continuous.

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$$ f(x)=\int_0^1\left[1-\frac1{1+tx}\right]t^{-3/2}\,dt $$ Making the obvious change of variable $tx=s$, we get $$ f(x)=x^{1/2}\int_0^x\frac{s}{1+s}s^{-3/2}\,ds=x^{1/2}(\int_0^\infty\frac{s}{1+s}s^{-3/2}\,ds-\int_x^\infty\frac{s}{1+s}s^{-3/2}\,ds)=Kx^{1/2}+g(x)\,. $$

where $K= \int_0^\infty\frac{s}{1+s}s^{-3/2}\,ds=\int_0^1\frac{s}{1+s}s^{-3/2}\,ds+\int_1^\infty\frac{s}{1+s}s^{-3/2}\,ds$

We already know that the first expression is finite, and the second is not greater than $\int_1^\infty s^{-3/2}\,ds < \infty$. Thus, $K < \infty$.

Since $g(x)=-x^\frac{1}{2}\int_x^\infty\frac{s}{1+s}s^{-3/2}\,ds$, $$|g(x)|\le x^\frac{1}{2}\int_x^\infty s^{-3/2}\,ds =2$$ for all $x>0$. Thus, by Lemma 1 $$ (**) \, \, \|KA^{1/2}-f(A)\|_{op}=\|g(A)\|_{op}\le 2 $$ for an arbitrary positive definite self-adjoint $A$. Now it will suffice to show that $f$ is "operator Lipschitz", i.e $\|f(A)-f(B)\| \le \tilde C\|A-B\|_{op}$.

Indeed, this would imply

$$ \|A^{\frac{1}{2}}-B^{\frac{1}{2}}\|_{op}\le \|A^{\frac{1}{2}}-\frac{1}{K}f(A)\|_{op} + \|\frac{1}{K}f(A)-\frac{1}{K}f(B)\|_{op} + \|\frac{1}{K}f(B)-B^{\frac{1}{2}}\|_{op} $$ $$ \le \frac{4}{K}+\frac{\tilde C}{K} \|A-B\|_{op}.$$

The last inequality holds for any $A,B \in \operatorname{Psym}_n$. Assuming $\|A-B\|_{op} \le 1$, it becomes:

$$ \|A^{\frac{1}{2}}-B^{\frac{1}{2}}\|_{op}\le \frac{4}{K}+\frac{\tilde C}{K}:=C $$

This finishes the proof, according to lemma 2.

We now turn to prove Lipschitzity of $f$:

First, note that integration and matrix operation commute. Thus,

$$ f(A)=\int_0^1 \id-(\id+tA)^{-1}t^{-3/2}\,dt,$$ so $$ f(A)-f(B)=\int_0^1 \left[(\id+tB)^{-1}-(\id+tA)^{-1}\right]t^{-3/2}\,dt=\int_0^1(\id+tB)^{-1}(A-B)(\id+tA)^{-1}t^{-1/2}\,dt $$

(where in the last passage we have used the resolvent identity $X^{-1}-Y^{-1}=X^{-1}(Y-X)Y^{-1}$).

Finally, we get

$$ \|f(A)-f(B)\|_{op} =\| \int_0^1(\id+tB)^{-1}(A-B)(\id+tA)^{-1}t^{-1/2}\,dt \|_{op} $$ $$\le \int_0^1 \|(\id+tB)^{-1}(A-B)(\id+tA)^{-1}t^{-1/2}\|_{op}\,dt$$ $$ \le \int_0^1 \|(\id+tB)^{-1}\|_{op}\|A-B\|_{op}\|(\id+tA)^{-1}\|_{op}t^{-1/2}\,dt \le \|A-B\|_{op} \int_0^1 t^{-1/2} \, dt =2\|A-B\|_{op} $$

(since $\|(1+tX)^{-1}\|\le 1$ for any positive definite self-adjoint $X$ and $t\ge 0$. This can be proved easily for diagonal matrices, and then using orthogonal diagonalization to all positive matrices).

Answer 3

If vector $x$ with $\|x\|=1$ is an eigenvector of $\sqrt A - \sqrt B$ with eigenvalue $\mu$ then \begin{align*} x^T(A-B)x &= x^T(\sqrt A - \sqrt B)\sqrt A x + x^T\sqrt B (\sqrt A - \sqrt B) x \\&= \mu x^T(\sqrt A + \sqrt B)x. \end{align*} Now for any $\epsilon>0$, if $\|A-B\|_{op}\le \delta=\epsilon^2$, choose $\mu=\pm\|\sqrt A - \sqrt B\|_{op}$ and as in the scalar case, \begin{align*} \|\sqrt A - \sqrt B\|_{op}^2 & = (x^T(\sqrt A - \sqrt B)x)^2 \\& \le |x^T(\sqrt A - \sqrt B)x| ~x^T(\sqrt A + \sqrt B)x \\&=|x^T(A-B)x| \\&\le \delta = \epsilon^2. \end{align*} The same argument also leads to a simple proof that the matrix square root is Lipschitz: https://math.stackexchange.com/a/3968118/484640