Analyzing linear operator convergence through transformations Assume $A_n: C[0,1] \rightarrow C[0,1]$, $n \in \mathbb{N}$, is a linear operator, having the following form:
$$A_nx(t) = \int_{0}^{1} \sqrt{(t-s)^2 + \frac{1}{n}} \cdot x(s)ds$$
I'm trying to analyze its (strong) convergence through checking whether 
$$ ||A_n - A_m|| = \sup_{||x|| \leq 1} ||A_n x - A_m x|| \rightarrow 0  $$
holds, when $n,m \rightarrow \infty$. 
While it seems that there are a few possible ways one could continue from here, my question revolves around dealing with the squared root. The simplest way that comes to mind is analysing the operator $A_n^{2}$, from where it should follow that
$$ ||A_n^2 - A_m^2|| = \sup_{||x|| \leq 1} ||A_n^2 x - A_m^2 x||  \leq \sup_{||x|| \leq 1} ||x||^2  |\frac{1}{n} - \frac{1}{m} | \rightarrow 0 $$
My question: Does such convergence of squared operators $(A_n^{2})$ imply convergence of non-transformed operators $(A_n)$? I'm sure that it doesn't hold in every case, although here the square transformation seems as a safe transformation with being a bijection on $C[0,1]$. 
Is there anything else I'm missing? The main goal I'm looking for is to understand whether we can (and how, if so) transform the expressions of various linear operators to allow for an easier/cleaner analysis.
 A: Your approach has some issues: first you are assuming that $A_n^2$ is obtained by squaring the symbol, which I don't think is the case. Also, I don't think the estimate you got for the supposed $A_n^2$ is totally correct. 
In general, you can't deduce convergence of $A_n$ from that of $A_n^2$, because in extreme cases there are operators that are nilpotent ($T^2=0$) where you lose all information, or $T^2=I$, where you  lose almost all information. Even in dimension $1$, then sequence $a_n=(-1)^n$ has the property that $\{a_n^2\}$ is Cauchy and $\{a_n\}$ is not. 
Back to your concrete example, one can make the estimation directly, since 
$$
\left|\sqrt{(t-s)^2+\frac1m}-\sqrt{(t-s)^2+\frac1n}\right|=\frac{\left|\frac1m-\frac1n\right|}{\sqrt{(t-s)^2+\frac1m}+\sqrt{(t-s)^2+\frac1n}}
$$
and out of a small interval this can be bounded; in that small interval the integral will be small because $x$ is bounded. 
A: It is unclear how to get the above control of $A_n^2$, which should be given by an even "uglier" formula. Even so, such a formula "for the squares" (correctly, rather for the compositions) does not say too much about a formula "for the initial sequence". Consider for instance the sequence $1,-1,1,-1,\dots$. (Here, $1$ is the identity, if we really need operators acting on a functions space.)
A way to proceed is as follows.
Let $A_n,A_m$ be two operators in the sequence, fix an $x$ in $C[0,1]$. 
Let $I=[0,1]$.
Then:
$$
\begin{aligned}
((A_n-A_m)x)(t)
&=
\int_I
\left(
\sqrt{(t-s)^2+\frac 1n}
-
\sqrt{(t-s)^2+\frac 1m}
\right)
\; x(s)\; ds
\\
&=
\int_I
\frac{
\left(
\sqrt{(t-s)^2+\frac 1n}
-
\sqrt{(t-s)^2+\frac 1m}
\right)
\left(
\sqrt{(t-s)^2+\frac 1n}
+
\sqrt{(t-s)^2+\frac 1m}
\right)
}
{\phantom{
\left(
\sqrt{(t-s)^2+\frac 1n}
-
\sqrt{(t-s)^2+\frac 1m}
\right)
}
\left(
\sqrt{(t-s)^2+\frac 1n}
+
\sqrt{(t-s)^2+\frac 1m}
\right)
}
\; x(s)\; ds
\\
&=
\int_I
\frac{
(t-s)^2+\frac 1n
-
(t-s)^2-\frac 1m
}
{\sqrt{(t-s)^2+\frac 1n}
+
\sqrt{(t-s)^2+\frac 1m}
}
\; x(s)\; ds
\\
&=
\left(\frac 1n-\frac 1m\right)
\int_I
\frac{
1
}
{\sqrt{(t-s)^2+\frac 1n}
+
\sqrt{(t-s)^2+\frac 1m}
}
\; x(s)\; ds
\\[3mm]
|\ ((A_n-A_m)x)(t)\ |
&=
\left|\frac 1n-\frac 1m\right|
\int_I
\frac{
1
}
{\sqrt{(t-s)^2+\frac 1n}
+
\sqrt{(t-s)^2+\frac 1m}
}
\; x(s)\; ds
\\
&\le
\left|\frac 1n-\frac 1m\right|
\int_I
\frac{
1
}
{\sqrt{(t-s)^2+\frac 1M}
}
\; x(s)\; ds
\qquad
\text{ with }
M = \min(m,n)
\\
&\le
\left|\frac 1n-\frac 1m\right|\;\|x\|
\int_I
\frac{
1
}
{\sqrt{(t-s)^2+\frac 1M}
}
\; ds
\\
&=
\left|\frac 1n-\frac 1m\right|\;\|x\|
\cdot2\cdot
\frac 12
\Big(
\text{arcsinh} (t\sqrt M) 
+
\text{arcsinh} ((1-t)\sqrt M) 
\Big)
\\
&\le
\left|\frac 1n-\frac 1m\right|\;\|x\|
\cdot
2\cdot
\text{arcsinh} 
\left(
\frac 12
\left(
t\sqrt M
+
(1-t)\sqrt M
\right)
\right)
\\
&
\qquad\qquad \text{ concavity of $\text{arcsinh}$ on $(0,\infty)$}
\\
&=
\left|\frac 1n-\frac 1m\right|\;\|x\|
\cdot
2\text{arcsinh} \left(\frac 12\sqrt M\right)
\\
&\in \mathcal O\left(\frac 1M\ln M\right)\ .
\end{aligned}
$$
We have bounded the expression by a sequence converging to $0$ for $M\to\infty$, $M\le m,n$. 
