This symmetric finite-rank <integral operator> is not positive. For $f(t)=1-\tfrac32t$, set $A_j=\int_0^1t^jf(t)dt$. Directly,
$$
A_0=\frac14,
\qquad
A_1=0,
\qquad
A_2=-\frac1{24}.
$$
The associated <quadratic form> is
$$
\langle Uf,f\rangle
=\int_0^1\!\int_0^1(s+t)^2f(s)f(t)dsdt
=2A_0A_2+2A_1^2
=-\frac1{48}<0.
$$
Every <covariance operator> is a <positive operator>, so this operator is not a covariance operator.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 225 / 1 / a / iii / Solution

{title2=$T^{1/2}$}

Every bounded <positive operator> $T$ has a unique positive operator $T^{1/2}$ satisfying $(T^{1/2})^2=T$. The <spectral theorem for normal operators on a separable Hilbert space> constructs it by applying the scalar square-root function to the spectrum.

Positive square root of an operator

{title2=$T\geq0$}
{wiki=Positive_operator_(Hilbert_space)}

A bounded <Hermitian operator> $T$ on a <Hilbert space> is positive when $\langle Tx,x\rangle\geq0$ for every $x$. Its <spectrum> is contained in $[0,\infty)$.

= Positive semidefinite operator
{synonym}

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 Incoming links: Positive operator