= Solution
A <Lévy measure> $K$ on the real line is a nonnegative <Borel measure> satisfying
$$
K(\{0\})=0,\qquad \int_{\mathbb R}(1\wedge y^2)\,K(dy)<\infty.
$$
It records the intensity per unit time of jumps of a <Lévy process>. Under the additional small-jump first-moment condition, the <uncompensated Lévy–Khintchine formula> is
$$
\mathbb E e^{iuX_t}=\exp\{t\psi(u)\},\qquad \psi(u)=ibu-\frac12\sigma^2u^2+\int_{\mathbb R\setminus\{0\}}(e^{iuy}-1)\,K(dy),
$$
where $b\in\mathbb R$ and $\sigma^2\ge0$. The integral is absolutely convergent: near zero its integrand has absolute value at most $|u||y|$, while outside a neighborhood of zero it is at most $2$ and $K$ has finite mass. If the usual compensated <Lévy–Khintchine formula> uses drift $\gamma$, this convention has $b=\gamma-\int_{|y|\le1}yK(dy)$.
A <Lévy process> starts at zero and has stationary <independent increments> and stochastic continuity. Its <characteristic function> at time $t$ is $e^{t\psi(u)}$, with $\psi(0)=0$ and $\psi$ continuous. The given time-one <characteristic function> is strictly positive. It therefore has the unique continuous logarithm vanishing at zero,
$$
\psi(u)=-\log(1+u^2/2).
$$
Any other continuous exponent would differ by an integer multiple of $2\pi i$; continuity on the real line and the value at zero make that integer identically zero. This determines all increment laws. More explicitly, for $0=t_0<t_1<\cdots<t_m$,
$$
\mathbb E\exp\!\left(i\sum_{j=1}^m u_jX_{t_j}\right)=\exp\!\left\{\sum_{j=1}^m(t_j-t_{j-1})\psi\!\left(\sum_{k=j}^m u_k\right)\right\}.
$$
Thus it determines every <finite-dimensional distribution>. For the standard <càdlàg> version of a <Lévy process>, coordinates at rational times determine the path, so these laws determine the entire process law on its canonical path space as well.
To find the <Lévy measure>, try the symmetric density $e^{-\alpha|y|}/|y|$. Its jump exponent is
$$
J_\alpha(u)=2\int_0^\infty\frac{\cos(uy)-1}{y}e^{-\alpha y}\,dy.
$$
Differentiation under the integral is justified by the integrable bound $2e^{-\alpha y}$. The elementary <Fourier sine transform> gives
$$
J_\alpha'(u)=-2\int_0^\infty e^{-\alpha y}\sin(uy)\,dy=-\frac{2u}{\alpha^2+u^2}.
$$
Since $J_\alpha(0)=0$, integration in $u$ yields $J_\alpha(u)=-\log(1+u^2/\alpha^2)$. Choosing $\alpha=\sqrt2$ matches the prescribed exponent exactly. This is the <Lévy measure of a symmetric Laplace time-one law>:
$$
\boxed{K(dy)=\frac{e^{-\sqrt2|y|}}{|y|}\mathbf1_{\{y\ne0\}}\,dy,\qquad b=0,\quad\sigma^2=0.}
$$
The candidate satisfies the <Lévy measure> condition because its density is of order $1/|y|$ near zero and decays exponentially at infinity; moreover $\int_{|y|\le1}|y|K(dy)=2\int_0^1e^{-\sqrt2y}dy<\infty$. Uniqueness of the triplet in the <Lévy–Khintchine formula> proves that these are the required coefficients. The infinite mass near zero means infinitely many small jumps, not a Gaussian component; their total variation is finite on bounded time intervals because $\int(1\wedge|y|)K(dy)<\infty$.
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