= Solution
For <independent and identically distributed random variables> sampled from a <probability density function> $f$, a <kernel density estimator> with <kernel for density estimation> $K$ and <smoothing bandwidth> $h>0$ is
$$
\widehat f_h(x)=\frac1n\sum_{i=1}^n K_h(x-X_i),\qquad K_h(u)=h^{-1}K(u/h),\qquad \int K=1.
$$
The usual <kernel for density estimation> is nonnegative; finite $\int K^2$ makes the integrated <variance> finite. The <mean integrated squared error> and its <bias-variance decomposition of mean squared error> are
$$
\operatorname{MISE}(\widehat f_h)=\mathbb E\int_{\mathbb R}(\widehat f_h(x)-f(x))^2\,dx
=\int\operatorname{Bias}(\widehat f_h(x))^2\,dx+\int\operatorname{Var}(\widehat f_h(x))\,dx.
$$
The interchange is justified by the <Tonelli theorem>.
For the <exponential distribution> and the specified unit-width <uniform kernel>, put $g_h(x)=\mathbb E\widehat f_h(x)$. The <expected value> is $h^{-1}\mathbb P\{x-h/2\leq X\leq x+h/2\}$. Intersecting this interval with the positive half-line gives
$$
\boxed{g_h(x)=\frac{e^{-\max(x-h/2,0)}-e^{-\max(x+h/2,0)}}h.}
$$
On $[-h/4,0)$ the true <probability density function> vanishes, whereas
$$
g_h(x)=\frac{1-e^{-(x+h/2)}}h\geq\frac{1-e^{-h/4}}h\geq\frac18\qquad(0<h\leq1).
$$
Here $1-e^{-z}\geq z/2$ for $0\leq z\leq1/4$. This leakage across the support boundary proves the <integrated squared bias from a density jump>:
$$
\boxed{B(h):=\int(g_h-f)^2\,dx\geq h/256\qquad(0<h\leq1).}
$$
In particular, $h_0=1$ and $c=1/256$ suffice. The phenomenon differs from a smooth interior <bias of a kernel density estimator>: the order-one boundary error persists over a region of width proportional to $h$.
By <independence>, $\operatorname{Var}(\widehat f_h(x))=n^{-1}\{\mathbb E K_h(x-X)^2-g_h(x)^2\}$. Integrating the first term and using $\int K^2=1$ yields
$$
\boxed{\int\operatorname{Var}(\widehat f_h(x))\,dx=\frac1{nh}-\frac{I(h)}n,\qquad I(h)=\int g_h(x)^2\,dx.}
$$
Also $I(h)\leq\int f^2=1/2$: the <kernel density estimator> mean is an average of translates of $f$, so the <Jensen inequality> followed by the <Tonelli theorem> bounds its squared integral.
To optimize over every $h>0$, we must establish that useful <smoothing bandwidths> approach zero, rather than optimize only a formal small-$h$ expression. An exact calculation supplies that justification. The autocorrelation integral of the <exponential distribution> density is
$$
\int f(x-u)f(x-v)\,dx=\tfrac12e^{-|u-v|}.
$$
Averaging over the two uniform windows, or integrating the piecewise expression for $g_h$, gives
$$
I(h)=\frac{h-1+e^{-h}}{h^2},\qquad
\int f g_h=\frac{1-e^{-h/2}}h,
\qquad B(h)=\frac12+\frac{h-1+e^{-h}}{h^2}-\frac{2(1-e^{-h/2})}h.
$$
These formulas show that $B$ is continuous on $(0,\infty)$, is strictly positive there because $g_h>0$ on $(-h/2,0)$, and tends to $1/2$ as $h\to\infty$. Consequently, $B$ is bounded away from zero on $[\delta,\infty)$ for every $\delta>0$. Its expansion agrees with the supplied $B(h)=h/12+o(h)$.
The choice $h_n=\sqrt{12/n}$ gives
$$
\sqrt n\operatorname{MISE}(\widehat f_{h_n})\longrightarrow\frac1{\sqrt3}.
$$
Choose <smoothing bandwidths> within $1/n$ of the <infimum>. Their <mean integrated squared error> tends to zero, and nonnegativity of the integrated <variance> gives $B(h_n)\to0$, hence $h_n\to0$. For any $\varepsilon\in(0,1/12)$, eventually
$$
\operatorname{MISE}(\widehat f_{h_n})
\geq(1/12-\varepsilon)h_n+\frac1{nh_n}-\frac1{2n}
\geq2\sqrt{\frac{1/12-\varepsilon}{n}}-\frac1{2n}.
$$
The last step is the <arithmetic-geometric mean inequality>. Taking the lower limit and then $\varepsilon\downarrow0$ matches the upper bound. Thus
$$
\boxed{\sqrt n\inf_{h>0}\operatorname{MISE}(\widehat f_h)\longrightarrow\frac1{\sqrt3}.}
$$
The original PDF has this constant; the TeX aid's $1/\sqrt2$ is a transcription error.
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