= Solution
The <Brownian reflection principle> and <Brownian scaling> give
$$
\mathbb P\left(\sup_{s\le K^2}B_s>r\right)
=2\mathbb P(B_{K^2}>r)=2\mathbb P(Z>r/K),
$$
where $Z$ has the standard <normal distribution>. Combine part (c) with the allowed <Gaussian tail bound> to obtain
$$
\boxed{\mathbb P(|f(X)-\mu|>r)\le4\exp\left(-\frac{r^2}{2K^2}\right).}
$$
This is the <Brownian martingale proof of Gaussian concentration>. It proves the intended <Gaussian concentration inequality>. The denominator printed in the initial bound is $2K$, whereas the gradient hypothesis and part (c) give $2K^2$. The stronger printed bound is false in general. For a concrete counterexample, take $d=1$, $K=2$, and $f(x)=2x$. At $r=10$,
$$
\mathbb P(|f(Z)|>10)=2\mathbb P(Z>5)
>\frac{2}{\sqrt{2\pi}}e^{-18}>4e^{-25},
$$
where the first strict inequality follows by integrating the decreasing normal density over $[5,6]$. The rightmost term is the printed bound. Thus the square on $K$ is a necessary source correction, rather than a change of proof technique.
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