= Solution
The supplied derivative identity and the <Cauchy-Schwarz inequality> give
$$
U_x(t,x)^2
\leq4U(t,x)\,\mathbb E[f'(x+B_{1-t})^2].
$$
Therefore
$$
\frac12\mathbb E\int_0^1\frac{U_x(t,W_t)^2}{U(t,W_t)}dt
\leq2\int_0^1\mathbb E[f'(W_t+B_{1-t})^2]dt.
$$
The sum $W_t+B_{1-t}$ is standard normal for every $t$, so the right side is $2\mathbb E f'(W_1)^2$. Part c proves the <Gaussian logarithmic Sobolev inequality>
$$
\boxed{
\mathbb E[f(W_1)^2\log f(W_1)^2]
\leq\mathbb E[f(W_1)^2]\log\mathbb E[f(W_1)^2]
+2\mathbb E[f'(W_1)^2].}
$$
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