How small can $\max_{x \ge 0} e^x f'(x)$ be in terms of $\max_{x \ge 0} e^x f(x)$ if $f \ge 0$, $f(0) = 0$ and $\int_0^\infty f(x) dx = 1$

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Assume that $f$ is a continuously differentiable function on $[0,\infty)$ such that

  • $f$ is nonnegative and compactly supported,
  • $f(0) = 0$,
  • $\int_0^\infty f(x) dx = 1$.

Denote $$ a = \max_{x \ge 0} e^x f(x) $$

How small can $$b = \max_{x \ge 0} |e^x f'(x)|$$ be in terms of $a$?

Since $g(x) := e^x f(x)$ is continuous, we have $a = g(y)$ for some $y > 0$. Clearly $g'(y) = 0$, hence $$ a = g(y) = g(y) - g'(y) = e^yf(y) - e^yf(y) - e^y f'(y) = -e^y f'(y) \le \max_{x \ge 0} |e^x f'(x)| = b. $$ This shows that $b \ge a$. Is it possible to obtain sharper estimate of $b$ in terms of $a$?