Assuming that $f$ is bounded, continuous, and non-negative, is it true that $$\sup_{x \in \mathbb R} k^2e^{−kx^2}f(x)≤\sup_{x \in \mathbb R} (k+1)^2e^{−(k+1)x^2}f(x)$$
I have a hard time proving this rigorously for a general $f(x)$ that meets the above criteria I mentioned.
I'm only able to see it intuitively, like this for $f(x)=e^{-x}$ (okay, I know, $e^{-x}$ is not bounded on $\mathbb R$, but thankfully it does not affect anything significantly here):
Is there a way to prove this rigorously, however? Not for any specific function $f(x)$, but for any arbitrary $f(x)$ that are bounded, continuous, and non-negative? I tried to prove it rigorously by taking the derivative and setting it equal to $0$, but with the appearance of $f(x)$ and $f'(x)$ in the terms, it's hard to solve for the critical value of $x$. Perhaps there might be a way around this.
(This question is somewhat related to my previous question.)
The inequality is not exact in general.
In fact, as it can be written under the form:
$$\sup_{x \in \mathbb R} e^{−kx^2}f(x)≤\left(\dfrac{k+1}{k}\right)^2 \sup_{x \in \mathbb R} e^{−kx^2}f(x)e^{−x^2},$$
setting $g(x):=e^{−kx^2}f(x)$, the issue is reduced to the following one: prove that:
$$\sup_{x \in \mathbb R} g(x)≤\left(\dfrac{k+1}{k}\right)^2 \sup_{x \in \mathbb R} e^{−kx^2}g(x) \ \ \ (1)$$
for all functions $g$ which are a subset of functions $f$ ($>0$ continuous bounded functions).
But inequality (1) is not exact in general. It suffices to take $g_k(x)=\dfrac{1}{1+(x+3k)^2}$ (shifted Cauchy or Lorentzian function):
the LHS of (1) has value 1, whereas
the RHS is less than $1$ as can be readily observed by plotting a curve of the function with equation $y=\varphi_k(x):=g_k(x)e^{−kx^2}$ for any value of $k$.
Remark: A rigorous proof of the fact that the RHS of (1) is (whatever $k$) less than 1 would involve the computation of the roots of the derivative $\varphi_k'(x)$ i.e., i;e., the roots of the third degree polynomial $$k x^3 + 6 k^2 x^2 + (9k^3 + k + 1)x + 3k.$$