Is true $\int\limits_{-\infty}^{\infty} f(t)\,g(t)\,dt \leq \sqrt{\int\limits_{-\infty}^{\infty} |f(t)|^2 dt}\,\cdot \sup\limits_t|g(t)|$?

Question

Does holds $\int\limits_{-\infty}^{\infty} f(t)\,g(t)\,dt \leq \sqrt{\int\limits_{-\infty}^{\infty} |f(t)|^2 dt}\,\cdot \sup\limits_t|g(t)|$ true for every real valued functions $f(t),\, g(t)$?

I want to know if is possible to mix the following inequalities (Hölder and Cauchy-Schwartz):

1. $\int\limits_{-\infty}^{\infty} f(t)\,g(t)\,dt \leq \left| \int\limits_{-\infty}^{\infty} f(t)\,g(t)\,dt \right| \leq \int\limits_{-\infty}^{\infty} |f(t)\,g(t)|\,dt$
2. $\int\limits_{-\infty}^{\infty} |f(t)\,g(t)|\,dt \leq \int\limits_{-\infty}^{\infty} |f(t)|\,dt \cdot \sup\limits_t |g(t)|$
3. $\int\limits_{-\infty}^{\infty} |f(t)\,g(t)|\,dt \leq \sqrt{\int\limits_{-\infty}^{\infty} |f(t)|^2\,dt}\,\cdot \sqrt{\int\limits_{-\infty}^{\infty} |g(t)|^2\,dt}$
4. $\left( \int\limits_{-\infty}^{\infty} f(t)\,g(t)\,dt \right)^2 \leq \int\limits_{-\infty}^{\infty} |f(t)|^2\,dt\,\cdot \int\limits_{-\infty}^{\infty} |g(t)|^2\,dt$

If the answer is false in general, There are conditions for $f(t), g(t) \neq 0\,\, \forall\, t$ where the required inequality is true?

Answer 1

Put $g = 1$ and $f = \frac 1{1 + |x|}$. If the inequality held, we would have $$ 2\int_0^\infty \frac 1{1 + x} \le \sqrt{2\int_0^\infty \frac 1{(1+x)^2}} = \sqrt 2 $$ but the left hand side is clearly infinite. In fact, by passing $g = 1$, your inequality would imply that $L^1 \subset L^2$, which is clearly false whenever the underlying space is $\Bbb R^n$.

One possible strengthening of the conditions could be to have $g$ have compact support. In this case, we can recover a version of your estimate. Indeed, if $X = \text{supp}(g)$ we have, by Jensen's Inequality: \begin{align} \int_X |f(t)|\text d t &\le \sqrt{m(X)}\left(\int_X |f(t)|^2 \text d t\right)^{1/2} \end{align} so that, if $f \in L^2$: \begin{align} \left|\int_\Bbb R f(t)g (t) \text d t\right| &\le ||g||_\infty \int_X |f(t)| \text d t \le ||g||_\infty \sqrt{m(X)}\left(\int_X |f(t)|^2 \text d t\right)^{1/2} \\ &\le \sqrt{m(X)}||g||_\infty ||f||_2 \end{align}