I have been working on the following minimization problem:
Minimize $I(f) := \int_{-\infty}^{\infty}{|f(x)|^2+|f'(x)|^2dx},$ where $f\in\mathcal{A} = \{f:\mathbb{R}\to\mathbb{R}$ is differentiable and $\underset{x\in\mathbb{R}}\sup{|f(x)|}=1\}$
In fact, I am suspecting that $I(f) \ge 2$ for all $f\in\mathcal{A}.$ To show this, I tried $g(x) = f(kx)$ for some $k>0, f\in\mathcal{A}.$ As $I(g) = \int_{-\infty}^{\infty}{|f(kx)|^2+|kf'(kx)|^2dx} = \frac{1}{k}\int_{-\infty}^{\infty}{|f(y)|^2dy} + k\int_{-\infty}^{\infty}{|f'(y)|^2dy},$ we must have $\int_{-\infty}^{\infty}{|f(x)|^2dx}\int_{-\infty}^{\infty}{|f'(x)|^2dx} \ge 1$ for all $f\in\mathcal{A}.$ Applying Holder inequality, I realized that showing $\int_{-\infty}^{\infty}{|f(x)f'(x)|dx} \ge 1$ for all $f\in\mathcal{A}$ is equivalent to the original minimization problem. Noting $2f(x)f'(x) = \frac{d(f(x))^2}{dx},$ I am trying to understand the term $ \int_{-\infty}^{\infty}{|f(x)f'(x)|dx}$ associated with the total variation of $\frac{1}{2}(f(x))^2$ over $\mathbb{R}.$ However, we only know $|f(x)| \le 1$, and this does not help in showing the above. How can I proceed further in this particular direction? Any hint or comment would be really helpful. Thanks in adavance.