An unusual integral inequality

Question

Is the following true when $f \in C^1([0,1])$ and $f(0) =0$

$$\int_0^1|(f^2)'| \leq \int_0^1|f'|^2$$

I can't come up with a counterexample but have not been able to prove this either. Tried Holder, Cauchy-Schwarz, etc.

Answer 1

It is true.

Taking $g(x) = \int_0^x |f'(s)| \, ds$

$$|f(x)| = \left|\int_0^xf'(s) \, ds \right| \leqslant \int_0^x |f'(s)| \, ds = g(x),$$

and, by the FTC, $g'(x) = |f'(x)|$.

Thus,

$$\int_0^1 |(f^2)'| \, dx = 2 \int_0^1|f(x)|\, |f'(x)| \, dx \leqslant 2 \int_0^1g(x)\, g'(x) \, dx = \int_0^1(g^2)' \,dx \\ = g^2(1) - g^2(0) \\ = \left(\int_0^1 |f'(x)| \, dx \right)^2 $$

Applying the Cauchy-Schwartz inequality to the RHS we get

$$\int_0^1 |(f^2)'| \, dx \leqslant \left(\int_0^1 |f'(x)| \, dx \right)^2\leqslant \int_0^1|f'(x)|^2 \, dx $$

Answer 2

EDIT: Turns out I was wrong and the inequality is indeed true. @RRL has posted a proof.

I believe you're inequality is not quite true, but nearly. It can be proven if you can put up with an extra factor of $2$. This is reminiscent of a Poincare-type inequality. Writing out the derivative on the left hand side and using Cauchy-Schwarz, we have $$\int^1_0 \lvert (f^2)' \rvert = 2\int^1_0 \lvert f' \rvert \cdot \lvert f \rvert \le 2 \left(\int^1_0 \lvert f' \rvert^2 \right)^{1/2} \left(\int^1_0 \lvert f \rvert^2 \right)^{1/2}.$$ But note that by the fundamental theorem of calculus, for any $t \in [0,1]$, $$\lvert f(t) \rvert = \left \lvert\int^t_0 f'(x) dx \right \rvert \le \int^t_0 \lvert f' \rvert \le \int_0^1 \lvert f' \rvert \le \left(\int^1_0 1^2\right)^{1/2}\left(\int^1_0 \lvert f ' \rvert^2\right)^{1/2} =\left(\int^1_0 \lvert f ' \rvert^2\right)^{1/2}.$$ Squaring both sides, integrating and taking the $1/2$ power, we have $$\left(\int^1_0 \lvert f \rvert^2\right)^{1/2} \le \left(\int^1_0 \lvert f ' \rvert^2\right)^{1/2}.$$ Plugging this bound into the first line above, we have $$\int^1_0 \lvert (f^2)'\rvert \le 2 \int^1_0 \lvert f ' \rvert^2.$$

Answer 3

If you want a tighter constant than $2$ you can prove using Fourier analysis that $$\int_{0}^1 |f(t)|^2 \mathrm d t \le \frac4{\pi^2} \int_0^1 |f'(t)|^2\mathrm d t$$ and using $$2|f'(t) f(t)| \le |f'(t)|^2 + |f(t)|^2$$ you prove that $$\int_0^1 \left|(f^2)'\right| = \int_0^1 2|f'(t)||f(t)|\mathrm d t \le \left(1+\frac4{\pi^2}\right) \int_0^1 |f'(t)|^2\mathrm d t.$$

Now to prove the first inequality assume that $$g(t) = \sum_{n\in \mathbb Z} c_n e^{in\pi t},\, \forall t\in \mathbb R$$ with $c_n$ such that $g \in C^1$. By differentiating you have $$g'(t) = \sum_{n\in \mathbb Z} in\pi c_n e^{in\pi t}$$ and by Parseval theorem $$\int_{-1}^{1} |g(t)|^2\mathrm d t = \sum_{n\in \mathbb Z} |c_n|^2$$ and $$\int_{-1}^{1} |g'(t)|^2\mathrm d t = \sum_{n\in \mathbb Z} \pi^2n^2|c_n|^2$$ this proves that $$\int_{-1}^1 |g(t)|^2 \mathrm d t \le \frac1{\pi^2} \int_{-1}^{1} |g'(t)|^2\mathrm d t$$ Now with the function $f$ let $h$ be the 2-periodic extension of the following function $$h(x) = \left\{\begin{array}{cc}f(2x) & x \in \left[0,\frac12\right]\\ f(2(1-x)) & x \in \left[\frac12, 1 \right] \\ -f(-2x) & x \in \left[-\frac12,0\right] \\ -f(2(1+x)) & x \in \left[-1,-\frac12\right]\end{array}\right.$$ $h$ is $C^1$ on $\mathbb R$, then using the above result on $h$, we have $$\int_{-1}^1 h(t)^2 \mathrm d t \le \frac1{\pi^2} \int_{-1}^1 |h'(t)|^2 \mathrm d t$$ since $$\int_{-1}^1 h(t)^2 \mathrm dt = \int_0^\frac12 f(2t)^2 \mathrm dt + \int_{-\frac12}^0 f(-2t)^2\mathrm dt + \int_{-1}^{-\frac12} f(2(1+t))^2 \mathrm d t +\int_{\frac12}^1 f(2(1-t))^2 \mathrm d t\,\; = 2\int_0^{1} |f(u)|^2 \mathrm d u$$ and with the same idea you have $$\int_{-1}^1 |h'(t)|^2 \mathrm d t = 8\int_{0}^1 |f'(t)|^2 \mathrm d t$$