a slight variant of "Prove $\int_0^1 |f'(x)|dx \leq 2\int_0^1 |f(x)|dx + \int_0^1 |f''(x)|dx.$"

Question

Suppose $f(x)$ is twice differentiable on $[0,1]$ (with $f''$ being continuous) and $f(0)f(1)\ge 0$. Prove $\int_0^1 |f'(x)|dx \leq 2\int_0^1 |f(x)|dx + \int_0^1 |f''(x)|dx.$

I have some new questions regarding this similar question's solution:

1. Why is $f(x)-f(0) > 2\int_0^1 f(t)dt$? We can conclude $f(x)-f(0) > 2x\int_0^1 f(t)dt$ but this doesn't give us the bounds we want.
2. Does the claim still hold if $f''$ is just integrable? The example where $f(x) = -(x-1/2)^2/2$ for $x\leq 1/2$ and $(x-1/2)^2/2$ doesn't work.

Here's the proof for the similar claim:

We have that $\int_a^1 |f'(x)|dx \leq \int_a^1 |\int_a^x f''(t)dt| dx \leq \int_a^1 \int_a^x |f''(t)|dt dx \leq \int_a^1 \int_t^1 |f''(t) |dxdt \leq \int_a^1 |f''(t)|dt $ and adding give $\int_0^1 |f'(x)|dx \leq \int_0^1 |f''(x)|dx.$

Suppose $|f'(x)|$ takes a minimum value $m$ at $x=x_1$ and a maximum value $M$ and $x=x_2$. These values are indeed attained since $f'$ is continuous (and hence so is $|f'|$). One can also apply the EVT to $f$. We then have $m \leq \int_0^1 |f'(x)|dx\leq M.$ If $f(0) = f(1)$, by Rolle's theorem we may choose $x\in(0,1)$ with $f'(x)=0$. If $f(0) < f(1),$ by the MVT there exists some $c\in (0,1)$ with $f'(c) > 0$. If $f(1) < f(0)$, by the MVT, there exists some $c\in (0,1)$ with $f'(c) < 0$. There has to be some significance to the condition that $f(0)f(1) \ge 0$. For instance, consider the function $f(x) = -x+\dfrac{1}2.$ Then we get $\int_0^1 |f'(x)|dx = 1, 2\int_0^1 |f(x)|dx + \int_0^1 |f''(x)|dx = 2(\int_0^{1/2} \dfrac{1}2 -x dx + \int_{1/2}^1 x-\dfrac{1}2 dx) = 4\int_0^{1/2} \dfrac{1}2 - xdx = 4(\dfrac{1}4 - \dfrac{1}8) = \dfrac{1}2$ and so the inequality does not hold in this case. It also doesn't matter whether $f''$ is continuous, so theorems involving continuity of functions should probably just be applied to f or $f'$.

Below is a list of things I think might help but that don't seem to lead to much progress.

For any $a,x\in \mathbb{R}$, Taylor's theorem ensures there exists $c$ strictly between $a$ and $x$ so that $f(x) = f(a) + f'(a)(x-a) + \dfrac{f''(c)(x-a)^2}2$. If there exists $c,d\in\mathbb{R}$ so that $f(x) = cx$, then the inequality in the question holds with equality. First consider the case where $x_1 < x_2.$ Then $m(x_2-x_1)\leq \int_0^1 |f'(x)|dx \leq \int_0^{x_1} |f'(x)| dx + M(x_2-x_1) + \int_{x_2}^1 |f'(x)|dx$.

Answer 1

Why is $f(x)-f(0) > 2\int_0^1 f(t)dt$? We can conclude $f(x)-f(0) > 2x\int_0^1 f(t)dt$ but this doesn't give us the bounds we want.

You just pointed out an error in the that answer. It should have been what you proposed, $f(x)-f(0)>2x\int_0^1f(t)dt$, which in fact, will still lead to a contradiction. Once this error has been corrected, that answer can go on as before.

therefore $$\int_0^1f(x)\,dx-f(0) =\int_0^1(f(x)-f(0))\,dx >\int_0^12x\,dx\int_0^1f=\int_0^1f,$$ which is false as $f(0)\ge0$ is assumed. This contradiction implies the existence of the wanted $\xi$.

2. Since the arguments are, in fact, about positive functions, all integrations and change of integration order are valid if $|f''|$ is just integrable.