Let $ f(x)>0$, $f''(x)>0$, and $ \int_0^1 f(x)\,dx=1 $, for $t\in \mathbb R $, prove that: $$ \int_0^1 |f(x)-t| \, dx \le \frac{(1-t)^2+1}{2}.$$
This inequality maybe is very interesting. But I can't prove this.
sorry,everyone ,This Problem $f''(x)>0$ edit $f''(x)<0$,and other didn't change. I think this is true.
I mean :Let $ f(x)>0$, $f''(x)<0$, and $ \int_0^1 f(x)\,dx=1 $, for $t\in \mathbb R $, then we have $$ \int_0^1 |f(x)-t| \, dx \le \frac{(1-t)^2+1}{2}.$$
Edit: This is true for sure when $\min f\geq t$, $\max f\leq t$, or $\min f < t< \max f$ and $t\geq 4$. Note that it always holds for $t\leq 0$ and $t\geq 4$ for trivial reasons, as observed by Ivan Loh. Other than that...? It is apparently false in general for $0<t<2$. If Ivan Loh's example works, it remains to decide whether this is true for $2\leq t<4$.
Case 1: $\min f\geq t$. Then $$\int_0^1|f(x)-t|dx=\int_0^1(f(x)-t)dx=\int_0^1f(x)dx-\int_0^1tdx=1-t\leq\frac{(1-t)^2+1}{2} .$$
Case 2: $\max f\leq t$. Then $$ \int_0^1|f(x)-t|dx=\int_0^1(t-f(x))dx=\int_0^1tdx-\int_0^1f(x)dx=t-1\leq\frac{(1-t)^2+1}{2} . $$
Case 3: $\min f < t< \max f$. By the intermediate value theorem there exists $x\in[0,1]$ such that $f(x)=t$. Since $f$ is strictly convex and since $t>\min f$, there exist actually $0\leq x_1< x_2\leq 1$ such that $f\geq t$ on $[0,x_1]\cup[x_2,1]$ and $f\leq t$ on $[x_1,x_2]$. Note that we may have $x_1=0$ or $x_2=1$. Now $$ \int_0^1|f(x)-t|dx=\int_0^{x_1}(f(x)-t)dx+\int_{x_2}^1(f(x)-t)dx+\int_{x_1}^{x_2}(t-f(x))dx $$ $$ =1-t+2t(x_2-x_1)-2\int_{x_1}^{x_2}f. $$ Now this is $\leq \frac{(1-t)^2+1}{2}$ if and only if $$ t^2-4(x_2-x_1)t+4\int_{x_1}^{x_2}f\geq 0. $$ One recovers the fact that it holds for $t\geq 4$. But I don't really know what to do with that next...
By triangle inequality, we have
\begin{align} \int_{0}^{1}{|f(x)-t| dx} \leq \int_{0}^{1}{|f(x)| dx}+\int_{0}^{1}{|t| dx}=\int_{0}^{1}{f(x) dx}+\int_{0}^{1}{|t| dx}=1+|t| \end{align}
If $t \leq 0$, $1+|t|=1-t \leq \frac{(1-t)^2+1}{2}$. If $t \geq 4$, $1+|t|=1+t \leq \frac{(1-t)^2+1}{2}$.
Thus the inequality is true for $t \leq 0$ and $t \geq 4$. It is false for $0<t<2$, as shown below.
Edit: As @gerw points out, fixing the point $\frac{1}{2}$ isn't a good idea. The counterexample can probably be improved to show that the inequality fails for $0<t<4$, by replacing $\frac{1}{2}$ with a variable $c$ which tends to $1$ from below. I don't feel like redoing the calculations at the moment, so I will provide an intuitive explanation why the inequality should fail (which can be easily made rigorous, if you please, by explicitly constructing the counterexample).
The reason is that we cannot do much better than the bound $1+|t|$ achieved earlier; In other words, if $0<t$, then $\forall \epsilon >0$ ($\epsilon$ sufficiently small), we can find a function $f(x)$ satisfying the given conditions and such that $1+t \geq \int_{0}^{1}{|f(x)-t| dx}>1+t-\epsilon$.
To do this, note that it is easy to see that if we fix $\epsilon$, then fix $a$ s.t. $1-\frac{\epsilon}{2t}<a<1$, we can find a strictly convex and positive function $f(x)$ s.t. $\int_{0}^{a}{f(x) dx}<\frac{\epsilon}{2}-t(1-a)$. (So the main contribution in the integral $\int_{0}^{1}{f(x) dx}$ comes from $[a, 1]$)
Then for $0<t$:
\begin{align} \int_{0}^{1}{|f(x)-t| dx}& =\int_{0}^{a}{|f(x)-t| dx}+\int_{a}^{1}{|f(x)-t| dx} \\ & \geq \int_{0}^{a}{(t-f(x)) dx}+\int_{a}^{1}{(f(x)-t) dx} \\ & =t(2a-1)+1-2\int_{0}^{a}{f(x) dx} \\ & >t(2a-1)+1-2(\frac{\epsilon}{2}-t(1-a)) \\ & =t+1-\epsilon \end{align}
Counterexample for $0<t<2$: $$f(x)=\begin{cases} m+\frac{6}{n}(x-\frac{1}{2})^2+\frac{3}{n}(x-\frac{1}{2})^3 & \text{if} \; 0\leq x \leq \frac{1}{2} \\ m+\frac{6}{n}(x-\frac{1}{2})^2+(64(1-m)-\frac{29}{n})(x-\frac{1}{2})^3 & \text{if}\; \frac{1}{2}<x \leq 1 \end{cases}$$ is a counterexample for $0<t<2$, where $n$ is sufficiently large and $m>0$ is sufficiently small depending on $t$.
Proof: Clearly $f$ is twice differentiable for $x \in [0,1], x\not=\frac{1}{2}$. We check that $f$ is also twice differentiable at $\frac{1}{2}$. Indeed,if we differentiate the 2 pieces of $f(x)$ twice, and then evaluate at $\frac{1}{2}$, the 2 values agree.
If $0\leq x \leq \frac{1}{2}$, then $f(x)=m+(x-\frac{1}{2})^2(\frac{6}{n}+\frac{3}{n}(x-\frac{1}{2}))>0$.
If $\frac{1}{2}<x \leq 1$, then $f(x)=m+(x-\frac{1}{2})^2(\frac{6}{n}+(64(1-m)-\frac{29}{n})(x-\frac{1}{2}))>0$.
If $0\leq x \leq \frac{1}{2}$, then $f''(x)=\frac{12}{n}+\frac{18}{n}(x-\frac{1}{2})>0$.
If $\frac{1}{2}<x \leq 1$, then $f''(x)=\frac{12}{n}+6(64(1-m)-\frac{29}{n})(x-\frac{1}{2})>0$.
\begin{align} &\int_{0}^{1}{f(x) dx}\\ &=m+\frac{6}{n}\int_{0}^{1}{(x-\frac{1}{2})^2 dx}+\frac{3}{n}\int_{0}^{\frac{1}{2}}{(x-\frac{1}{2})^3 dx}+(64(1-m)-\frac{29}{n})\int_{\frac{1}{2}}^{1}{(x-\frac{1}{2})^3 dx} \\ &=1 \end{align}
\begin{align} &\int_{0}^{1}{|f(x)-t| dx}\\ &=\int_{0}^{\frac{1}{2}}{|m+\frac{6}{n}(x-\frac{1}{2})^2+\frac{3}{n}(x-\frac{1}{2})^3-t| dx}+\int_{\frac{1}{2}}^{1}{|m+\frac{6}{n}(x-\frac{1}{2})^2+(64(1-m)-\frac{29}{n})(x-\frac{1}{2})^3-t| dx} \\ & \geq \int_{0}^{\frac{1}{2}}{(t-(m+\frac{6}{n}(x-\frac{1}{2})^2+\frac{3}{n}(x-\frac{1}{2})^3)) dx}+\int_{\frac{1}{2}}^{1}{(m+\frac{6}{n}(x-\frac{1}{2})^2+(64(1-m)-\frac{29}{n})(x-\frac{1}{2})^3-t) dx} \\ & =(64(1-m)-\frac{26}{n})\frac{(\frac{1}{2})^4}{4} \\ & =1-m-\frac{13}{32n} \\ & >\frac{(1-t)^2+1}{2} \end{align}
for $0<t<2$, when we take sufficiently large $n$ and sufficiently small $m$.
I will present a counter-example for $t \in (0,4)$. Set $$f(x) = \epsilon + \begin{cases}c_1 \, (x-x_0)^4 & x \le x_0, \\ c_2 \, (x-x_0)^4 & x > x_0.\end{cases}$$ Obviously, we have $f \in C^2([0,1])$. For $c_1,c_2,\epsilon > 0$ and $x_0 \in (0,1)$, you have $f > 0$, $f'' > 0$. Now, let $c_1, \epsilon$ small and $x_0$ big be given, such that $f(0) < t$ (and hence, $f < t$ on $[0,x_0]$). We have $$\int_0^1 f \, d x = \epsilon + \frac{c_1}{5} \, x_0^5 + \frac{c_2}{5}\,(1-x_0)^5,$$ hence, we choose $c_2 = \frac{5}{(1-x_0)^5}\big(1 - \epsilon - \frac{c_1}5 \, x_0^5\big)$, such that the integral equals one.
Let us look for $f(\tilde x) = t$ for $\tilde x \in (x_0,1)$. We have $$\tilde x = x_0 + (1-x_0)^{5/4} \, \frac{t-\epsilon}{5\,(1-\epsilon-c_1\,x_0^5/5)} \le x_0 + (1-x_0)^{5/4} < 1$$ for $c_1, \epsilon$ small enough.
Now, we have \begin{align*}\int_0^1 |f-t|\,dx &\ge \int_0^{x_0} | f - t | \, dx+ \int_{\tilde x}^1 f -t \, d x. \end{align*} For the first integral, we obtain \begin{align*}\int_0^{x_0} | f - t | \, dx= t\,x_0 - \epsilon - \frac{c_1}{5} \, x_0^5. \end{align*} For the second, we have \begin{align*} \int_{\tilde x}^1 f - t \, d x &\ge \int_{x_0 + (1-x_0)^{5/4} }^1 c_2 \, (x-x_0)^4 + \epsilon \, d x - t\,(1-\tilde x) \\ &= \frac15 \, c_2 \, \Big( (1-x_0)^5 - (1-x_0)^{25/4} \Big) + \epsilon\,(1-x_0-(1-x_0)^{5/4}) - t \,(1-\tilde x) \\ &\ge \Big( 1 - (1-x_0)^{5/4} \Big) \, (1-\epsilon-\frac{c_1}5\,x_0^5) + \epsilon\,(1-x_0-(1-x_0)^{5/4}) - t \,(1-\tilde x) \end{align*} Alltogether, we have \begin{align*} \int_0^1 |f-t| \, d x &\ge t\,x_0 - \epsilon - \frac{c_1}{5} \, x_0^5 \\&\quad+ \Big( 1 - (1-x_0)^{5/4} \Big) \, (1-\epsilon-\frac{c_1}5\,x_0^5) + \epsilon\,(1-x_0-(1-x_0)^{5/4}) - t \,(1-\tilde x) \\ &= t\,(x_0+\tilde x - 1) + \Big( 2 - (1-x_0)^{5/4} \Big) \, (-\epsilon-\frac{c_1}5\,x_0^5) \\&\quad+ \Big( 1 - (1-x_0)^{5/4} \Big) + \epsilon\,(1-x_0-(1-x_0)^{5/4}) \end{align*} Note that the right-hand side converges to $t + 1$ if $x_0, \epsilon, c_1$ are chosen appropriate. Since $t+1 > \frac{(1-t)^2 +1}{2}$ for $t \in (0,4)$, we have $$\int_0^1 | f - 1| \, dx > \frac{(1-t)^2+1}2.$$
