# If $\int_0^x f^2(t)dt \le f(x)$ for all $x \in [0,1]$, then $\min_{[0,1]} f(x) \le 1$?

Suppose that $$f$$ is a continuous function on $$[0,1]$$ and $$\int_0^x [f(t)]^2dt \le f(x) \quad \text{for all} \quad x \in[0,1].$$ Prove or disprove $$\min_{0\le x\le 1} f(x) \le 1.$$ In case the desired inequality does not hold, what is the best upper bound?

Thanks.

• I'm assuming $f^2(t)=\left(f(t)^2\right)$ not $f(f(t))$, right? Commented Oct 6, 2018 at 20:17

I assume that $$f^2(t)$$ means $$\big(f(t)\big)^2$$. I have a very weak bound $$\min_{x\in[0,1]}\,f(x)<2\sqrt{2}\,,$$ and do not know how to improve it. Maybe somebody can use my proof to get a better bound.

Suppose on the contrary that there exists a function $$f:[0,1]\to\mathbb{R}$$ satisfying $$\int_0^x\,\big(f(t)\big)^2\,\text{d}t\leq f(x)\text{ for all }x\in[0,1]\tag{*}$$ such that $$k:=\min_{x\in[0,1]}\,f(x)\geq 2\sqrt{2}\,.$$ From (*), we get that $$f(x)\geq \max\big\{k^2x,k\}\text{ for all }x\in[0,1]\,.$$ We use (*) once again and find that $$f(x)\geq \int_0^{\frac1k}\,k^2\,\text{d}x+\int_{\frac{1}{k}}^x\,(k^2t)^2\,\text{d}t=\frac{2k}{3}+\frac{k^4x^3}{3}$$ for $$x\in\left[\dfrac1k,1\right]$$.

Define polynomials $$P_0$$, $$P_1$$, $$P_2$$, $$\ldots$$ as follows: $$P_0(z):=1\,,\,\,P_1(z):=z\,,\,\,P_2(x):=\frac{2}{3}+\frac{z^3}{3}\,,$$ and $$P_r(z)=1+\int_{1}^z\,\big(P_{r-1}(\zeta)\big)^2\,\text{d}\zeta$$ for $$r=3,4,5,\ldots$$. It can be proven by induction that, for each $$r=0,1,2,\ldots$$, $$f(x)\geq k\,P_r(kx)$$ for all $$x\in\left[\dfrac1k,1\right]$$.

We can prove by induction that $$0\leq P_r(z)\leq 1$$ for every $$z\in[0,1]$$. That is, the constant term of $$P_r$$ is nonnegative for every $$r=0,1,2,\ldots$$. It is obvious that the coefficients of higher-order terms in $$P_r$$ are nonnegative. Furthermore, the degree of $$P_r$$ is $$2^r-1$$, and the coefficient $$\lambda_r$$ of the $$(2^r-1)$$-st degree term of $$P_r$$ is given by the recurrence relation $$\lambda_r=\frac{1}{2^r-1}\lambda_{r-1}^2\,.$$ In other words, $$\lambda_r=\frac{1}{\prod\limits_{j=1}^r\,(2^j-1)^{2^{r-j}}}\geq \frac{1}{\prod\limits_{j=2}^r\,2^{j\cdot2^{r-j}}}=\frac{1}{2^{3\cdot 2^{r-1}-r-2}}\,.$$ That is, $$f(1)\geq k\,P_r(k)\geq \frac{k^{2^r}}{2^{3\cdot2^{r-1}-r-2}}$$ for every $$r=0,1,2,\ldots$$. However, as $$k\geq 2\sqrt{2}$$, $$k^{2^r}$$ grows faster than $$2^{3\cdot2^{r-1}-r-2}$$, namely, $$\lim_{r\to\infty}\,\frac{k^{2^r}}{2^{3\cdot2^{r-1}-r-2}}=\infty\,.$$ This yields a contradiction.

Remark. I believe that the polynomials $$P_r$$ converge pointwise, as $$r\to\infty$$, to $$P$$ on $$(-2,+2)$$, where $$P(z):=\frac{1}{2-z}\text{ for }z\in(-2,+2)\,.$$ I conjecture also that, for $$z\geq 2$$, $$\lim\limits_{r\to\infty}\,P_r(z)=\infty$$. If this is true, then it follows immediately that $$\min\limits_{x\in[0,1]}\,f(x)<2$$.

Counterexample. Interestingly, let $$k\in(1,2)$$ and define $$f(x):=\max\left\{k,\frac{k}{2-kx}\right\}\text{ for all }x\in[0,1]\,.$$ Then, we see that $$\min\limits_{x\in[0,1]}\,f(x)=k$$. Furthermore, $$f(x)=k\geq k^2x=\int_0^x\,k^2\,\text{d}t=\int_0^x\,\big(f(t)\big)^2\,\text{d}t$$ for $$x\in\left[0,\dfrac1k\right]$$. For $$x\in\left[\dfrac1k,1\right]$$, we have $$f(x)=\frac{k}{2-kx}=k+\int_{\frac1k}^x\,\left(\frac{k}{2-kt}\right)^2\,\text{d}t=\int_0^x\,\big(f(t)\big)^2\,\text{d}t\,.$$ This shows that, if $$\min\limits_{x\in[0,1]}\,f(x) for every such function $$f$$, then $$c\geq 2$$.

• @Nuno I think it is better that you accept Martin R's solution instead of mine. He gave a complete (and excellent) proof for your question. Commented Oct 7, 2018 at 10:15

We can show that $$\min_{x\in[0,1]}\,f(x) < 2 \, ,$$ and that is the best possible result, as Batominovski demonstrated with his counter-example.

Assume that $$f(x) \ge 2$$ in the interval. For $$0 < x \le 1$$ we define $$g(x) = \frac{1}{\int_0^x f^2(t) \, dt} \, .$$ Then $$g'(x) = -\frac{f^2(x)}{\left( \int_0^x f^2(t) \, dt\right)^2} \le -1$$ and therefore $$g(x) \ge g(1) + (1 - x) > 1 - x \, .$$ This implies $$4 x \le \int_0^x f^2(t) \, dt = \frac{1}{g(x)} < \frac{1}{1-x}$$ and setting $$x = \frac 12$$ gives a contradiction.

• This looks correct. With my counterexample, you have found the best bound for $\min\limits_{x\in[0,1]}\,f(x)$. Commented Oct 6, 2018 at 21:07
• But I would suggest that you make the assumption that $f(x)\geq 2$ for all $x$ at the beginning, so you don't have to say "we can assume that $f$ is strictly positive in the interval." Commented Oct 6, 2018 at 21:10
• @Batominovski: Good suggestion, thanks for the feedback! Commented Oct 6, 2018 at 21:19
• Great solution by the way, wish I could vote it up twice. Commented Oct 6, 2018 at 21:20