# Bounding integral of square root by square root of integral

Let $$f(x)\geq 0$$ be a function over $$[0,\infty)$$. How can I lower bound $$\int_{x=0}^{u}\sqrt{f(x)}dx$$ by $$c \sqrt{\int_{x=0}^{u}f(x)dx}$$ where $$\sqrt{\int_{x=0}^{u}f(x)dx}<\infty$$ and $$c>0$$ is a constant?

Is that in general possible? If not, what conditions can we assume for $$f(x)$$ to make the bound work? I think if I wanted to upper bound (instead of lower bound), I could use Jensen's inequality.

• An upper bound is not possible. For example take $f(x)=\frac1{x^2+1}$. – Peter Foreman Apr 9 at 13:20
• @PeterForeman thanks. I edited the question. – Susan_Math123 Apr 9 at 13:32

## 1 Answer

I'm afraid you won't be able to find any general lower bound since the family of functions $$f_{a} : x \mapsto x^{2a}$$ for $$a>0$$ would contradict any such bound. Indeed $$\int_{0}^{1} \sqrt{x^{2a}} \, \mathrm{d}x = \frac{1}{a+1} \, ,$$ and $$\sqrt{\int_{0}^{1} x^{2a} \, \mathrm{d}x } = \frac{1}{\sqrt{2a+1}} \, ,$$ so any lower bound of the form you suggest would imply $$\frac{1}{a+1} > \frac{C}{\sqrt{2a+1}}, \quad \forall a$$ for some absolute constant $$C$$. This cannot be true since taking $$a \to \infty$$ would give a contradiction.

Your problem does not end there however... the type of lower bound that you want does not scale properly. Indeed, suppose that we have the lower bound estimate. Let f be some function $$f : [0,1] \to [0,\infty)$$. For $$a>1$$ define the family of rescaled functions $$f_{a} : [0,1] \to [0,\infty)$$ $$f_{a}(x) = \begin{cases} f(ax) \quad &x \leq 1/a, \\ 0 \quad &x>1/a, \end{cases}$$ then you will see that $$\int_{0}^{1} \sqrt{f_{a}(x)} \, \mathrm{d}x = \frac{1}{a} \int_{0}^{1} \sqrt{f(x)} \, \mathrm{d}x \, ,$$ and that $$\sqrt{ \int_{0}^{1} f_{a}(x) \, \mathrm{d}x } = \frac{1}{\sqrt{a}} \sqrt{\int_{0}^{1} f(x) \, \mathrm{d}x} \, ,$$ so then using the lower bound estimate, we have that \begin{align*}\int_{0}^{1} \sqrt{f_{a}(x)} \, \mathrm{d}x > C \sqrt{ \int_{0}^{1} f_{a}(x) \, \mathrm{d}x }\,,\quad \forall a>1 \end{align*} and substituting in the above equalities we have \begin{align*}\frac{1}{a} \int_{0}^{1} \sqrt{f(x)} \, \mathrm{d}x > \frac{C}{\sqrt{a}} \sqrt{\int_{0}^{1} f(x) \, \mathrm{d}x}\,,\quad \forall a>1\,. \end{align*} Then upon taking the limit $$a\to \infty$$, (assuming $$\int_{0}^{1}f(x)\,\mathrm{d}x > 0$$) we must have that C = 0, and so there cannot be such an estimate.