If $f(x)$ is positive and continuous on $[0,1]$ and $f^2(t) \le 1+2\int_0^tf(s)\mathrm{d}s$, prove that $f(t)\le 1+t$.

Here's my thinking.

$$f^2(t) \le 1+2\int_0^tf(s)\mathrm{d}s \Rightarrow f(t)\le \sqrt{1+2\int_0^tf(s)\mathrm{d}s} $$ $$\Rightarrow \int_0^t\frac{f(t)}{\sqrt{1+2\int_0^tf(s)\mathrm{d}s}}\mathrm{d}t = \sqrt{1+2\int_0^tf(s)\mathrm{d}s}-1 \le t$$ $$ \Rightarrow 2\int_0^tf(s)\mathrm{d}s\le t^2+2t$$

This comes from the integration of two sides of $f(t)\le 1+t$. But I don't know what to do next.


2 Answers 2


You are almost done. You correctly derived that $$ \sqrt{1+2\int_0^tf(s)\mathrm{d}s} - 1 \le t \, . $$ Now use the initial given inequality to conclude that $$ f^2(t) \le 1+2\int_0^tf(s)\mathrm{d}s \le (1+t)^2 \implies f(t) \le 1+t \, . $$


Then $$f^2(t)\leq t^2+2t+1 = (t+1)^2\implies |f(t)|\leq |t+1|$$


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