I have this problem and I don't know where to start. Can you give me some hint, please?
Suppose that $f:[a,b]\to [0,\infty]$ is a lebesgue integrable function. Let $B=\int_a^b fd\mu.$ Prove that $\sqrt{1+B^2}\leq \int_a^b\sqrt{1+f^2}d\mu\leq 1+B$.
Upper bound and lower bound of an integral
2
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lebesgue-integral
1 Answers
2
This is only true if $b-a=1$.
If $b-a=1$, then using Jensen inequality with $\phi(x)=\sqrt{1+x^2}$ which is convex, we get that $\phi\big(\int_a^b f(x)dx\big)\leq \int_a^b \phi(f(x))dx$ which implies the first inequality.
For the second inequality, just use the fact that $1+f^2\leq (1+f)^2$, since $f$ is nonegative.
To see that the inequality may not be true if $b-a\neq 1$, take $f(x)=x$, $a=1$ and $b=1.5$. Use matlab to see that $\sqrt{1+B^2}=1.9080$ and $\int_1^{1.5}\sqrt{1+x^2}dx=0.8017$.