Prove a inequality involving integrals

Question

Let $f$ be continuous on $[0,\infty)$ and let $t>0$. Show that $$\int_0^t(f(x))^2 dx\geq \frac{1}{t} \left(\int_0^t f(x)dx\right)^2$$ The hint is to consider $\displaystyle \int_0^t \left(f(x)-\frac{1}{t}\int_0^tf(x) dx \right)^2dx\geq 0$, but I have no idea how to, except for expanding the hint and get the LHS. Any kind soul please help!

Answer 1

If you do take the hint given to you: Let $c = \frac 1t\int_0^t f(x)dx$. $$ \int_0^t \left(f(x) -c\right)^2 dx = \int_0^t f(x)^2dx + c^2dx -2cf(x)dx \\ = \int_0^t f(x)^2dx + tc^2 - 2c\int_0^t f(x)dx = \int_0^t f(x)^2dx + tc^2 -2tc^2 \\ = \int_0^t f(x)^2dx - tc^2 $$

Of course, the initial LHS is non-negative, therefore $\int_0^t f(x)^2dx \geq tc^2$. Now substitute for $tc^2$ to get the result.

Answer 2

Jensen's inequality: \begin{align*} \left(\int_{0}^{t}f(x)\dfrac{dx}{t}\right)^{2}\leq\int_{0}^{t}f(x)^{2}\dfrac{dx}{t} \end{align*} applied to $u\rightarrow u^{2}$ on $[0,\infty)$.

Answer 3

$$\displaystyle \int_0^t \left(f(x)-\frac{1}{t}\int_0^tf(u) \,du \right)^2\,dx\geq 0$$

\begin{align}\displaystyle \int_0^t \left(f(x)\right)^2\,dx &\ge \frac2t \int_0^tf(x) \, dx \int_0^t f(u) \, du-\frac{1}{t^2}\int_0^t\left(\int_0^tf(x) du\right)^2 \, dx \\ &=\frac2t \left(\int_0^tf(x) \, dx\right)^2-\frac{1}{t^2}\left(\int_0^tf(x) du\right)^2 \int_0^t\, dx \\ &=\frac2t \left(\int_0^tf(x) \, dx\right)^2-\frac{1}{t}\left(\int_0^tf(x) du\right)^2\\ &=\frac1t \left(\int_0^tf(x) \, dx\right)^2\end{align}

Answer 4

$$\int_0^tf(x)dx$$ $$=\int_0^{\infty}1_{[0,t]}(x)f(x)dx$$ Now you can apply Cauch-Schwartz as claimed by the above comment.This is another way you can solve your problem without using you hint.

Answer 5

Some other good answers — especially the application of Jensen's inequality. Wanted to chime in with an approach rooted in basic probability. Sort of what you'd see if you took an introductory course.

Start with the hint and do some algebra:

$$\int_{0}^t(f(x) - \frac{1}{t}\int_0^tf(x)dx)^2dx = t\int_{0}^t(f(x) - \frac{1}{t}\int_0^tf(x)dx)^2\frac{1}{t}dx$$

If we tilt our heads to the side, we can look at x as a uniform random variable taking values on $[0, t]$. Then, the above can be rewritten as:

$$ \int_{0}^t(f(x) - \frac{1}{t}\int_0^tf(x)dx)^2\frac{1}{t}dx= t\cdot E[(f(x) - E[f(x))^2] $$

$$= t\cdot Var(f(x)) > 0$$

Therefore, we know the expression given in the hint is at least $0$.

We also know that $t\cdot Var(f(x)) = t\cdot (E[f(x)^2]-E[f(x)]^2) > 0$.

$$t\cdot(E[f(x)^2]-E[f(x)]^2) = t\cdot\int_{0}^t\frac{f(x)^2}{t}dx - t\cdot (\int_{0}^t\frac{f(x)}{t}dx)^2 $$

$$= \int_{0}^tf(x)^2dx - \frac{1}{t}(\int_{0}^tf(x)dx)^2 \geq 0$$