If $f>0$ then $\int_{0}^{t}f^{2}(u)du$ strictly increasing?

Question

Let $f:[0,T]\to \mathbb{R}$ a Riemann integrable function that satisfies $f(t)>0$ for all $t\in [0,T]$.

Consider the integral $$g(t)=\int_{0}^{t}f^{2}(u)du$$

Is it true that $g(t)$ is strictly increasing?

Answer 1

The question is not as simple as it sounds. It is based on the following result:

If $f$ is Riemann integrable on $[a, b] $ and $f(x) >0$ for all $x\in[a, b] $ then $\displaystyle \int_{a} ^{b} f(x) \, dx>0$.

This can be proved using the following deep result :

If $f$ is Riemann integrable on $[a, b] $ then it is continuous at some point in $[a, b] $.

For your function $g$ note that $$g(y) - g(x) =\int_{x}^{y}f^{2}(u)\,du>0$$ if $x<y$ so that $g$ is indeed strictly increasing.

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Update: On request of user gimusi via comment I am providing some explanation for the common approach to this problem and its limitations.

Let's relax our conditions and assume that $f$ is discontinuous only at a finite number of points say $t_{1}, t_{2}, \dots, t_{n}$ in interval $[0,T]$. Then $f^{2}$ is Riemann integrable on $[0,T]$ and $g$ is continuous on $[0,T]$. These points $t_{i} $ together with $0,T$ form a partition of interval $[0,T]$ and let this partition be $$P=\{x_{0}=0,x_{1},x_{2},\dots,x_{k}=T\} $$ Clearly $f^{2}$ is continuous on each open interval $(x_{i-1},x_{i})$ and therefore via Fundamental Theorem of Calculus $g'(t) =f^{2}(t)>0$ on each of these open intervals. Since $g$ is continuous on the closed interval $[x_{i-1}, x_{i}] $ it follows via mean value theorem that $g$ is strictly increasing in each closed interval $[x_{i-1},x_{i}]$ and hence it is strictly increasing on the whole interval $[x_{0},x_{k}]=[0,T]$.

Thus the usual approach based on sign of derivative can be used to show that $g$ is strictly increasing but only under the less general condition that $f$ has a finite number of discontinuities in $[0,T]$. For a general Riemann integrable function $f$ it is possible that the number of discontinuities is infinite (countably or uncountably). And then the above approach doesn't work.

On the other hand instead of counting the discontinuities of $f$ the proper approach is to measure them in the manner specified by Lebesgue and by doing so we get another deep theorem (with some effort):

If $f$ is bounded on $[a, b] $ then it is Riemann integrable on $[a, b] $ if and only if the set of discontinuities of $f$ on $[a, b] $ is of measure zero.

A trivial implication of the above result is that a Riemann integrable function must be continuous at infinitely many points which brings us back to the result mentioned (and linked) earlier.

Answer 2

If $f$ is continuos then:

$$g(t)=\int_{0}^{t}f^{2}(u)du \implies g'(t)=f^2(t)>0$$

Thus $g$ is strictly increasing

NOTE

it also applies to finite number of discontinuities (see Paramanand Singh answer and comments)

the converse is not necessary true