A calculus problem by Alapan Das

Firstly $0 \le f(x) \le K = \int_0^1f(t)\,dt$ . We are assuming that $A$ only contains integrable functions, of course! Then $\begin{aligned} f(x) & \le \; \int_0^x K\,dt \; = \; Kx \\ f(x) & \le \; \int_0^x Kt\,dt \; = \; \tfrac12Kx^2 \\ f(x) & \le \; \int_0^x \tfrac12Kt^2\,dt \; = \; \tfrac{1}{6}Kx^3 \end{aligned}$ and so on, deducing by induction that $0 \le f(x) \le \tfrac{1}{n!}K x^n \hspace{2cm} 0 \le x \le 1 \;,\; n \in \mathbb{N}$ Letting $n \to \infty$ , we deduce that $f(x) = 0$ for all $0 \le x \le 1$ , and hence $A$ contains only $\boxed{1}$ element.

Let $F = \int_0^xf(t)dt - f(x)$ .

$\frac{dF}{dx} = f(x) - \frac{df(x)}{dx}$ .

We are interested in the minima of F because we want it to be $\geq 0$ all the time.

So, By Definition Of Extrema ( $\frac{dF}{dx} = 0$ ),

$f(x) = \frac{df(x)}{dx}$ .

But this happens for only two functions.

1) $f(x) = 0$

2) $f(x) = Ce^x$

For C $\neq 0$ , $f(x) = Ce^x$ is not an eligible function for this problem as it doesn’t equal 0 at x = 0. After checking for $f(x) = 0$ , we conclude that cardinality of A = 1.

It's easily notable that $f(0)=0$ . Because if we use $x=0$ in the inequality then we get $0≥f(0)≥0$ . So, $f(0)=0$ .

Now take any point $x=h\rightarrow 0, h>0$ and use the inequality $\int_{0}^h f(t) dt$ .

We get that $\int_{0}^h f(t) dt=\frac{{f(h)}{h}}{2}<f(h)$ as $h\rightarrow0$ , doesn't satisfy the inequality $\int_{0}^h f(t) dt≥f(h)$ .

So, $f(h)=0$ is must and following this above process we get $f(x)=0$ for all $0≤x≤1$ .

And again if we take $x=2h$ then we get $I(2h)$ = $\int_{h}^{2h} f(t) dt$ as $I(h)=0$ . So again get the an triangle of area $\frac{{f(2h)}{h}}{2}$ .So, $f(2h)=0$ and so on.

Again if we let $x=3h$ then we get $I(3h)$ = $\int_{2h}^{3h} f(t) dt$ as $I(2h)=0$ . So again get the an triangle of area $\frac{{f(3h)}{h}}{2}$ . So, $f(3h)=0$ and so on.

And in this way get $f(rh)=0$ for all r such that $0\leq rh \leq1$ .

Hence, $f(x)=0$ is the only function in $A$ . Cardinality 1.