Which continuous function satisfies $\int\limits_0^x f(t) dt \ge f(x)$ for all $x$ in $[0,1] $?

Question

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Let $f:[0,1] \to [0,\infty)$ be continuous. Suppose
$$ \int\limits_0^x f(t) dt \ge f(x), \text{ for all } x \in [0,1]$$ Then which of the following is true
A. No such function exists
B. There are infinitely many such functions
C. There is only one such function
D. There are exactly two such functions

See the end of the question for correct answer.
I have been able to figure out that, perhaps $e^x$ is one such function so A is NOT the answer.
Correct answer : - C
Source - Tata Institute of Fundamental Research, Graduate School Admissions 2014

Answer 1

$e^x$ does not satisfy the condition: try $x=0$.

Hint: if $0 < x < 1$, $$\int_0^x f(t)\; dt \le x \sup_{0 \le t \le x} f(t)$$

Answer 2

Let $g(x) = \int_0^x f(t)\, dt$. Then $g'(x) = f(x) \le g(x)$ for all $x\in [0,1]$. Show that $\frac{d}{dx}(g(x)e^{-x}) \le 0$, and hence $g \le 0$. Since $f \ge 0$, then $g \ge 0$. So $g \equiv 0$, and consequently $f = 0$.

Answer 3

Given any $f : [0,1] \to [0,\infty)$ continuous which satisfies

$$f(x) \le \int_0^x f(t) dt,\quad \forall x \in [0,1]\tag{*1}$$

Since $[0,1]$ is compact, $f$(x) achieves it maximum value $M \ge 0$ somewhere.
We have:

$$\begin{align} & f(x) \le M,\quad \forall x \in [0,1]\\ \implies & f(x) \le \int_0^x f(t) dt \le \int_0^x M dt = Mx, \quad\forall x \in [0,1]\\ \implies & f(x) \le \int_0^x f(t) dt \le \int_0^x Mt dt \le M\frac{x^2}{2} \le \frac{M}{2},\quad\forall x \in [0,1] \end{align} $$ Since $M$ is the maximum of $f(x)$ over $[0,1]$, the last inequality implies $$M = \max_{x\in [0,1]} f(x) \le \frac{M}{2}$$
Together with $M \ge 0$, this forces $M = 0$ and hence $f(x)$ is the zero function.

So the answer is $C$. There is only one function, the zero function, which can satisfy $(*1)$.

Answer 4

Less elegant than the other answers, but maybe it will help someone...

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Following Robert Israel's hint, note $$f(x) \le \int_0^x f(t)\mathop{dt} \le x \cdot \sup_{0 \le t \le x} f(t)$$ for all $x \in [0,1]$.

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Hint 2: What happens when you apply the above hint at $x=0$?

For $x=0$, you get $f(0) \le 0$ and thus $f(0)=0$ since $f$ is nonnegative.

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Hint 3: Can you use the above to conclude that $f =0$ on $[0,1]$?

Suppose $f$ is not identically zero. Then $f(x) > 0 $ for some $x \in (0,1]$. By continuity (intermediate value theorem), we can actually choose $x \in (0,1)$. Then there exists some $c$ such that $x<c<1$.

By continuity, the maximum of $f$ on $[0,c]$ is attained at some $x^* \in [0,c]$, and we have $0<f(x) \le f(x^*)$. But then, applying hint 1 gives $$f(x^*) \le \int_0^{x^*} f(t) \mathop{dt} \le x^*\cdot \sup_{0 \le t \le x^*} f(t) = x^* \cdot f(x^*) \le c \cdot f(x^*) < f(x^*),$$ a contradiction.