# fixed constant function [closed]

let $f: [a,b]\longrightarrow \mathbb{R}$ be a Continuous and nonnegative function, so that $f(x) \leq \int_{a}^{x} f(t) dt$ ,$\forall x \in [ a, b]$.

Please tell me how to show that " $f$ is a fixed constant function".

## closed as off-topic by Adrian Keister, Theoretical Economist, Xander Henderson, Namaste, Deepesh MeenaSep 14 '18 at 1:24

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Have a look at https://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality#Integral_form_for_continuous_functions and the proof there.

In fact, in the spirit of amsmath's answer, we can, with a little more work, show that

$f(x) = 0, \; x \in [a, b]; \tag 1$

for we are given that

$0 \le f(x) \le \displaystyle \int_a^x f(t) \; dt; \tag 2$

now set

$g(x) = \displaystyle \int_a^x f(t) \; dt; \tag 3$

we note that $0 \le f(x)$ continuous implies

$g(x) \ge 0, \; x \in [a, b]; \tag 4$

also,

$g(a) = \displaystyle \int_a^a f(t) \; dt = 0, \tag 5$

and

$g'(x) = f(x) \le \displaystyle \int_a^x f(t) \; dt = g(x), \; x \in [a, b]; \tag 6$

in the light of (4) and (6), Gronwall's inequality implies

$0 \le g(x) \le g(a) e^{x - a}, \; x \in [a, b]; \tag 7$

now (5) and (7) force

$g(x) = 0, \; x \in [a, b]; \tag 8$

thus via (3),

$f(x) = g'(x) = 0, \; x \in [a, b]. \tag 9$