Integral of function smaller equal to the absolute value of the function?

Question

my question is if following inequality is true?

$$ \int_{0}^{t} f(t) dt \le |f(t)|$$

Is that a true statement? I cannot find a counter example to disprove that inequality.

Thank for help.

Answer 1

I think you mean $\int_{0}^{t} f(x) dx \le |f(t)|$.

Look at $f(x)=x$ and $t=3$.

Answer 2

Take $t=1$, and let $f(x)=x$ and $g(x)=2-x$.

In both cases, $|f(1)|=|g(1)|=1$.

However:

$$\int_0^1f(x)dx = \frac12<1<\frac32=\int_0^1g(x)dx.$$

Without placing any further restrictions on $f$, it appears this inequality can go either way.

Answer 3

No. Consider the simple example of $f(t)\equiv 1$. Then $\int_0^tf(t)\; dt =t$, and $|f(t)|\equiv 1$, so you are asking whether $$t\leq 1$$ is always true. Clearly not.

Considering positive functions, this is asking whether the area under a curve above an interval is bounded by the height of the curve at the right end of the interval. You should be able to see that if the curve dips down further out, the area will still increase but the height on the right end could be as close to zeros as you wish.

Another instructive example might be to consider $f(x) =e^{-x}$.