Show that $\int_{a}^{c}f(x)dx = 0$ for all $c\in [a,b]$ if and only if $f(x) = 0$ for all $x\in [a,b]$.

Question

Exercise:

Suppose that $a<b$ and that $f:[a,b]\rightarrow R$ is continuous. Show that $\int_{a}^{c}f(x)dx = 0$ for all $c\in [a,b]$ if and only if $f(x) = 0$ for all $x\in [a,b]$.

attempt of proof:

Suppose that $a<b$ and that $f:[a,b]\rightarrow R$ is continuous. Let $m$ and $M$ be the infimum and supremum of f. Since $f(x) = 0 $ for all $x\in [a,b]$ $m = 0$ since $f(x) = 0$. Thus, for all any partition $P$ of $[a,b]$, $L(f,P) = m(b-a) = 0$ implies $L(f,P) = 0$. Hence, $(L)\int_{a}^{c}f(x)dx$ = $sup{L(f,P)}$ = $0$.

In a similar way we can working with the supremum. Thus, if both the lower integral of f and the upper integral of f have the same value , we can conclude then the value of the

$\int_{a}^{c}f(x)dx = 0$ for all $c\in [a,b]$

The converse is trivial, since $\forall c\in [a,b]$ $f(c) = 0$ since $f(x)=0$ , so $\int_{a}^{c}f(x)dx = 0$

Can someone please help me? I don't know if this is a way to prove it. Any feedback/hint or better way would be really appreciated.Thank you in advance.

Answer 1

The basic idea is that, if $f$ is continuous and there is a point $z$ such that $f(z) \ne 0$, then there is a neighborhood of $z$ such that $f(x) \ne 0$ and has the same sign as $f(z)$ in that neighborhood. We then look at the integral of $f$ in that neighborhood and prove that the integral of $f$ over that neighborhood is non-zero.

Suppose $f(z) \ne 0$ for some $z$. Then, since $f$ is continuous, for any $\epsilon > 0$ there is a $\delta$ such that $|f(z)-f(x)| < \epsilon$ for all $x$ such that $|z-x| < \delta$.

In what follows, assume that $f(z) > 0$. If $f(z) < 0$, reverse the sign of $f$.

Now choose $\epsilon = |f(z)/2|$. Let $d$ be the $\delta$ for this $\epsilon$. Then $|f(z)-f(x)| < |f(z)/2|$ for all $x$ such that $|z-x| < d$.

Therefore, by the triangle inequality, for $|x-z| \le d $, $|f(z)| =|f(z)-f(x) + f(x)| \le |f(z)-f(x)| + |f(x)| $ or $|f(x)| \ge |f(z)|-|f(z)-f(x)| \ge |f(z)|-|f(z)|/2 = |f(z)|/2 $.

Since $|f(x)-f(z)| \le f(z)/2 $, $f(x) \ge f(z)/2$.

Therefore, $\int_{z-d}^{z+d} f(x)dx \ge \int_{z-d}^{z+d} f(z)/2\ dx = (2d)(f(z)/2) =d f(z) > 0 $.

Therefore, since $\int_a^{z+d} f(x) dx =\int_a^{z-d} f(x) dx +\int_{z-d}^{z+d} f(x) dx $, since $\int_{z-d}^{z+d} f(x) dx > 0 $, if $\int_a^{z+d} f(x) dx = 0 $, then $\int_a^{z-d} f(x) dx < 0 $.

Therefore, if $f$ is continuous and there is a $z$ such that $f(z) \ne 0$, it is not true that $\int_a^c f(x) dx = 0 $ for all $c \in [a, b]$.

Taking the contrapositive, if $f$ is continuous and $\int_a^c f(x) dx = 0 $ for all $c \in [a, b]$, then $f(x) = 0$ for all $c \in [a, b]$.

Answer 2

You have proved that if $f(x)=0$, then $\int_a^c f(x)\;dx=0$ just fine - just clean it up a bit. However, the other direction you have not shown. You want to show that if $\int_a^c f(x)\;dx=0$ for all $c\in[a,b]$ then $f(x)=0$. You do not yet know that $f(x)=0$. I shall give a hint:

You know that $\int_a^cf(x)\;dx=0$ for $c \in [a,b]$. But what of the area for the curve between $[c,b]$? You know that $f$ is continuous. Use the fact that it has a well defined inf and sup. How large/small can the integral be for $f$ on $[c,b]$? Can you make this very small despite any choice of partition by choosing $c$ very close to $b$?