0
$\begingroup$

I need help in this particular question due to the reason that I am not very comfortable in questions when integrals are to be proved identically zero.

Question Let $f:[0,1]\times[0,1]\longrightarrow[0,\infty)$ be a continuous function . Suppose that $$\int_{0}^{1}\bigg(\int_{0}^{1} f(x,y)dy\bigg)dx=0.$$ Then prove that $f$ is an identically zero function.

I think this would be due to the fact that range of $f$ is non-negative. But In such kind of questions I am not able to rigorously prove these. Actually, I was taught analysis by a really poor instructor and problem solving/assignments in our university are minimal. So, I use to try exercises by myself and ask questions here.

Can anyone please tell on how should I attempt this particular problem.

I shall be really thankful.

$\endgroup$
10
  • 1
    $\begingroup$ This is a standard theorem in integration. See math.stackexchange.com/questions/777827/… $\endgroup$ Aug 15 '20 at 7:02
  • 3
    $\begingroup$ Suppose there is some $(x_0,y_0) \in [0,1] \times [0,1]$ such that $f(x_0,y_0) = c > 0$. Can you show that continuity implies that $f(x,y) > c/2$ on some neighborhood of $(x_0, y_0)$? $\endgroup$
    – user169852
    Aug 15 '20 at 7:03
  • 3
    $\begingroup$ Does this answer your question? Proof that the Integral of a Positive Function is Positive? $\endgroup$
    – Romain S
    Aug 15 '20 at 7:04
  • 1
    $\begingroup$ I don't see how the links help. $\endgroup$
    – user169852
    Aug 15 '20 at 7:05
  • 2
    $\begingroup$ I guess we'll agree to disagree. You just need the simple theorem that if $f \geq g$ then $\int f \geq \int g$. $\endgroup$
    – user169852
    Aug 15 '20 at 7:22
1
$\begingroup$

Hint :

  1. Prove the result in the one-dimensional case : that is, if $\varphi : [0,1] \rightarrow \mathbb{R}$ is a non-negative and continuous function such that $$\int_0^1 \varphi(x) \mathrm{dx} = 0$$ then $\varphi = 0$ (maybe you already know how to prove that).

  2. Apply this result to the function $x \mapsto \int_0^1 f(x,y) \mathrm{dy}$ (check all the hypothesis correctly !) to deduce that for all $x \in [0,1]$, you have $\int_0^1 f(x,y) \mathrm{dy} = 0$.

  3. Re-apply the result to the function $y \mapsto f(x,y)$ with $x$ fixed to deduce finally that for all $x,y$, you have $f(x,y)=0$.

$\endgroup$
0
0
+50
$\begingroup$

For continuous functions on $S=[0,1]\times [0,1],$ the iterated integral is the same as the double integral. So your expression is the same as $\int_S f\,dA,$ where $A$ denotes area measure.

Now if $f(a,b)>0$ for some $(a,b)\in S,$ there is a subsquare of positive area $S'\subset S$ containing $(a,b)$ such that $f>f(a,b)/2$ on $S'$. This follows from the continuity of $f.$ Since $f\ge 0$ in $S,$ we have

$$\int_S f\,dA \ge \int_{S'} f\,dA \ge f(a,b)/2\cdot A(S')>0.$$

That's a contradiction, hence $f=0$ everywhere in $S.$

$\endgroup$
0
$\begingroup$

By contradiction, suppose there is an $(x_0, y_0)$ such that $f(x_0, y_0) \neq 0 $. Since $f$ is continous then $\forall k: 0<k<f(x_0, y_0)$ there is a $U = (x_0+\varepsilon, x_0-\varepsilon) \times (y_0+\delta, y_0-\delta)$ such that $f \geq k$ in $U$. We can conclude that then $$\int_0^1\int_0^1f(x, y)dxdy \geq \int_{x_0-\varepsilon}^{x_0+\varepsilon}\int_{y_0-\delta}^{y_0+\delta} f(x, y)dxdy \geq \int_{x_0-\varepsilon}^{x_0+\varepsilon}\int_{y_0-\delta}^{y_0+\delta} k dxdy =k\cdot 2\varepsilon\cdot2\delta>0$$ Which is a contradiction with the assumption that $\displaystyle \int_0^1\int_0^1f(x, y)dxdy=0$.

$\endgroup$
-2
$\begingroup$

Since the function is strictly non-negative we can interpret this integral as the volume under the graph of the function on the square $[0,1]\times [0,1]$. Now assume there exists some point $(x_0,y_0)$ such that $f(x_0,y_0) \neq 0$. Since the function is continuous there exists some small neighborhood of $(x_0,y_0)$ where all $f(x,y) > 0$ in that neighborhood, by the intermediate value theorem. Since this neighborhood is strictly positive, it must have some positive, non-zero volume. This is impossible because the volume is given as zero.

$\endgroup$
5
  • 2
    $\begingroup$ horrible answer $\endgroup$ Sep 30 '20 at 19:35
  • $\begingroup$ @mathworker21 what requires elaboration? $\endgroup$ Sep 30 '20 at 19:37
  • 2
    $\begingroup$ OP asked for rigorous solution. you're just throwing around the word "volume" all willy-nilly. $\endgroup$ Sep 30 '20 at 19:42
  • $\begingroup$ @mathworker21 the volume is motivating but doesn't relate to the argument. The key property is the intermediate value theorem ensuring the existence of some small open neighborhood that is strictly positive. Should I set up the Reimann sum to make it more transparent? $\endgroup$ Sep 30 '20 at 21:20
  • 1
    $\begingroup$ sorry, I should be kinder. The intermediate value theorem does not ensure the existence of some small open neighborhood. Continuity does that by itself. Please let me know if anything doesn't make sense. $\endgroup$ Oct 1 '20 at 7:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.