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Let $f\in C^0(\mathbb R^2, \mathbb R)$, so that there exists an $M>0 \in \mathbb R$ with $f(x,y) \geq M$ for all $(x,y)\in\mathbb R^2$.

Does it follow for any $g \in C^0(\mathbb R,\mathbb R)$ that there exsists a unique $ h \in C^0(\mathbb R,\mathbb R)$ so that $$\int_0^{h(a)} f(a,x)\,\text{d}x=g(a)$$ for all $a\in\mathbb R$?

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  • $\begingroup$ By the way, the function $h$ exists and is unique, but continuity is not so clear off the bat. $\endgroup$
    – s.harp
    Jan 22, 2017 at 18:13
  • $\begingroup$ Can you explain, why the function $h$ exists necessarily. Thanks. $\endgroup$
    – Dattier
    Jan 22, 2017 at 18:17
  • $\begingroup$ Note that $K:y\mapsto \int_0^yf(a,x)\,dx$ is continuous. Since $f≥M>0$ you have that $K$ is a strict monotone function with $\lim_{y\to\infty}K(y)=\infty$ and $\lim_{k\to-\infty}K(y)=-\infty$. These limits and continuity of $K$ imply that for any value $z$ there exists a $y$ with $K(y)=z$. From strict monotony of $K$ there follows the uniqueness of such a $y$. This means for any $a$ there exists a unique value $y=:h(a)$ so that $\int_0^{h(a)} f(a,x)\, dx = g(a)$. $\endgroup$
    – s.harp
    Jan 22, 2017 at 18:33
  • $\begingroup$ Maybe by using a uniform approximation of f by a polynomial function with 2 variables $\endgroup$
    – Dattier
    Jan 25, 2017 at 6:56
  • $\begingroup$ Theorem M-R : $F:\mathbb R^2\rightarrow \mathbb R$ continous with $F$ derivable to the first composante, with the function dervied continuous and uniformly minored by M>0. Then $\exists h$ function unique with : $F(h(a),a)=0,\forall a\in\mathbb R$ and $h$ is continuous. $\endgroup$
    – Dattier
    Jun 18, 2017 at 17:33

2 Answers 2

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Existence and unicity have been prooved in comments, let's prove continuity.

Take $a\in \mathbf R$ and assume $h(a) \neq 0$. The function $f$ is continuous on $\mathbf R^2$ and thus uniformly continuous on the compact set $K=[a-1;a+1]\times [0;h(a)]$. This means that for every $\varepsilon>0$ there exists some $1>\delta>0$ such that $|a-b|<\delta \Rightarrow |f(a,x)-f(b,x)|<\varepsilon/|h(a)|$ for every $x$ in $[0;h(a)]$ and thus

$$|a-b|<\delta \Rightarrow \left|\int_0^{h(a)}f(a,x)dx-\int_0^{h(a)}f(b,x)dx\right|<\varepsilon.$$

Now take $\delta$ to be small enough so that $|g(a)-g(b)|<\varepsilon$ and notice that

$$|g(a)-g(b)|=\left|\int_0^{h(a)}f(a,x)dx-\int_0^{h(b)}f(b,x)dx\right|=\left|\int_0^{h(a)}f(a,x)dx-\int_0^{h(a)}f(b,x)dx -\int_{h(a)}^{h(b)}f(b,x)dx\right|.$$

Since $f\geq M>0$ everywhere we get that $$|h(a)-h(b)|> 2\varepsilon/M \Rightarrow \left|\int_{h(a)}^{h(b)}f(b,x)dx \right|>2\varepsilon$$ which in turn would imply $|g(a)-g(b)|>2\varepsilon-\varepsilon=\varepsilon$, contradicting the assumption made earlier. So the only possibility is that $|h(a)-h(b)|<2\varepsilon/M$ and thus $h$ is continuous. The case $h(a)=0$ is treated in the exacte same way except that we don't have to bound the term $\left|\int_0^{h(a)}f(a,x)dx-\int_0^{h(a)}f(b,x)dx\right|$ since it's equal to $0$.

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  • $\begingroup$ For my part I think that it is missing a lot of step to understand your proof, that a specialist like Harp can understand without difficulty. In short, if Harp is pleased with your proof I validate it, otherwise I would abstain, as long as I did not have a proof which I understand, and I am very slow. $\endgroup$
    – Dattier
    Jan 26, 2017 at 19:46
  • $\begingroup$ I wouldn't say you need to be a specialist to understand, being familiar with $\varepsilon$/$\delta$ proofs should be enough. Anyway, i'll rewrite my answer later to make it more easily readable. $\endgroup$
    – Renart
    Jan 26, 2017 at 23:05
  • $\begingroup$ " i'll rewrite my answer later to make it more easily readable." Thanks. $\endgroup$
    – Dattier
    Jan 27, 2017 at 7:47
  • $\begingroup$ I'm not a specialist! In your proof you are also assuming $h(a)>0$, but the case $h(a)<0$ can be done in the same way except for minus signs appearing in some places :) $\endgroup$
    – s.harp
    Jan 27, 2017 at 13:05
  • $\begingroup$ Compare to me, everyone here is specialist $\endgroup$
    – Dattier
    Jan 27, 2017 at 14:09
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As Renart, I'll ignore existence and uniqueness, and focus on continuity. However, I'll use a different (but equivalent) definition of continuity : a function is continuous if and only if the pre-image of any open ball is open. That avoids $\varepsilon-\delta$ arguments, although this proof can be adapted to use the $\varepsilon-\delta$ definition if needed.


First point : for all $M \in \mathbb{R}$, the function $I_M : a \mapsto \int_0^M g(a,x) dx$ is continuous.

Proof(s) : I'll give two proofs.

1) Let $a$, $M \in \mathbb{R}$. The subspace $[a-1, a+1]\times [0,M]$ is compact and $f$ is continuous, so $f$ is uniformly continuous on $[a-1, a+1]\times [0,M]$. Hence, there exists a modulus of continuity $\omega$ for $f$ on this subspace ; that is, for all $x$, $y \in [a-1, a+1]\times [0,M]$,

$$|f(x)-f(y)| \leq \omega (\|x-y\|).$$

Then, we get, for all $a' \in [a-1,a+1]$:

$$\left| \int_0^M f(a,x) dx - \int_0^M f(a',x) \right| \leq \int_0^M |f(a,x)-f(a',x)| dx \leq M \omega (|a-a'|).$$

In particular, the differences converges to $0$ when $a'$ converges to $a$.

2) Since $[a-1, a+1]\times [0,M]$ is compact and $f$ is continuous, $f$ is bounded on $[a-1, a+1]\times [0,M]$. Let $(a_n)_n$ be a sequence which converges to $a$. Then:

$$\lim_{n \to + \infty} \int_0^M f(a_n,x)dx = \int_0^M \lim_{n \to + \infty} f(a_n,x)dx =\int_0^M f(a,x)dx,$$

where we used first the dominated convergence theorem, and then the continuity of $f$.


Now, I'll prove the claim. Let $y \in \mathbb{R}$ and $\varepsilon >0$. Then:

$$h^{-1} (B(y,\varepsilon)) = \left\{a : y-\varepsilon < h(a) \right\} \cap \left\{a : h(a) < y+\varepsilon \right\}$$

But, since $M \mapsto \int_0^M f(a,x) dx$ is strictly increasing, we have $\left\{a : y-\varepsilon < h(a) \right\} = \left\{a : \int_0^{y-\varepsilon} f(a,x) dx -g(a) < 0 \right\} = (I_{y-\varepsilon}-g)^{-1} (\mathbb{R}_-^*)$.

Since $I_{y-\varepsilon}-g$ is continuous by the fist point, $\left\{a : y-\varepsilon < h(a) \right\}$ is open. By the same argument, $\left\{a : y+\varepsilon > h(a) \right\}$ is open. Hence, as the intersection of two open sets is open, $h^{-1} (B(y,\varepsilon))$ is open. Since this is true for all $y \in \mathbb{R}$ and all $\varepsilon >0$, it follows that $h$ is continuous.

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  • $\begingroup$ I think this is very nice, so I made another bounty. Lucky for you it required me to double the amount! ;) $\endgroup$
    – s.harp
    Jan 27, 2017 at 13:07
  • $\begingroup$ I valide your proof, because Harp says "this is very nice", and Renart got the bounty. but I don't know, what's a "modulus of continuity". $\endgroup$
    – Dattier
    Jan 27, 2017 at 14:16

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