Skip to main content
added 462 characters in body
Source Link
Andrey Rekalo
  • 22.3k
  • 12
  • 89
  • 122

Let $E$ and $E'$ be Banach spaces. Mappings $f:E\to E'$, which satisfy the inequality $$\|f(x + y) − f(x) − f(y)\| \leq\epsilon$$ for all $x, y \in E$, are called $\epsilon$-additive (or approximately additive). The main result concerning approximately additive functions is due to D. Hyers (link).

Theorem. Let $f(x)$ be an $\epsilon$-additive mapping of a Banach space $E$ into another Banach space $E'$. Then $l(x)=\lim f(2^nx)/2^n$ exists for each $x$ in $E$, $l(x)$ is linear, and the inequality $$\|f(x)-l(x)\|\leq\epsilon$$ is true for all $x$ in $E$. Moreover, $l(x)$ is the only linear mapping satisfying this inequality. If $f(x)$ is continuous at at least one point, then $l(x)$ is continuous everywhere in $E$.

So in your case $E=L^2(\Omega)$, $E'=\mathbb R$.

Edit. As Yemon Choi indicated, finite dimensional versions of the result had been discovered earlier and independently. Check out, for instance, the Pólya and Szegö problem book (Ch 3, Problem 99):

Assume that the terms of the sequence $a_1,a_2,a_3,\dots$ satisfy the condition $$a_m+a_n-1 < a_{m+n} < a_m+a_n+1.$$ Then $$\lim\limits_{n\to\infty}\frac{a_n}{n}=\omega$$ exists; $\omega$ is finite and we have $$\omega n-1 < a_n < \omega n +1.$$

Let $E$ and $E'$ be Banach spaces. Mappings $f:E\to E'$, which satisfy the inequality $$\|f(x + y) − f(x) − f(y)\| \leq\epsilon$$ for all $x, y \in E$, are called $\epsilon$-additive (or approximately additive). The main result concerning approximately additive functions is due to D. Hyers (link).

Theorem. Let $f(x)$ be an $\epsilon$-additive mapping of a Banach space $E$ into another Banach space $E'$. Then $l(x)=\lim f(2^nx)/2^n$ exists for each $x$ in $E$, $l(x)$ is linear, and the inequality $$\|f(x)-l(x)\|\leq\epsilon$$ is true for all $x$ in $E$. Moreover, $l(x)$ is the only linear mapping satisfying this inequality. If $f(x)$ is continuous at at least one point, then $l(x)$ is continuous everywhere in $E$.

So in your case $E=L^2(\Omega)$, $E'=\mathbb R$.

Let $E$ and $E'$ be Banach spaces. Mappings $f:E\to E'$, which satisfy the inequality $$\|f(x + y) − f(x) − f(y)\| \leq\epsilon$$ for all $x, y \in E$, are called $\epsilon$-additive (or approximately additive). The main result concerning approximately additive functions is due to D. Hyers (link).

Theorem. Let $f(x)$ be an $\epsilon$-additive mapping of a Banach space $E$ into another Banach space $E'$. Then $l(x)=\lim f(2^nx)/2^n$ exists for each $x$ in $E$, $l(x)$ is linear, and the inequality $$\|f(x)-l(x)\|\leq\epsilon$$ is true for all $x$ in $E$. Moreover, $l(x)$ is the only linear mapping satisfying this inequality. If $f(x)$ is continuous at at least one point, then $l(x)$ is continuous everywhere in $E$.

So in your case $E=L^2(\Omega)$, $E'=\mathbb R$.

Edit. As Yemon Choi indicated, finite dimensional versions of the result had been discovered earlier and independently. Check out, for instance, the Pólya and Szegö problem book (Ch 3, Problem 99):

Assume that the terms of the sequence $a_1,a_2,a_3,\dots$ satisfy the condition $$a_m+a_n-1 < a_{m+n} < a_m+a_n+1.$$ Then $$\lim\limits_{n\to\infty}\frac{a_n}{n}=\omega$$ exists; $\omega$ is finite and we have $$\omega n-1 < a_n < \omega n +1.$$

added 52 characters in body
Source Link
Andrey Rekalo
  • 22.3k
  • 12
  • 89
  • 122

Let $E$ and $E'$ be Banach spaces. Mappings $f:E\to E'$, which satisfy the inequality $$\|f(x + y) − f(x) − f(y)\| \leq\epsilon$$ for all $x, y \in E$, are called $\epsilon$-additive (or approximately additive). The main result concerning approximately additive functions is due to D. Hyers (link).

Theorem. Let $f(x)$ be an $\epsilon$-additive mapping of a Banach space $E$ into another Banach space $E'$. Then $l(x)=\lim f(2^nx)/2^n$ exists for each $x$ in $E$, $l(x)$ is linear, and the inequality $$\|f(x)-l(x)\|\leq\epsilon$$ is true for all $x$ in $E$. Moreover, $l(x)$ is the only linear mapping satisfying this inequality. If $f(x)$ is continuous at at least one point, then $l(x)$ is continuous everywhere in $E$.

So in your case $E=L^2(\Omega)$, $E'=\mathbb R$.

Let $E$ and $E'$ be Banach spaces. Mappings $f:E\to E'$, which satisfy the inequality $$\|f(x + y) − f(x) − f(y)\| \leq\epsilon$$ for all $x, y \in E$, are called $\epsilon$-additive (or approximately additive). The main result concerning approximately additive functions is due to D. Hyers (link).

Theorem. Let $f(x)$ be an $\epsilon$-additive mapping of a Banach space $E$ into another Banach space $E'$. Then $l(x)=\lim f(2^nx)/2^n$ exists for each $x$ in $E$, $l(x)$ is linear, and the inequality $$\|f(x)-l(x)\|\leq\epsilon$$ is true for all $x$ in $E$. Moreover, $l(x)$ is the only linear mapping satisfying this inequality. If $f(x)$ is continuous at at least one point, then $l(x)$ is continuous everywhere in $E$.

Let $E$ and $E'$ be Banach spaces. Mappings $f:E\to E'$, which satisfy the inequality $$\|f(x + y) − f(x) − f(y)\| \leq\epsilon$$ for all $x, y \in E$, are called $\epsilon$-additive (or approximately additive). The main result concerning approximately additive functions is due to D. Hyers (link).

Theorem. Let $f(x)$ be an $\epsilon$-additive mapping of a Banach space $E$ into another Banach space $E'$. Then $l(x)=\lim f(2^nx)/2^n$ exists for each $x$ in $E$, $l(x)$ is linear, and the inequality $$\|f(x)-l(x)\|\leq\epsilon$$ is true for all $x$ in $E$. Moreover, $l(x)$ is the only linear mapping satisfying this inequality. If $f(x)$ is continuous at at least one point, then $l(x)$ is continuous everywhere in $E$.

So in your case $E=L^2(\Omega)$, $E'=\mathbb R$.

added 144 characters in body
Source Link
Andrey Rekalo
  • 22.3k
  • 12
  • 89
  • 122

FunctionsLet $E$ and $E'$ be Banach spaces. Mappings $f:E\to E'$, which satisfy the inequality $$|f(x + y) − f(x) − f(y)| \leq\epsilon$$$$\|f(x + y) − f(x) − f(y)\| \leq\epsilon$$ for all $x, y \in \mathbb R^n$$x, y \in E$, are called $\epsilon$-additive (or approximately additive). The main result concerning approximately additive functions is due to D. Hyers (link).

Theorem. Let $f: \mathbb R^n \to \mathbb R$$f(x)$ be an  $\epsilon$-additive functionmapping of a Banach space $E$ into another Banach space $E'$. Then there$l(x)=\lim f(2^nx)/2^n$ exists for each $x$ in $E$, $l(x)$ is a unique additive functionlinear, and the inequality $g: \mathbb R^n \to \mathbb R$ such that $$|f(x)-g(x)|\leq\epsilon $$ for$$\|f(x)-l(x)\|\leq\epsilon$$ is true for all $x\in \mathbb R^n$$x$ in $E$. Moreover, $l(x)$ is the only linear mapping satisfying this inequality. If $f$$f(x)$ is continuous at at least one point, then $g$$l(x)$ is continuous everywhere in $\mathbb R^n$$E$.

Functions , which satisfy the inequality $$|f(x + y) − f(x) − f(y)| \leq\epsilon$$ for all $x, y \in \mathbb R^n$, are called $\epsilon$-additive (or approximately additive). The main result concerning approximately additive functions is due to D. Hyers (link)

Theorem. Let $f: \mathbb R^n \to \mathbb R$ be an  $\epsilon$-additive function. Then there is a unique additive function $g: \mathbb R^n \to \mathbb R$ such that $$|f(x)-g(x)|\leq\epsilon $$ for all $x\in \mathbb R^n$. If $f$ is continuous at at least one point, then $g$ is continuous everywhere in $\mathbb R^n$.

Let $E$ and $E'$ be Banach spaces. Mappings $f:E\to E'$, which satisfy the inequality $$\|f(x + y) − f(x) − f(y)\| \leq\epsilon$$ for all $x, y \in E$, are called $\epsilon$-additive (or approximately additive). The main result concerning approximately additive functions is due to D. Hyers (link).

Theorem. Let $f(x)$ be an $\epsilon$-additive mapping of a Banach space $E$ into another Banach space $E'$. Then $l(x)=\lim f(2^nx)/2^n$ exists for each $x$ in $E$, $l(x)$ is linear, and the inequality $$\|f(x)-l(x)\|\leq\epsilon$$ is true for all $x$ in $E$. Moreover, $l(x)$ is the only linear mapping satisfying this inequality. If $f(x)$ is continuous at at least one point, then $l(x)$ is continuous everywhere in $E$.

added 7 characters in body
Source Link
Andrey Rekalo
  • 22.3k
  • 12
  • 89
  • 122
Loading
added 411 characters in body; deleted 7 characters in body
Source Link
Andrey Rekalo
  • 22.3k
  • 12
  • 89
  • 122
Loading
Source Link
Andrey Rekalo
  • 22.3k
  • 12
  • 89
  • 122
Loading