Here is a persuasive advertisement for ghostwriting services for the theme:
**Unlock the Secrets of Additive Functions with Our Expert Ghostwriting Services**
Are you struggling to prove that an additive function f(x) is either f(x)≡0 or f(x)≡x? Do you need help utilizing the properties of being bounded from below and the theorem of additive functions to crack this complex mathematical problem?
Look no further! Our team of experienced ghostwriters specializes in crafting top-notch mathematical proofs that will impress even the most discerning professors. With our expertise, you can:
* Get a clear and concise proof that showcases your understanding of additive functions
* Save time and effort by letting our experts handle the complex mathematical derivations
* Boost your confidence and grades with a well-written solution that meets the highest academic standards
Our ghostwriting services cover:
* Mathematical proofwriting for additive functions and beyond
* Customized solutions tailored to your specific needs and requirements
* Expert guidance and support throughout the entire writing process
Why choose our ghostwriting services?
* **Unparalleled expertise**: Our team of mathematicians and writers have years of experience in crafting high-quality mathematical proofs.
* **Fast turnaround**: We understand the importance of meeting deadlines, and we guarantee timely delivery of your proof.
* **Confidentiality**: Your privacy is our top priority, and we ensure that all our services are completely confidential.
Don’t let mathematical frustrations hold you back. Let us help you unlock the secrets of additive functions and achieve academic success. Contact us today to get started!
for all \(x,y\in\mathbb{R}\). Then, either \(f(x)\equiv 0\) or \(f(x)\equiv x\).
Note that \(f\) is bounded from below in \([0,1]\), since for \(x\geq 0\), we have
\[f(x)=f(\sqrt{x}\sqrt{x})=f^{2}(\sqrt{x})\geq 0.\]
Since \(f\) is additive, using the above theorem, we can conclude that \(f(x)=cx\) for some constant \(c\). Then \(f(1)=f^{2}(1)\) gives \(c=c^{2}\), thus either \(c=0\) or \(c=1\).
Next we show that not every additive function is of the form \(f(x)=cx\). We need the following concept called a _Hamel basis._
**Definition 80**.: A set \(H\subset\mathbb{R}\) is called a _Hamel basis_ if for every real number \(x\), there exists a natural number \(n\in\mathbb{N}\), \(r_{i}\in\mathbb{Q}\) and \(h_{i}\in H\) where \(i=1,2,\ldots n\), such that \(x\) can be written uniquely as:
\[x=r_{1}h_{1}+r_{2}h_{2}+\cdots+r_{n}h_{n}.\]
Observe that this definition is just a restatement of the concept of an algebraic basis for a vector space. The proof of the following proposition depends on the existence of a Hamel basis.
**Proposition 30**.: _Not every additive function is necessarily of the form \(f(x)=cx\)._
Proof
: We claim that if \(a\) and \(b\) are nonzero real numbers such that \(\dfrac{a}{b}\notin\mathbb{Q}\), then we can find an additive function \(f\) with the properties that \(f(a)\neq 0\) and \(f(b)=0\). Thus, such a function cannot have the form \(f(x)=cx\). Since \(\dfrac{a}{b}\) is irrational, the set \(\{a,b\}\) is a linearly independent set over the rationals and it is contained in any Hamel basis \(H\). We define the function \(f\) as:
\[f(x)=\begin{cases}0&\text{ if $a$ does not occur among $h_{1},h_{2},\ldots,h_{n}$}\\ \text{coefficient of $a$ }&\text{ in the unique representation of $x$ given above.}\end{cases}\]
Then the function \(f\) is well defined and moreover for \(x,y\in\mathbb{R}\), if
\[y=t_{1}h_{1}+t_{2}h_{2}+\cdots+t_{n}h_{n},\quad\text{then}\]
\[x+y=(r_{1}+t_{1})h_{1}+\cdots+(r_{n}+t_{n})h_{n}\]
is the unique representation of \(x+y\). We now consider the following cases:* If \(a=h_{i}\), then \(f(x+y)=r_{i}+t_{i}=f(x)+f(y)\).
* If \(a\) does not occur among \(h_{i}\)’s for \(1\leq i\leq n\), then \[f(x)=0=f(y)=f(x+y).\]
In either case \(f\) satisfies Cauchy’s equation \(f(x+y)=f(x)+f(y)\).
**Remark 54**.: In 1906 G. Hamel was the first mathematician to prove the existence of discontinuous additive functions. Discontinuous additive functions are sometimes called _Hamel functions_. Since the very first theorem we proved in this section tells us that every additive function \(f:\mathbb{R}^{n}\to\mathbb{R}\) is a homomorphism from \(\mathbb{R}^{n}\) to \(\mathbb{R}\) where both \(\mathbb{R}^{n}\) and \(\mathbb{R}\) are considered as linear spaces over \(\mathbb{Q}\), we have the following theorem as an immediate consequence.
**Theorem 103**.: _Let \(H\) be an arbitrary Hamel basis for \(\mathbb{R}^{n}\). Then for every function \(g:H\to\mathbb{R}\) there exists a unique additive function \(f:\mathbb{R}^{n}\to\mathbb{R}\) such that \(f|_{H}=g\) (\(f\) restricted to \(H\) is equal to \(g\))._
As a corollary to this theorem, we can prove:
**Corollary 17**.: _Let \(H\) be an arbitrary Hamel basis for \(\mathbb{R}\), and let \(g:H\to\mathbb{R}\) be an arbitrary function. Let \(f:\mathbb{R}\to\mathbb{R}\) be the unique extension of \(g\). Then \(f\) is continuous if and only if \(\dfrac{g(x)}{x}=\) constant for \(x\in H\)._
Proof
: Continuity of \(f\) implies that \(f(x)=cx\) for \(x\in\mathbb{R}\). In particular, for \(x\in H\), we have \(g(x)=f(x)=cx\). Conversely, if \(g(x)=cx\) for \(x\in H\), then the function \(cx\) is an additive extension of \(g\), and by the uniqueness of such extensions \(f(x)=cx\). Thus \(f\) is continuous.
Now we see that it is enough to take any function \(g:H\to\mathbb{R}\) such that
\[\dfrac{g(x)}{x}\neq\text{constant}\]
on \(H\), and the additive extension of \(g\) defines a discontinuous additive function.
**Remark 55**.: If \(f\) is additive and \(f(x)\neq cx\) for any constant \(c\), then we can find \(x_{1},x_{2}\) such that
\[\dfrac{f(x_{1})}{x_{1}}\neq\dfrac{f(x_{2})}{x_{2}},\]
which means that the determinant of the following matrix
\[\begin{pmatrix}x_{1}&f(x_{1})\\ x_{2}&f(x_{2})\end{pmatrix}\]is equal to \(x_{1}f(x_{2})-x_{2}f(x_{1})\neq 0\). Thus the vectors \(\{(x_{1},f(x_{1})),\;(x_{2},f(x_{2}))\}\) form a linearly independent set over \(\mathbb{Q}\). However, if we look at a linear combination of these vectors, namely
\[r_{1}(x_{1},f(x_{1}))+r_{2}(x_{2},f(x_{2}))=(r_{1}x_{1}+r_{2}x_{2},f(r_{1}x_{1}+ r_{2}x_{2}))\]
for \(r_{1},r_{2}\in\mathbb{Q}\), we get a point belonging to the graph \(G(f)\) of \(f\). This gives an idea why the following theorem is valid:
**Theorem 104** (Hamel, 1906 and San Juan, 1946).: _If \(f:\mathbb{R}\to\mathbb{R}\) is additive and discontinuous, then_
\[\overline{G(f)}=\mathbb{R}^{2},\]
_i.e., the closure of the graph of \(f\) in \(\mathbb{R}^{2}\) is equal to \(\mathbb{R}^{2}\)._
### Jensen’s Equation
Jensen’s equation for convex functions is the functional equation
\[f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}.\]
If we assume Jensen’s equation is valid for all real \(x,y\), then it is reduced to Cauchy’s functional equation. If we set \(y=0\) in Jensen’s equation, we obtain
\[f\left(\frac{x}{2}\right)=\frac{f(x)+f(0)}{2}=\frac{f(x)+c}{2},\]
where \(f(0)=c\). Thus using Jensen’s equation again we obtain
\[\frac{f(x)+f(y)}{2}=f\left(\frac{x+y}{2}\right)=\frac{f(x+y)+c}{2},\]
that is
\[f(x+y)=f(x)+f(y)-c.\]
Setting \(g(x)=f(x)-c\), we see that the function \(g\) satisfies Cauchy’s equation, \(g(x+y)=g(x)+g(y)\). Thus if \(f\) is continuous at a point then \(g(x)=cx\). We just proved the following theorem.
**Theorem 105**.: _The function \(f(x)=ax+c\) is the most general solution of the functional equation_
\[f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}\]
_that holds for all real \(x,y\) and that is continuous at one point._Note that above argument can be repeated to conclude the following corollary.
**Corollary 18**.: _The most general solution of Jensen’s functional equation is of the form_
\[f(x)=h(x)+a,\]
_where \(a\) is an arbitrary constant and \(h\) is an additive function satisfying \(h(x+y)=h(x)+h(y)\)._
If we assume that \(f\) is continuous only on a closed interval without paying attention to the behavior of the function outside this interval, we can still get information about the solution of Jensen’s functional equation.
发表回复