Chapter
Theory
A functional equation is an equation where the unknown is a function rather than a number. At the start one almost always substitutes special values: \(0\), \(1\), \(x\), \(-x\), \(x=y\), \(y=0\). These substitutions give the first anchor values and properties of the function.
The main goal of this introductory module is to avoid guessing the function immediately. Instead, extract information step by step: find \(f(0)\), \(f(1)\), check injectivity or surjectivity, and then move toward a linear or additive structure.
If \(f(x+y)=f(x)+f(y)\), one usually first gets \(f(0)=0\). On \(\mathbb Z\) and \(\mathbb Q\), additivity implies \(f(n)=nf(1)\) and \(f\left(\frac{p}{q}\right)=\frac{p}{q}f(1)\).
On \(\mathbb R\), additivity alone is not enough to conclude \(f(x)=cx\). Additional regularity is needed: monotonicity, boundedness on an interval, nonnegativity on positive numbers, or a similar condition.
Substitution is almost always needed. Injectivity is useful when \(f(A)=f(B)\) can give \(A=B\). Surjectivity is useful when \(f(y)\) needs to be replaced by an arbitrary number. Testing a linear form is useful when \(f(x)=ax+b\) is expected but must still be justified.
If the equation contains \(x+y\), try \(y=0\) and \(x=0\). If it contains \(f(f(x))\), look for injectivity or invertibility. If it contains \(x+f(y)\), first fix \(y\), then choose \(y\) so that \(f(y)\) becomes a convenient argument.
Do not assume the function is linear without proof. Do not divide by \(f(x)\) before proving this value is nonzero. Do not use “for every \(t\) there is \(y\) with \(f(y)=t\)” before surjectivity has been proved.
1. Have \(0\) and \(1\) been substituted? 2. Are \(f(0)\), \(f(1)\) known? 3. Is there injectivity or surjectivity? 4. Can one pass to \(g(x)=f(x)+c\) or \(g(x)=f(x)-x^2\)? 5. Have the found functions been checked in the original equation?
Examples
The first move is to substitute the simplest argument.
Problem. Let \(f(x+y)=f(x)+f(y)\) for all integers \(x,y\). Prove that \(f(0)=0\).
Substitute \(x=0\), \(y=0\): \(f(0)=f(0)+f(0)\). Hence \(f(0)=0\).
On \(\mathbb Z\), additivity is completely determined by the value at one.
Problem. If \(f:\mathbb Z\to\mathbb Z\), \(f(m+n)=f(m)+f(n)\), \(f(1)=3\), find \(f(n)\).
For \(n>0\), \(f(n)=nf(1)=3n\). Also \(0=f(0)=f(n+(-n))=f(n)+f(-n)\), so \(f(-n)=-3n\). Answer: \(f(n)=3n\) for all \(n\in\mathbb Z\).
Sometimes one value immediately determines the recurrence.
Problem. Let \(f(n+1)=f(n)+2n+1\) for \(n\ge0\), \(f(0)=0\). Find \(f(n)\).
Add the equalities for \(n=0,1,\ldots,k-1\): \(f(k)=1+3+\cdots+(2k-1)=k^2\). Hence \(f(n)=n^2\).
A linear form may be tested, but it cannot simply be assumed without conditions.
Problem. Find all linear functions \(f(x)=ax+b\) such that \(f(x+y)=f(x)+f(y)+4\).
Substitute \(ax+b\). We get \(a(x+y)+b=ax+b+ay+b+4\). The coefficients of \(x,y\) match automatically, and the constant terms give \(b=2b+4\), so \(b=-4\). Answer: \(f(x)=ax-4\), where \(a\) is any real number.
Injectivity allows one to remove \(f\) from both sides of an equality.
Problem. Let \(f\) be injective and \(f(x+1)=f(y+1)\). Prove that \(x=y\).
By injectivity, \(x+1=y+1\). Therefore \(x=y\).
Surjectivity lets us replace a function value by an arbitrary number.
Problem. Let \(f:\mathbb R\to\mathbb R\) be surjective. Prove that there exists \(t\) such that \(f(t)=0\).
By surjectivity, every real number is a value of the function. In particular, \(0\) is a value: there exists \(t\) such that \(f(t)=0\).
On rational numbers, the Cauchy equation does not require regularity.
Problem. Let \(f:\mathbb Q\to\mathbb Q\), \(f(x+y)=f(x)+f(y)\), \(f(1)=5\). Find \(f\left(\frac{7}{3}\right)\).
From \(3\cdot\frac{1}{3}=1\), we get \(3f\left(\frac{1}{3}\right)=f(1)=5\), hence \(f\left(\frac{1}{3}\right)=\frac{5}{3}\). Then \(f\left(\frac{7}{3}\right)=7f\left(\frac{1}{3}\right)=\frac{35}{3}\).
Sometimes one subtracts a known part to obtain ordinary additivity.
Problem. Let \(f:\mathbb Q\to\mathbb Q\), \(f(x+y)=f(x)+f(y)+2xy\), \(f(1)=1\). Find \(f(x)\).
Set \(g(x)=f(x)-x^2\). Then \(g(x+y)=f(x+y)-(x+y)^2=f(x)-x^2+f(y)-y^2=g(x)+g(y)\). Also \(g(1)=0\). On \(\mathbb Q\), an additive function with \(g(1)=0\) is zero. Hence \(f(x)=x^2\).
Problems
A function \(f:\mathbb R\to\mathbb R\) satisfies \(f(x)+f(-x)=2\) for all \(x\). Find \(f(0)\).
Substitute \(x=0\).
At \(x=0\), we get \(f(0)+f(0)=2\), so \(2f(0)=2\). Hence \(f(0)=1\).
Let \(f(x+1)=f(x)+3\) for all real \(x\), and \(f(0)=2\). Find \(f(5)\).
Apply the equality five times.
\(f(1)=5\), \(f(2)=8\), \(f(3)=11\), \(f(4)=14\), \(f(5)=17\). Answer: \(17\).
Let \(f(x+y)=f(x)+f(y)\) for all integers \(x,y\). Prove that \(f(0)=0\).
Substitute \(x=y=0\).
We get \(f(0)=f(0)+f(0)\). Subtracting \(f(0)\), we obtain \(f(0)=0\).
A function \(f:\mathbb R\to\mathbb R\) satisfies \(f(xy)=xf(y)+yf(x)\) for all \(x,y\). Find \(f(0)\) and \(f(1)\).
First set \(x=0\), then \(x=1\).
For \(x=0\): \(f(0)=y f(0)\) for all \(y\). Taking \(y=2\), we get \(f(0)=2f(0)\), hence \(f(0)=0\).
For \(x=1\): \(f(y)=f(y)+y f(1)\). Taking \(y=1\), we get \(f(1)=f(1)+f(1)\), hence \(f(1)=0\).
Find all linear functions \(f(x)=ax+b\) such that \(f(x+1)=f(x)+2\) for all \(x\).
Substitute \(ax+b\) and compare the constant parts.
We have \(a(x+1)+b=ax+b+2\). After canceling \(ax+b\), we get \(a=2\). The number \(b\) is arbitrary. Answer: \(f(x)=2x+b\).
Let \(f:\mathbb Z\to\mathbb Z\), \(f(m+n)=f(m)+f(n)\), and \(f(1)=4\). Find \(f(n)\) for all \(n\in\mathbb Z\).
First find \(f(n)\) for positive \(n\), then for negative \(n\).
For \(n>0\), \(f(n)=nf(1)=4n\). Also \(0=f(0)=f(n+(-n))=f(n)+f(-n)\), hence \(f(-n)=-4n\). Therefore \(f(n)=4n\) for all integers \(n\).
Let \(f:\mathbb Q\to\mathbb Q\), \(f(x+y)=f(x)+f(y)\), \(f(1)=3\). Prove that \(f(q)=3q\) for all \(q\in\mathbb Q\).
For \(q=\frac{m}{n}\), use \(nq=m\).
For an integer \(m\), \(f(m)=3m\). Let \(q=\frac{m}{n}\), \(n>0\). Then \(n f(q)=f(nq)=f(m)=3m\), so \(f(q)=\frac{3m}{n}=3q\).
Let \(f(0)=0\) and \(f(n+1)=f(n)+2n+1\) for all integers \(n\ge0\). Prove that \(f(n)=n^2\) for all \(n\ge0\).
Add the odd numbers \(1,3,\ldots,2n-1\).
Adding the equalities from \(0\) to \(n-1\), we get \(f(n)-f(0)=1+3+\cdots+(2n-1)=n^2\). Since \(f(0)=0\), \(f(n)=n^2\).
Find all linear functions \(f(x)=ax+b\) satisfying \(f(x+y)=f(x)+f(y)+5\) for all real \(x,y\).
Compare the constant terms after substitution.
Substitution gives \(a(x+y)+b=ax+b+ay+b+5\). The coefficients of \(x,y\) match. For constants, \(b=2b+5\), so \(b=-5\). Answer: \(f(x)=ax-5\), where \(a\) is arbitrary.
Let \(f:\mathbb R\to\mathbb R\) be additive, meaning \(f(x+y)=f(x)+f(y)\), and injective. Prove that if \(f(a)=0\), then \(a=0\).
First find \(f(0)\).
By additivity, \(f(0)=0\). If \(f(a)=0\), then \(f(a)=f(0)\). By injectivity, \(a=0\).
A function \(f:\mathbb R\to\mathbb R\) satisfies \(f(x+f(y))=x+y\) for all \(x,y\). Prove that \(f\) is injective.
Assume \(f(a)=f(b)\) and substitute \(y=a\), \(y=b\) with the same \(x\).
Let \(f(a)=f(b)\). For any \(x\), we have \(f(x+f(a))=x+a\) and \(f(x+f(b))=x+b\). The left sides are equal because \(f(a)=f(b)\). Hence \(x+a=x+b\), so \(a=b\). The function is injective.
Let \(f:\mathbb Z\to\mathbb Z\), \(f(m+n)=f(m)+f(n)+2mn\), \(f(0)=0\), \(f(1)=1\). Find \(f(n)\).
Set \(g(n)=f(n)-n^2\).
Let \(g(n)=f(n)-n^2\). Then \(g(m+n)=f(m+n)-(m+n)^2=f(m)-m^2+f(n)-n^2=g(m)+g(n)\). Thus \(g\) is additive on \(\mathbb Z\).
Also, \(g(1)=f(1)-1=0\). Hence \(g(n)=0\) for all \(n\in\mathbb Z\), and \(f(n)=n^2\).
Let \(f:\mathbb Z\to\mathbb Z\), \(f(m+n)=f(m)+f(n)+mn\), \(f(0)=0\), \(f(1)=0\). Find \(f(n)\).
Compare with the function \(\frac{n(n-1)}{2}\).
Set \(g(n)=f(n)-\frac{n(n-1)}{2}\). Since \(\frac{(m+n)(m+n-1)}{2}=\frac{m(m-1)}{2}+\frac{n(n-1)}{2}+mn\), we get \(g(m+n)=g(m)+g(n)\).
Because \(g(1)=0\), \(g(n)=0\) for all integers \(n\). Therefore \(f(n)=\frac{n(n-1)}{2}\).
Find all linear functions \(f(x)=ax+b\) such that \(f(x)+f(1-x)=1\) and \(f(x+1)=f(x)+1\) for all \(x\).
The second condition first finds \(a\), the first finds \(b\).
From \(f(x+1)=f(x)+1\), we get \(a=1\). Thus \(f(x)=x+b\). The first condition gives \(x+b+1-x+b=1\), so \(1+2b=1\), hence \(b=0\). Answer: \(f(x)=x\).
Find all linear functions \(f(x)=ax+b\) such that \(f(f(x))=4x+6\) for all \(x\), with the additional condition \(f(0)>0\).
Compute \(f(f(x))\) in terms of \(a,b\).
\(f(f(x))=a(ax+b)+b=a^2x+b(a+1)\). Hence \(a^2=4\) and \(b(a+1)=6\).
If \(a=2\), then \(3b=6\), so \(b=2\). If \(a=-2\), then \(-b=6\), so \(b=-6\), but then \(f(0)<0\). Only \(f(x)=2x+2\) works.
Let \(f:\mathbb Q\to\mathbb Q\), \(f(x+y)=f(x)+f(y)\), \(f(2)=5\). Find \(f\left(\frac{7}{3}\right)\).
First find \(f(1)\), then \(f\left(\frac{1}{3}\right)\).
Since \(f(2)=2f(1)=5\), we have \(f(1)=\frac{5}{2}\). Next, \(3f\left(\frac{1}{3}\right)=f(1)=\frac{5}{2}\), so \(f\left(\frac{1}{3}\right)=\frac{5}{6}\). Therefore \(f\left(\frac{7}{3}\right)=7\cdot\frac{5}{6}=\frac{35}{6}\).
Let \(f:\mathbb R\to\mathbb R\), \(f(x+y)=f(x)+f(y)\), and suppose \(f\) is nondecreasing. Prove that there exists \(c\ge0\) such that \(f(x)=cx\) for all \(x\).
First prove the formula for rationals, then squeeze a real number between rationals.
Let \(c=f(1)\). On rational numbers, \(f(q)=cq\). Since \(1>0\), monotonicity gives \(c=f(1)\ge f(0)=0\).
Let \(x\) be real. For any rational \(r
Let \(f:\mathbb R\to\mathbb R\) be additive and suppose \(f(t)\ge0\) for all \(t\ge0\). Prove that \(f(x)=cx\) for some \(c\ge0\).
First show that \(f\) is nondecreasing.
If \(x
Let \(f:\mathbb Q\to\mathbb Q\) be additive and satisfy \(f(x^2)=f(x)^2\) for all \(x\in\mathbb Q\). Find all such functions.
First use additivity on \(\mathbb Q\): \(f(x)=cx\).
By additivity on \(\mathbb Q\), \(f(x)=cx\), where \(c=f(1)\). The condition gives \(c x^2=c^2 x^2\) for all \(x\). At \(x=1\), \(c=c^2\), so \(c=0\) or \(c=1\).
Both functions work: \(f(x)=0\) and \(f(x)=x\).
A function \(f:\mathbb Z\to\mathbb Z\) satisfies \(f(m+n)+f(m-n)=2f(m)+2f(n)\), \(f(0)=0\), \(f(1)=1\). Prove that \(f(n)=n^2\) for all \(n\in\mathbb Z\).
Substitute \(m=n\), then \(m=n\), \(n=1\), to get a recurrence.
At \(m=0\), we get \(f(n)+f(-n)=2f(n)\), so \(f(-n)=f(n)\). At \(n=1\): \(f(m+1)+f(m-1)=2f(m)+2\).
This recurrence with \(f(0)=0\), \(f(1)=1\) determines all values. We prove by induction that \(f(k)=k^2\) for \(k\ge0\). It is true for \(0,1\). If it is true for \(k\) and \(k-1\), then \(f(k+1)=2f(k)+2-f(k-1)=2k^2+2-(k-1)^2=(k+1)^2\). For negative \(n\), use \(f(-n)=f(n)\).
Let \(f:\mathbb R\to\mathbb R\) be additive and take integer values on the whole interval \([0,1]\). Prove that \(f(x)=0\) for all \(x\).
For \(t\in[0,1]\), consider \(f\left(\frac{t}{n}\right)\).
Let \(t\in[0,1]\). Then \(\frac{t}{n}\in[0,1]\), so \(f\left(\frac{t}{n}\right)\) is an integer. But \(n f\left(\frac{t}{n}\right)=f(t)\). If \(f(t)\neq0\), then for \(n>|f(t)|\), the integer \(f\left(\frac{t}{n}\right)=\frac{f(t)}{n}\) cannot be an integer. Hence \(f(t)=0\) on \([0,1]\).
For any real \(x\), choose a positive integer \(N>|x|\). Then \(\frac{x}{N}\in[-1,1]\). If \(x<0\), use \(f(-u)=-f(u)\), so \(f\left(\frac{x}{N}\right)=0\). Therefore \(f(x)=N f\left(\frac{x}{N}\right)=0\).
Let \(f:\mathbb R\to\mathbb R\) be increasing and satisfy \(f(x+f(y))=f(x)+y\) for all \(x,y\). Find \(f\).
First substitute \(y=0\), then \(x=0\), and then replace \(y\) by \(f(t)\).
Let \(a=f(0)\). At \(y=0\), we have \(f(x+a)=f(x)\). Since \(f\) is increasing, it is injective, so \(a=0\). At \(x=0\), we get \(f(f(y))=y\).
Now put \(y=f(t)\) in the original equation. Then \(f(x+t)=f(x)+f(t)\), so \(f\) is additive. An increasing additive function has the form \(f(x)=cx\), where \(c>0\). From \(f(f(y))=y\), we get \(c^2y=y\) for all \(y\), hence \(c=1\). Answer: \(f(x)=x\).
Let \(f:\mathbb Q\to\mathbb Q\) satisfy \(f(x+y)=f(x)+f(y)+2xy\) for all \(x,y\in\mathbb Q\), and \(f(1)=1\). Find \(f\).
Subtract \(x^2\): set \(g(x)=f(x)-x^2\).
Let \(g(x)=f(x)-x^2\). Then \(g(x+y)=f(x+y)-(x+y)^2=f(x)-x^2+f(y)-y^2=g(x)+g(y)\). Thus \(g\) is additive on \(\mathbb Q\).
Since \(g(1)=f(1)-1=0\), \(g(q)=0\) for all rational \(q\). Therefore \(f(x)=x^2\). The check is immediate: \((x+y)^2=x^2+y^2+2xy\).
Let \(f:\mathbb Z\to\mathbb Z\) be surjective and satisfy \(f(n+1)\ge f(n)+1\) for all integers \(n\). Prove that there exists an integer \(c\) such that \(f(n)=n+c\) for all \(n\).
If some step is greater than \(1\), which integer value is skipped?
The condition implies that \(f\) is strictly increasing. If for some \(n\), \(f(n+1)\ge f(n)+2\), then the integer \(f(n)+1\) cannot be a value of the function: for \(k\le n\), \(f(k)\le f(n)\), while for \(k\ge n+1\), \(f(k)\ge f(n+1)\ge f(n)+2\). This contradicts surjectivity.
Hence \(f(n+1)=f(n)+1\) for all \(n\). Then \(f(n)-n\) is constant. Denoting this constant by \(c\), we get \(f(n)=n+c\).
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