Practice
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 \(f(n)=n+c\).