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Chapter 9: Problem 4

Welche stetigen Funktionen \(f: \mathbb{R} \rightarrow \mathbb{R}\) erf眉llen die Funktionalgleichung $$ f(x+y)=f(x)+f(y), \quad x, y \in I ? $$

Short Answer

Expert verified
Answer: The general form of a continuous function that satisfies the given functional equation is \(f(x) = cx\), where \(c\) is a constant belonging to the set of real numbers.

Step by step solution

01

Prove \(f(0)=0\)

First, let's apply the functional equation by setting \(x=y=0\). We get: $$ f(0+0) = f(0) + f(0) $$ Since \(0+0=0\), we get: $$ f(0) = 2f(0) $$ Taking \(f(0) = 2f(0)\), we can divide both sides by 2, thus \(f(0) = 0\).
02

Prove \(f\) is an additive function

Next, we want to prove that \(f\) is linear. First, we use the functional equation to show that \(f\) is an additive function, meaning that \(f(x+y) = f(x) + f(y)\). The functional equation is already given in this form, so we know that \(f\) is an additive function for all \(x,y\in I\).
03

Prove \(f\) is homogeneous

Now we want to prove that \(f\) is a homogeneous function of degree 1, meaning that \(f(rx) = rf(x)\) for all \(r \in \mathbb{Q}\) and \(x \in I\). Using the functional equation and the fact that \(f(0) = 0\), we have: For \(n\in\mathbb{Z}\): $$ f(nx) = f(x+\cdots+x) = f(x)+\cdots+f(x) = nf(x) $$ Now let \(r = \frac{m}{n}\) with \(m\in\mathbb{Z}\) and \(1\leq n \in\mathbb{Z}\), then $$ nf(rx) = f(m(rx)+\ldots+rx) = f((m+n(rx))) = nf(m(x)+\ldots + mx) = nf(mx) $$ Thus, \(f(rx) = f(mx)\) for all \(r\in\mathbb{Q}\).
04

Prove \(f\) is continuous

Since \(f\) is continuous, we can use the property that if \(f\) is continuous and homogeneous of degree 1, then it can be extended to all real numbers. Thus, for any \(r\in\mathbb{R}\), we can find a sequence of rational numbers \((q_n)\) such that \(q_n\to r\) as \(n\to\infty\). Therefore, using the homogeneity of \(f\) and its continuity, we get: $$ f(rx) = \lim_{n\to\infty} f(q_nx) = r\lim_{n\to\infty} f(x) = rf(x) $$
05

Conclusion

In conclusion, we found that the continuous function \(f\) is an additive function and a homogeneous function of degree 1 for all real numbers. Therefore, the function \(f\) that satisfies the given functional equation is a linear function of the form: $$ f(x) = cx, \text{ for some constant } c \in \mathbb{R} $$

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Most popular questions from this chapter

\bullet Welche der folgenden Teilmengen von \(\mathrm{C}\) sind beschr盲nkt, abgeschlossen und/oder kompakt? (a) \(\quad\\{z \in \mathbb{C}|| z-2 \mid \leq 2\) und \(\operatorname{Re}(z)+\operatorname{Im}(z) \geq 1\\}\) (b) \(\quad\left\\{\left.z \in \mathbb{C}|| z\right|^{2}+1 \geq 2 \operatorname{Im}(z)\right\\}\) (c) \(\\{z \in \mathbb{C} \mid 1>\operatorname{Im}(z) \geq-1\\}\) $$ \begin{aligned} &\cap\\{z \in \mathbb{C} \mid \operatorname{Re}(z)+\operatorname{Im}(z) \leq 0\\} \\ &\cap\\{z \in \mathbb{C} \mid \operatorname{Re}(z)-\operatorname{Im}(z) \geq 0\\} \end{aligned} $$ (d) \(\\{z \in \mathbb{C}|| z+2 \mid \leq 2\\} \cap\\{z \in \mathbb{C}|| z-\mathrm{i} \mid<1\\}\)

Betrachten Sie die beiden Funktionen \(f\), \(g: \mathbb{R} \rightarrow \mathbb{R}\) mit $$ f(x)= \begin{cases}4-x^{2}, & x \leq 2 \\ 4 x^{2}-24 x+36, & x>2\end{cases} $$ und $$ g(x)=x+1 $$ Zeigen Sie, dass die Graphen der Funktionen mindestens vier Schnittpunkte haben.

\bullet Bestimmen Sie die globalen Extrema der folgenden Funktionen. (a) \(f:[-2,2] \rightarrow \mathbb{R}\) mit \(f(x)=1-2 x-x^{2}\) (b) \(f: \mathbb{R} \rightarrow \mathbb{R}\) mit \(f(x)=x^{4}-4 x^{3}+8 x^{2}-8 x+4\)

Gegeben sind zwei stetige Funktionen \(f\), \(g: \mathbb{R} \rightarrow \mathbb{C}\) mit \(f(x)=g(x)\) f眉r alle \(x \in \mathbb{Q} .\) Zeigen Sie \(f(x)=g(x)\) f眉r alle \(x \in \mathbb{R}\)

Zeigen Sie: Jedes Polynom ungeraden Grades mit reellen Koeffizienten hat mindestens eine reelle Nullstelle.
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