Find all continuous functions f: ℝ → ℝ such that f(x+y) = f(x) + f(y) for all x,y ∈ ℝ.

Step 1: Identify the type of equation

The given equation is a functional equation known as Cauchy's functional equation. We are looking for continuous functions that satisfy this property.

Step 2: Evaluate at specific points

First, let's substitute $x=0$ and $y=0$ into the equation. This gives $f(0+0) = f(0) + f(0)$, which simplifies to $f(0) = 2f(0)$. Subtracting $f(0)$ from both sides yields $f(0) = 0$.

Step 3: Prove for natural numbers

We can prove by induction that $f(nx) = nf(x)$ for any natural number $n$. For $n=1$, $f(1x) = 1f(x)$ is true. Assuming $f(kx) = kf(x)$ for some $k \in \mathbb{N}$, we have $f((k+1)x) = f(kx+x) = f(kx) + f(x) = kf(x) + f(x) = (k+1)f(x)$. Thus, it holds for all natural numbers.

Step 4: Extend to integers

Since $f(0)=0$, we have $f(x + (-x)) = f(x) + f(-x)$, which means $f(0) = f(x) + f(-x)$, so $0 = f(x) + f(-x)$, implying $f(-x) = -f(x)$. This extends the property to negative integers, so $f(qx) = qf(x)$ for all integers $q$.

Step 5: Extend to rational numbers

Let $r = p/q$ where $p, q \in \mathbb{Z}$ and $q \neq 0$. Then $f(qx) = qf(x)$. Also, $f(x) = f(q \cdot \frac{x}{q}) = qf(\frac{x}{q})$. So $f(\frac{x}{q}) = \frac{1}{q}f(x)$. Combining these, $f(rx) = f(\frac{p}{q}x) = p f(\frac{x}{q}) = p \frac{1}{q} f(x) = \frac{p}{q} f(x) = rf(x)$.

Step 6: Extend to real numbers using continuity

Since $f$ is continuous and $f(rx) = rf(x)$ for all rational numbers $r$, we can let $x=1$ to get $f(r) = rf(1)$. Let $c = f(1)$. Then $f(r) = cr$ for all rational numbers $r$. Because $f$ is continuous and the rational numbers are dense in the real numbers, this property extends to all real numbers. Thus, $f(x) = cx$ for some constant $c \in \mathbb{R}$.

More from this chapter

• 2. Consider the function f(x) = 3cos(e^(5x-3)) + (4)/(x) Find f'(x) [(d)/(dx)(cos x) = -sin x]
• **प्रश्नावली 5.2** प्रश्न 1 से 8 में x के सापेक्ष निम्नलिखित फलनों का अवकलन कीजिए: 1. sin(x² + 5) 2. cos(sin x) 3. sin(ax + b) 4. sec(tan(√x)) 5. (sin(ax + b))/
• 1. Prove that the function f(x) = 2x² - 3x + 2 is continuous at x = 1. 2. The function f(x) is defined as follows: f(x) = 2x - 3, & x < 2 x - 1, & x ≥ 2 Prove t
• Find dy/dx if y = (ln(x))^(cos(x)).
• Differentiate the following w.r.t. x. (i) cos^-1(sin x) (ii) tan^-1(sin x / (1 + cos x)) (iii) sin^-1(2^(x+1) / (1 + 4^x))
• If x = t² and y = t³, then (d² y)/(dx²) is equal to:
• Find (d y)/(d x), if x=a(θ+sin θ), y=a(1-cos θ).
• Differentiate the following w.r.t. x. (i) cos ^(-1)(sin x) (ii) tan ^(-1)((sin x)/(1+cos x)) (iii) sin ^(-1)((2^(x+1))/(1+4^x))

All Continuity and Differentiability solutions →