Find all continuous functions f: ℝ → ℝ such that f(x+y) = f(x) + f(y) for all x,y ∈ ℝ.
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Step-by-Step Solution
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∈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∈Z and q=0. Then f(qx)=qf(x). Also, f(x)=f(q⋅qx)=qf(qx). So f(qx)=q1f(x). Combining these, f(rx)=f(qpx)=pf(qx)=pq1f(x)=qpf(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∈R.