Real Analysis Live - Problem Solving - Continuous Functions (Problems here: https://tbsom.de/live)

A piecewise “rational vs. irrational” function can look wildly discontinuous everywhere—yet still be continuous at a single point. The stream’s first problem uses the sequence definition of continuity to prove that the function f(x)=x for rational x, and f(x)=0 for irrational x is continuous at x=0. The key move is to take an arbitrary sequence (x_n)→0 and show f(x_n)→f(0)=0. Because each x_n is either rational (then f(x_n)=x_n) or irrational (then f(x_n)=0), the absolute difference |f(x_n)−f(0)| is always ≤|x_n|, which goes to 0 by construction. That inequality works uniformly across all possible sequences, including those that mix rationals and irrationals infinitely often—so the “jumps” shrink to nothing at the origin.

The session then escalates to a parameter-fitting continuity problem. A new piecewise function is defined using three formulas on (-∞,-2], [-2,4], and [4,∞): a quartic x^4−4 on the left, a linear ax+b in the middle, and √x on the right. Continuity on the open intervals is automatic because polynomials, linear functions, and √x are continuous there; the only possible failures occur at the “gluing points” x=-2 and x=4. Continuity at x=-2 forces the middle line’s right-limit value to match the quartic’s value: (-2)^4−4=12 must equal -2a+b. Continuity at x=4 forces 4a+b to match √4=2. Solving the resulting 2×2 linear system yields a=-5/3 and b=26/3, producing a function continuous across all real x.

Next comes a classic existence proof: any continuous function f:[0,1]→[0,1] has at least one fixed point. The argument uses geometry and the Intermediate Value Theorem by defining g(x)=f(x)−x. Since g is continuous and g(0)=f(0)−0≥0 while g(1)=f(1)−1≤0, the IVT guarantees some x̃ with g(x̃)=0, i.e., f(x̃)=x̃. The stream stresses that existence is guaranteed, not uniqueness—multiple intersections (hence multiple fixed points) are possible.

A counterexample shows what breaks when the domain is no longer compact: on ℝ, a function like f(x)=x+1 has no fixed point because f(x)=x would imply 1=0. The final segments shift from continuity definitions to stronger notions and limits: an ε–δ continuity exercise is handled by explicitly choosing δ in terms of ε, then a pointwise-limit example demonstrates that pointwise convergence of continuous functions can fail to preserve continuity (the limit function has a jump at 0). The session closes with Lipschitz continuity: if |f(x)−f(y)|≤L|x−y| for all x,y, then continuity follows immediately by choosing δ=ε/L (with a brief note on the L=0 case, which forces f to be constant). The throughline is consistent: continuity is controlled either by sequences, ε–δ estimates, or inequalities strong enough to prevent “jumps,” and the exact definition dictates the proof strategy.