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Continuity

Continuity means nearby inputs produce outputs approaching the actual function value. Compare a continuous call-payoff corner with a fixed-fee jump, and check endpoints using the stated domain.

Continuity means that a function's values approach its actual value at a point as the input approaches that point within its domain. In a financial model, it separates a payoff that changes without a jump from a fee that switches on abruptly when trading begins. A continuous function can still have a sharp corner.
For a real-valued function ff, continuity at an interior point aa is expressed by
lim⁡x→af(x)=f(a).\lim_{x\to a}f(x)=f(a).
The limit must exist, the value f(a)f(a) must be defined, and the two must agree. Knowing only what happens near aa does not tell you what the function does at aa.

More precisely, for every output tolerance ε>0\varepsilon>0, there must be an input tolerance δ>0\delta>0 such that every valid input xx satisfying ∣x−a∣<δ\lvert x-a\rvert<\delta also satisfies ∣f(x)−f(a)∣<ε\lvert f(x)-f(a)\rvert<\varepsilon. The tolerances use the respective output and input units. This is a local condition: it does not assert that every small-looking change has a small financial effect at every scale.

The call joins up, even at its corner

Consider the payoff of a call option at expiry, per underlying unit, with strike USD 100. Let S≥0S\ge0 denote the terminal underlying price in USD:
f(S)=max⁡(S−100,0).f(S)=\max(S-100,0).

Below 100, the payoff is zero. Above 100, it increases dollar for dollar with the terminal price. At the join,

lim⁡S→100−f(S)=0,lim⁡S→100+f(S)=0,f(100)=0.\lim_{S\to100^-}f(S)=0, \qquad \lim_{S\to100^+}f(S)=0, \qquad f(100)=0.

All three agree, so the payoff is continuous there. For instance, terminal prices 99.99, 100 and 100.01 give payoffs 0, 0 and 0.01 dollars. These are hypothetical terminal scenarios, not option prices observed before expiry. Premiums and trading costs are excluded: this is payoff, not profit.

The limiting argument does not depend on a few selected prices. For every admissible SS,

0≤f(S)≤∣S−100∣.0\le f(S)\le\lvert S-100\rvert.

As SS approaches 100, the upper bound shrinks to zero. That proves the payoff approaches its value at the strike.

A call expiry payoff joins at strike USD 100 despite slopes zero and one. A divisible-order fee equals zero at quantity zero but approaches USD 2 from positive quantities.
A call expiry payoff joins at strike USD 100 despite slopes zero and one. A divisible-order fee equals zero at quantity zero but approaches USD 2 from positive quantities.
Original hypothetical functions. The call payoff is in USD per underlying unit; the fee is a total USD charge. The fee's open circle marks the value approached by positive orders, not the charge at zero.
The payoff is not differentiable at 100. Its left slope is 0 and its right slope is 1, so there is no single derivative at the strike. Differentiability implies continuity at an interior point; continuity does not imply differentiability.
Keep that terminal-payoff statement separate from a live option's delta, which differentiates the pre-expiry option price with respect to today's spot, holding the other pricing inputs fixed. In the standard Black–Scholes model, a European vanilla call with positive time remaining and positive volatility has a smooth price as a function of positive spot, including at the strike. Its expiry payoff corner does not make that pre-expiry delta undefined.

The first trade can switch on a charge

Now model a hypothetical transaction cost. Let q≥0q\ge0 be a continuously divisible order quantity. Placing no order costs nothing; any positive order incurs a USD 2 entry charge plus USD 0.01 per unit:
F(q)={0,q=0,2+0.01q,q>0.F(q)= \begin{cases} 0, & q=0,\\ 2+0.01q, & q>0. \end{cases}

A quantity of 1 costs USD 2.01, while 0.01 costs USD 2.0001. Reducing a positive order makes the variable charge disappear, but leaves the entry charge:

lim⁡q→0+F(q)=2≠0=F(0).\lim_{q\to0^+}F(q)=2\ne0=F(0).

The model is discontinuous at zero: an arbitrarily small positive trade still triggers almost USD 2. A smooth approximation that connects the origin to the positive-order curve can hide this participation cost and change an optimiser's choice between trading and staying out.

The domain matters. There are no negative quantities in this model, so continuity at zero uses only the right-hand limit. For a function on a closed interval [a,b][a,b], continuity at aa is checked from the right and at bb from the left; interior points use both sides. On a disconnected domain, only approaches through valid inputs count.

If quantity were restricted to whole units, zero would be isolated in the domain: there would be no sequence of positive admissible quantities shrinking to zero. The entry charge would remain economically relevant, but the continuous-domain limit above would no longer describe that integer decision problem.

What the assumption buys you

Continuity makes substitution into a limit valid at the point. It also supports existence results. A continuous function on a closed interval takes every value between its endpoint values; with opposite endpoint signs, this justifies a bracket for bisection. A continuous objective on a nonempty compact feasible set attains a maximum and a minimum. Neither result establishes uniqueness, and continuity alone supplies no derivative or hedge.

For a longer treatment of the definitions and interval convention, the Calculus I continuity lesson has a public preview; the complete lesson requires membership. Start with “Continuity at a point” and “Continuity on an interval.”

Sources

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  • continuity
  • limits
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