Verified answer

For all x we have

f‵(x) ≤ f(x)

⇒ f‵(x) / f(x) ≤ 1

⇒ ∫_[0,x] (f‵(t) / f(t)) dt ≤ ∫_[0,x] dt

⇒ ln(f(x)) - ln(f(0)) ≤ x

⇒ exp(ln(f(x))) ≤ exp(ln(f(0)) + x)

⇒ f(x) ≤ f(0).exp(x).

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Hmm, I seem to have disregarded the possibility f(0) = 0.

I'll figure it out later. (In this case f must be the zero function,

assuming the conclusion holds).

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Let f(0) = 0.

Suppose that f(x) = 0 for all x ∈ [0,n], where n ∈ ℕ.

Now take 0 < b ≤ 1.

By the mean value theorem there exists c ∈ (n,n+b) such that

f‵(c) = (f(n + b) - f(n)) / (n + b - n) = f(n + b) / b.

Applying the inequality f‵ ≤ f we get

f(n + b) = b.f‵(c) ≤ b.f(c) ≤ f(c).

But c < n + b and f is increasing, so we must then have that f is constant on (c,n+b].

This holds for any b ∈ (0,1], so f is constant on (n,n+1].

Since f(x) = 0 for x ∈ [0,n] and f is continuous, f(x) = 0 for x ∈ [0,n+1].

By assumption f is 0 on the interval [0,0], so by induction f = 0.

f then trivially satisfies the inequality f(x) ≤ f(0).exp(x) for all x ∈ [0,∞).