Existence of $f_n(t)=o(f(t))$ where $(f_n)$ is a sequence of functions.

Question

Let $(f_n)$ a sequence of functions over $\mathbb{R}$ to $\mathbb{R}$,

Show that there exist $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f_n(t)=o(f(t))$

I suspect that $f_n$ converges pointwise to $f$,

Anyway, Does I have to found an example or can I prove the existence without finding an example ?

Answer 1

No, $f_n$ does not have to converge pointwise to anything. It can be any sequence of functions whatsoever.

You have to prove that $f$ exists. Proofs of existence can be constructive or non-constructive (e.g., by contradiction). That said, "constructive" does not mean having an explicit formula like "$f(x) = 23x^9$". You can't have such a formula since $f_n$ are not known. The best thing to hope for is an explicit formula in terms of $f_n$. This is what the suggestion below gives.

Hint. Define $f$ piecewise, e.g., $$ f(x) = m\cdot\max(f_1(x),\dots,f_m(x)) \quad \text{ when } \ m\le x<m+1 $$

Answer 2

This problem, with the solution the same as in the accepted answer, is historically important as the first published use of a diagonalization, by Paul Du Bois-Reymond.