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Proof. If (7) holds for each bounded sequence x, since L (x) ⊆ R is a nonempty set
then f (L (x)) ⊆ R is also nonempty, and therefore (6) holds. Of course, if (9) holds for each
bounded sequence x, then again (6) holds. If (8) holds for each bounded sequence x, since
L (f (x)) ⊆ (R ∪ {±∞}) is nonempty and f (L (x)) is a subset of R, then L (f (x)) ⊆ R is
nonempty and therefore (6) holds. Using now Theorem 3, we see that each of (ii), (iii) and (iv)
implies (i).

To finish the proof, it is sufficient to show that (i) implies (iv). So suppose that f is
continuous on R and consider a bounded sequence of real numbers x = (x n ) . If α ∈ L (x),
n≥1
) → α as n → +∞. Since f is
then α ∈ R and there exists a subsequence (x k n n≥1 such that x k n
) → f (α) as well. Thus f (α) ∈ L (f (x)). We have therefore proved
continuous at α, then f (x k n
that f (L (x)) ⊆ L (f (x)). For the reverse inclusion, let us observe first that the continuity
of f implies that f (x) is also a bounded sequence in R. So let β ∈ L (f (x)). Then β ∈ R,
) ) ⊆ R is
and there exists a subsequence (x j n n≥1 of x such that f(x j n ) → β. Since (x j n n≥1

) such that
also a bounded sequence, there exist α ∈ R and a subsequence x j k n n≥1 of (x j n n≥1

→ f (α).
x j k n → α as n → +∞. Then α ∈ L (x). Also, the continuity of f at α gives f x j k n

)) ) → β, then f (α) = β. Thus
n≥1
Since f x j k n is a subsequence of (f (x j n n≥1 and f(x j n
β ∈ f (L (x)). □
References

[1] R. Basu, V. Kannan, K. Sannyasi, N. Unnikrishnan, Functions preserving limit superior,
The College Mathematics Journal, Vol. 50 Issue 1 (2019), 58-60.