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with being a continuous function.
In view of x being a continuous function, we can get (3.2).
In view of u being a continuous function, there exists a point (t_{1}in[0,T]), such that d_{1}leq u(t_{1} leq d_{2}.
The semi-Fredholm index being a continuous function, we also have from the above that σ x ( S + R ) = σ x ( S ) ; σ x = σ w or σ a w or σ s w.
This theorem states that if { f n } n ∈ N ( f n : K → R, n ∈ N, being a sequence of continuous functions and K being a compact metric space) converges pointwise to f ( f : K → R being a continuous function) and if f n (x)≥fn+1 x) (∀x∈K and ∀ n ∈ N ), then { f n } n ∈ N converges uniformly to f.
Similar(55)
It's because it is a continuous function.
And assume that that is a continuous function.
If the function f is a continuous function, does the level set or surface have to be continuous?
More suggestions(17)
being a continuous presence
being a monotonous function
being a continuous source
being a continuous risk
being a continuous sore
being a necessary function
being a fixed function
being a continuous survey
being a binary function
being a continuous compact
being a unimodal function
being a discrete function
being a continuous narrative
being a continuous quasicontraction
being a linear function
being a continuous domain
being a continuous loop
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