Lebesgue space?

Question

find a continuous function f in Lp space (p=1) and f does not have compact support

Answer 1

How about e^(-x²)? This is continuous, and is easily seen to be in L¹ (in fact, we have [-∞, ∞]∫|e^(-x²)| dx = [-∞, ∞]∫e^(-x²) dx = √π), but is nonzero on the entire real line.

Answer 2

Let 1 ≤ p < ∞ and (S, μ) be a measure space. Consider the set of all measurable functions from S to C (or R) whose absolute value raised to the p-th power has a finite Lebesgue integral, or equivalently, that

The set of such functions form a vector space, with the following natural operations:

and, for a scalar λ,

That the sum of two pth power integrable functions is again pth power integrable follows from the inequality |f + g|p ≤ 2p (|f|p + |g|p). In fact, more is true. Minkowski's inequality says the triangle inequality holds for

Thus the set of pth power integrable functions, together with the function ||·||p, is a seminormed vector space, which we denote by

This can be made into a normed vector space in a standard way; one simply takes the quotient space with respect to the kernel of ||·||p. Since ||f||p = 0 if and only if f = 0 almost everywhere, in the quotient space two functions f and g are identified if f = g almost everywhere. The resulting normed vector space is, by definition,

For p = ∞, the space L∞(S, μ) is defined as follows. We start with the set of all measurable functions from S to C (or R) which are essentially bounded, i.e. bounded up to a set of measure zero. Again two such functions are identified if they are equal almost everywhere. Denote this set by L∞(S, μ). For f in L∞(S, μ), its essential supremum serves as an appropriate norm:

As before, we have

if f ∈ L∞(S) ∩ Lq(S) for some q < ∞.
For 1 ≤ p ≤ ∞, Lp(S, μ) is a Banach space. Completeness can be checked using the convergence theorems for Lebesgue integrals.

The above definitions generalize to Bochner spaces.

Edit:
I hope the above information can help you determine your answer...Your answer, I think, can be inferred from the above info....