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Is it possible for a sequence of functions in $\mathcal{C}\left[0,1\right]$ to converge to one function pointwise (not necessarily to a continuous function) and to a different function in average (i.e with respect to $\Vert f\Vert=\left(\int\limits _{0}^{1}\left|f\left(x\right)\right|^{2}dx\right)^{\frac{1}{2}}$ ?

I know neither convergence mandates the other but I'm wondering whether it's possible for both convergences to occur but to different functions.

Thanks in advance!

Following some of the replies let's take as an example the sequence of functions defined by $f_{n}\left(x\right)=x^{n}$ , pointwise it has a limit which is $$f\left(x\right)=\begin{cases} 0 & x\in\left[0,1\right)\\ 1 & x=1 \end{cases}$$ but on average I can see that it converges to the zero function as $$\Vert f_{n}\left(x\right)-0\Vert\overset{n\to\infty}{\longrightarrow}0$$ Obviously $f\left(x\right)$ and $0$ are a.e equivalent so $f_{n}$ also converges to $f$ on average. However, while the zero function is in $\mathcal{C}\left[0,1\right]$ the pointwise limit $f$ is not. So unless I'm missing something here, the limit only has to be same up to a.e equivalence even if it's a sequence of continuous functions. I assume that if the pointwise limit was also continuous then it would have to be equivalent everywhere?

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Keep in mind that the second norm is only a norm up to a.e. equivalence, i.e. the indicator function of the rationals must be considered equivalent to $0$. –  Alex Becker Jan 5 '13 at 23:15
I edited my answer addressing the issue of continuity. –  Ayman Hourieh Jan 6 '13 at 10:09
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1 Answer

up vote 1 down vote accepted

Convergence in average to a function $f$ implies the existence of a subsequence that converges pointwise a.e. to the same function $f$ (See Rudin's Real & Complex Analysis theorem 3.12). Thus, if the sequence also converges pointwise to a function, it must be the same function a.e.

The other issue you raise is continuity of the limit. Not all modes of convergence preserve limits. Pointwise convergence doesn't as your example illustrates. Convergence in the supremum norm does, however, and it is equivalent to uniform convergence.

The $L^2$ norm (or average as you call it) doesn't preserve continuity. In fact, one can prove that a class of continuous functions called continuous with compact support is dense in the space of integrable functions $L^p$. This means that for every integrable function in the $L^p$ sense (not necessarily continuous), one can find a sequence of continuous functions with compact support that converges to it in $L^p$. This is an important result in functional analysis.

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Excellent, thanks for the clarification. –  Serpahimz Jan 6 '13 at 12:47
@Serpahimz Happy to help! –  Ayman Hourieh Jan 6 '13 at 12:52
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