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Consider the interval $[0,1]$ and define $$ X_1:= \left[0, \frac{1}{2}\right], ~~~X_2 := \left[\frac{1}{2}, \frac{3}{4}\right], ~~ X_3 := \left[\frac{3}{4}, \frac{7}{8}\right], ...$$

Define a piecewise constant function as follow : $$ f(x) := \frac{1}{\sqrt{\mu(X_n)}\cdot n}$$ for $x \in X_n$. It seems to me that $f \in L_2[0,1]$ with respect to Lebesgue measure but $f \not\in L_1[0,1]$.

However, this can't be since $[0,1]$ is a finite measured space. Therefore we should have the following inclusion $L_1[0,1] \supseteq L_2[0,1]$. What am I getting wrong here ?

I had the idea of such a function while thinking about the inclusion $\ell_1 \subset \ell_2$. Since $\frac{1}{n} \in \ell_2 \backslash \ell_1$, I thought of adding trying to create a piecewise constant function taking values on this sequence (adding what seems to be appropriate weight) to see what's going on. But I ended up getting quite confused. Please illuminate me.

M.G
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2 Answers2

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This is a more expanded version of 251257's comment: Your function is bounded above by any convex function that satisfies $f(1-2^{-n})\geq 2^{(n+1)/2}/n$; in particular, by $f(x)=2(1-x)^{-1/2}$.

This integrates to $4\sqrt{1-x}$ (unless I screwed up the constant; calculus is hard), so you do have a finite $L^1$ norm.

[You don't really need to being in the convexity here; I just like integrals more than sums...]

Eric Stucky
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$\mu(X_n)=2^{-n}$ so for $x\in X_n$ we have $f(x)=2^{n/2}/n$ and $f(x)^2=2^n/n^2.$

So $\int_{X_N}f(x)\;dx=2^{-n/2}n^{-1}$ and $\int_{X_n}f(x)^2\;dx=n^{-2}.$

So $\|f\|_1=\sum_n 2^{-n/2}n^{-1}<\infty$ and $(\|f\|_2)^2=\sum_nn^{-2}<\infty.$

DanielWainfleet
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