Approximation of $L^1$ function with compactly supported smooth function with same mass

Question

Given $f\in L^1(\mathbb{R})$, I can find for any $\epsilon>0$ a $g\in C_c^\infty(\mathbb{R})$ such that $\|f-g\|_{L^1}\leq \epsilon$.

Can I assume wlog that $\|f\|_{L^1} = \|g\|_{L^1}$?

Answer 1

Yes, you can choose $g$ so that $\lVert f\rVert_{L^1} = \lVert g\rVert_{L^1}$.

If $\lVert f\rVert_{L^1} = 0$, it is immediate that $g \equiv 0$ does the desired, so in the following assume $\lVert f\rVert_{L^1} > 0$. Given $\varepsilon > 0$, choose $\varepsilon_1 \in \bigl(0, \min \bigl\{\frac{1}{2}\varepsilon,\, \frac{1}{2} \lVert f\rVert_{L^1}\bigr\}\bigr]$, and $g_1 \in C_c^{\infty}(\mathbb{R})$ with $\lVert f-g_1\rVert_{L^1} < \varepsilon_1$. Then

$$0 < \frac{1}{2}\lVert f\rVert_1 \leqslant \lVert f\rVert_{L^1} - \varepsilon_1 < \lVert g_1\rVert_{L^1}.$$

Let $g = \frac{\lVert f\rVert_{L^1}}{\lVert g_1\rVert_{L^1}}\cdot g_1$. Then $g \in C_c^{\infty}(\mathbb{R})$ with $\lVert g\rVert_{L^1} = \lVert f\rVert_{L^1}$, and

\begin{align} \lVert f-g\rVert_{L^1} &\leqslant \lVert f-g_1\rVert_{L^1} + \lVert g_1-g\rVert_{L^1}\\ &= \lVert f-g_1\rVert_{L^1} + \frac{\lvert \lVert f\rVert_{L^1} - \lVert g_1\rVert_{L^1} \rvert}{\lVert g_1\rVert_{L^1}}\lVert g_1\rVert_{L^1}\\ &\leqslant 2\lVert f-g_1\rVert_{L^1}\\ &< 2\varepsilon_1\\ &\leqslant \varepsilon, \end{align}

so $g$ has all desired properties.