Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here is another homework question that I did and I'd be glad if you could tell me if it's right.

We now strengthen the result of Question Two for $R$ where we have the notion of differentiability. Prove that for any open $Ω ⊂ R$ the set of smooth functions with compact support is dense in $L_1(Ω, λ)$ where $λ$ is the usual Lebesgue measure.

a) Define $J(x) = ke^{\frac{-1}{1−x^2}}$ for $|x| < 1$ and equal to zero elsewhere. Here, the constant $k$ is chosen such that $\int_R J = 1$. Prove that the mollifier $J_ε(x) = \frac{1}{\varepsilon}J(\frac{x}{\varepsilon})$ vanishes for $|x| ≥ ε$ and $\int J_\varepsilon = 1$.

For $f ∈ L_1$ define the regularization of f by convolving with $J_ε$: $$ f_ε(x) = J_ε \ast f(x) = \int_\Omega J_\varepsilon (x - y ) f(y) d \lambda(y)$$

b) Prove that $f_ε$ is integrable.

c) Prove that $f_ε$ is smooth.

d) Prove that if $f$ has compact support then so does $f_ε$.

e) Finish the proof: For any $f ∈ L_1(Ω)$ there exists $g ∈ C_C^\infty(Ω)$ such that $|f − g| < ε$.


a) $|x| > \varepsilon \implies |\frac{x}{e}| > 1$ by definition $J_\varepsilon = 0$.

$$ \int J_\varepsilon = \frac{1}{\varepsilon} \int_R J(\frac{x}{\varepsilon}) = \frac{1}{\varepsilon} \int J(y) \varepsilon d \lambda = 1$$ doing a variable substitution

b) $$ \int_R \int_\Omega J_\varepsilon (x - y) f(y) d \lambda(y) d\lambda(x) = \int_\Omega \int_R J_\varepsilon (x - y) dx f(y) dy = \int_\Omega f(y) dy \leq \int_R |f(y)| dy < \infty$$ using Fubini

c) $$ \frac{d}{dx^{(n)}} f_\varepsilon (x) = \frac{d}{dx^{(n)}} \int_\Omega J_\varepsilon(x -y) f(y) dy = \int_\Omega \frac{d}{dx^{(n)}} J_\varepsilon (x -y ) f(y) dy$$ where $J_\varepsilon(x-y) = e^{g(x)}$ is smooth.

d) $\int$ is linear $\implies $ continuous $\implies $ maps compact sets to compact sets

$f$ has compact supp. $A$, $J_\varepsilon$ has compact support $B$ then $f J_\varepsilon$ has compact support $A \cap B$

e) take $g$ to be $f_\varepsilon$

Many thanks for your help.

share|cite|improve this question
What do you mean with "$J_\varepsilon(x-y) = e^g(x)$"? – kahen Sep 25 '11 at 8:49
Thanks for pointing it out, that was a typo, I corrected it! – Rudy the Reindeer Sep 25 '11 at 8:54
up vote 6 down vote accepted

Directly proving that $f_{\varepsilon}$ is integrable is rather unnecessary since $C_c^\infty(\Omega) \subset C_c(\Omega) \subset L_1(\Omega)$.

It's a general fact of convolution that $f*g$ is at least as smooth as the smoothest of $f$ and $g$.

Your argument that $f * J_{\varepsilon}$ has compact support is flawed since you're not integrating $fJ_{\varepsilon}$, so you need to be a bit more explicit there.

share|cite|improve this answer

We can ignore the information that $J_\varepsilon$ is a mollifier. All we need is a smooth function with integral one. $J_\varepsilon$ is such a function as proven in a) in the question above.

We will use that $C_c(X)$ is dense in $L^1$ to show that $C_c^\infty(X)$ is also dense in $L^1$ where $X$ is an open subset of $\mathbb{R}$. Let $\epsilon > 0$ and $f \in L^1$. Then by density of $C_c(X)$ there is a $g$ in $C_c(X)$ such that $\| f - g \|_{L^1} < \epsilon$.

Now we need to turn $g$ into a smooth function by convolving it with $J_\varepsilon$. Let $$g_\varepsilon (x) := (J_\varepsilon \ast g ) (x) = \int_\mathbb{R} J_\varepsilon(x - y) g(y) dy$$

Then $g_\varepsilon$ is smooth because $\left ( f \ast g \right )^\prime = f^\prime \ast g = f \ast g^\prime$ and $J_\varepsilon$ is infinitely differentiable.

$g_\varepsilon$ has compact support because if $[-S,S]$ is the support of $g$ and $[-R,R]$ is the support of $J_\varepsilon$ then the support of $J_\varepsilon \ast g$ is contained in $[-S - R, S + R]$ and hence is also compact.

To finish the proof we claim that $\| f - g_\varepsilon \|_{L^1} < \epsilon$:

$$ \| f - g_\varepsilon \| \leq \| f - g \| + \|g - g_\varepsilon \| < \epsilon$$

Where $\| f - g \| < \frac{\epsilon}{2}$ holds because $C_c(X)$ is dense in $L^1$ and $\|g - g_\varepsilon \| < \frac{\epsilon}{2}$ holds because:

$$\begin{align} \|g - g_\varepsilon \|_{L^1} = \int_X \left | g(z) - g_\varepsilon (z)\right | dz &= \int_X \left | g(z) - \int_\mathbb{R} J_\varepsilon(z -y) g(y) dy \right | dz \\ &= \int_X \left | g(z)\int_\mathbb{R}J_\varepsilon(y)dy - \int_\mathbb{R}J_\varepsilon(z -y) g(y) dy \right | dz\\ &\stackrel{(*)}{=} \int_X \left | g(z)\int_\mathbb{R}J_\varepsilon(z - y)dy - \int_\mathbb{R}J_\varepsilon(z -y) g(y) dy \right | dz \\ &= \int_X \left | \int_\mathbb{R} g(z) J_\varepsilon(z - y)dy - \int_\mathbb{R}J_\varepsilon(z -y) g(y) dy \right | dz \\ &\leq \int_X  \int_\mathbb{R} | g(z) J_\varepsilon(z - y) |dy - \int_\mathbb{R} | J_\varepsilon(z -y) g(y) | dy dz \\ &= \int_X \int_\mathbb{R} |g(z) - g(y)| J_\varepsilon (z -y) dy dz \end{align}$$

Where the equality marked with (*) holds because the integral is over all of $\mathbb{R}$ so the shift by the constant $z$ doesn't change the integral and $J_\varepsilon$ is even hence $J_\varepsilon (y) = J_\varepsilon (-y)$.

$g$ is continuous and compactly supported hence it is uniformly continuous and so there exists a $\delta$ such that $|g(z) - g(y)| < \frac{\epsilon}{2 \lambda(X)}$ for all $z,y \in X$ hence by choosing $\varepsilon := \delta$ we get

$$ \int_X \int_\mathbb{R} |g(z) - g(y)| J_\delta (z -y) dy dz < \frac{\epsilon}{2} $$

Note that $\epsilon$ and $\varepsilon$ are not the same.

share|cite|improve this answer
This scheme can also be applied to $f\in L^p$ – Xavi Martinez Nov 6 '13 at 15:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.