# $f \in L^p$ implies there exists $h_0$ such that for all $h<h_0$, $\|f(x+h)-f(x)\|_{L^p} < \varepsilon$

So there is this statement:

$f \in L^p$ implies for all $\varepsilon > 0$ there exists $h_0$ such that for all $h$ with $|h|<h_0$, $\|f(x+h)-f(x)\|_{L^p} < \varepsilon$

I can show a weaker version of this statement using Dominated Convergence Theorem and it looks like

$f \in L^p$ implies for all $\varepsilon > 0$ there exists $h$ such that $\|f(x+h)-f(x)\|_{L^p} < \varepsilon$

and this is simply DCT on $f(x+h_n)-f(x)$ where $h_n$ is a sequence tending to $0$. The problem I have is is seeing how to get the first statement without assuming DCT works for nets but a quick search tells me there is no DCT for nets. The reason I need the first statement is because I will be integrating over $h$ so I need the bound to apply over an interval instead of over a sequence of $h$. Can anyone help?

## 1 Answer

Let $\epsilon \gt 0$. Compactly supported functions are dense in $L^p$. So for $f \in L^p$, there exists a $g \in C_c(\mathbb{R})$ such that $\|f-g\|_p \lt \frac{\epsilon}{3}$

Now since $g \in C_c(\mathbb{R})$, there exists a $\delta \gt 0$ such that whenever $|h| \lt \delta$, we have $\|g(x+h)-g(x)\|_p \lt \frac{\epsilon}{3}$ (Note: I am using the fact that translation is continuous.)

Now for all $|h| \lt \delta$, we have $$\|f(x+h)-f(x)\|_p \le \|f(x+h)-g(x+h)\|_p+\|g(x+h)-g(x)\|_p+\|g(x)-f(x)\|_p \lt \epsilon$$

• Ah, the ubiquitous $C_c^\infty$ approximation, I really should have known. Thanks. – E.Lim Aug 30 '17 at 0:53
• @E.Lim You are welcome:) – tattwamasi amrutam Aug 30 '17 at 0:54