Hello all I am stuck on the following small question in real analysis for practice in which we are given a function f of compact support and a measurable set A we are asked to prove the following is continuous:

$$ H(x)=\int_{\mathbb{R}^d}\chi_A(t)f(t-x)dm(t)$$

Could you please help me? Thanks to all of you


For all $x$, $Δx$, we have $$ {\left|H(x+Δx)-H(x)\right|} = {\left|\int_{\mathbb{R}^d}χ_A(t) \cdot \left(f(t-x-Δx)-f(t-x)\right) \,\mathrm{d}m(t) \right|} = {\left|\int_{\mathbb{R}^d}χ_A(t) \cdot \left(f(t-Δx)-f(t)\right) \,\mathrm{d}m(t) \right|} ≤ m(A) \cdot \sup_{t∈A}\left|f(t-Δx)-f(t)\right|. $$

Thus, if $f$ is uniformly continuous, $H$ is continuous.


Let me just provide some ideas: First consider: $$|H(x)-H(y)| \leq \int_A |f(t-x)-f(t-y)| dm(t)$$ What should happen now? If for given $\epsilon$ we find that $x$ and $y$ are $\delta$-close, then the integrand will be $\epsilon$ small. However, you still need to take the support of $f$ into account to obtain a small integral.

So maybe $\epsilon$ is not the best choice, but $\epsilon/C$, where $C$ is related to the support of $f$ in $A$, is.

So since $f$ is conitnuous, and particularly uniformly continuous since compactly supported (can you show this?), we find for any $\epsilon>0$ a $\delta>0$, such that $|f(\tilde x)-f(\tilde y)|< \epsilon$ if $|\tilde x -\tilde y|<\delta$.

So let us consider the integral again and use that $t-x$ is close to $t-y$ if $x$ is close to $y$:

$$|H(x)-H(y)| \leq \int_A \epsilon \chi_{\text{supp}{f(t-x)-f(t-y)}}\leq \epsilon C$$ Now choose $\varepsilon = \epsilon/C$ and everything works.

  • $\begingroup$ Thanks Quickbeam2k1 very much. I think I can proceed but as a novice in real analysis could you please go a step further with me? $\endgroup$
    – kroner
    Mar 20 '15 at 14:20

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