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The following question is related to the one in Is this convolution continuous?

Let $0 \leq \xi \leq T$, $Y_0 \in L^2([-T, T])$, $a_1 \in L^2([0,T])$ and consider the following function: $$ Y_{1}(\xi)=\int_{0}^{\xi} a_{1}(\alpha) Y_{0}(\alpha-\xi) d \alpha $$ Now if inside the integral $Y_0$ did not depend on $\xi$ then $Y_1$ would be absolutely continuous. Instead in the case I describe is $Y_1$ continuous? The difference with the other question is that here I only require $Y_0 \in \mathcal L^2([-T, T])$ and not $Y_0$ being continuous$.

Inspired by The convolution of two $L^2(\mathbb R)$ functions is continuous I could proceed in the following way by a density argument:
Let $Y_0^n \to Y_0$ in $L^2([-T, T])$ where $Y_0^n$ are continuos functions. Then by defining $Y_1^n(\xi)$ with $Y_0^n$ in place of $Y_0$ we have that $Y_1^n(\xi)$ is continuous (actually uniformly continuous, see Is this convolution continuous? ).

Finally we show that $Y_1^n \to Y_1$ uniformly on $\xi \in [0,T]$, i.e. \begin{align} |Y_1^n(\xi)-Y_1(\xi)|^2 &=\left |\int_0^\xi a_1(\alpha)[Y_0^n(\xi-\alpha)-Y_0(\xi-\alpha)] d \alpha \right|^2\\ & \leq \int_0^\xi a_1(\alpha)^2 d \alpha \int_0^\xi [Y_0^n(\xi-\alpha)-Y_0(\xi-\alpha)]^2 d \alpha \\ & \leq \int_0^T a_1(\alpha)^2 d \alpha \int_0^T [Y_0^n(\xi-\alpha)-Y_0(\xi-\alpha)]^2 d \alpha\\ & = |a_1|_{L^2} \int_{\xi-T}^{\xi} [Y_0^n(x)-Y_0(x)]^2 d \alpha\\ & \leq |a_1|_{L^2} \int_{-T}^{T} [Y_0^n(x)-Y_0(x)]^2 d \alpha \to 0 \end{align} where the convergence is uniform wrt $\xi \in [0,T]$.

Now we conclude that $Y_1$ is continuous as uniform limit of continuous functions.

Is everything correct?

carlos85
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  • Indeed, I did the exercise to quickly, and made some mistakes... – Surb May 23 '22 at 11:15
  • Anyway as I was saying should be correct, consider: \begin{align} \big|Y_1(\zeta) - Y_1(\xi)\big|&= \bigg| \int_0^\zeta a_1(\alpha)Y_0(\alpha+\zeta) \mathrm{d}\alpha - \int_0^\xi a_1(\alpha)Y_0(\alpha+\xi) \mathrm{d}\alpha \bigg| \ &\le \int_0^\zeta \big|a_1(\alpha)\big| \big|Y_0(\alpha+\zeta) - Y_0(\alpha+\xi)\big| \mathrm{d}\alpha + \int_\zeta ^ \xi |a_1(\alpha)||Y_0(\alpha+\xi)| \mathrm{d}\alpha \end{align} – carlos85 May 23 '22 at 11:17
  • The first term on the rhs goes to zero by dominated convergence and the second by Holder's inequality so it is continuous, do you agree? Uniform continuity I'm not sure – carlos85 May 23 '22 at 11:18
  • yes it's correct, however, without a continuity assumption on $Y_0$ unfortunately, you can't conclude. – Surb May 23 '22 at 11:53
  • You mean I cant conclude the uniform continuity without the contiuity assumption? – carlos85 May 23 '22 at 13:46
  • You can't conclude on the continuity of $Y_1$ without the assumption that $Y_0$ is continuous. – Surb May 23 '22 at 14:01
  • Hi @Surb you were right. I uptdated the question showing a possible way. Do you agree now? – carlos85 May 24 '22 at 16:53
  • Notice that continuous function are dense in $L^2$, so you can always take a sequence of continuous function $(Y_0^n)$ s.t. $Y^n_0\to Y_0$ in $L^2$ (it's not a supplementary assumption). I don't see any mistake in your proof, so it's correct to me (and indeed, no need continuity of $Y_0$). And congratulation for your determination ! Despite the fact that I said that it doesn't work, you proved me that I was wrong :-D (I should have been more careful and say that it may not work ^^). So (+1) – Surb May 24 '22 at 19:06
  • Great, thanks for the check and remark! – carlos85 May 24 '22 at 21:30

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