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Let $f \in L^{2}([0, 1])$, and extend it to be defined on all of $\mathbb{R}$ by setting it equal to $0$ outside of $[0, 1]$.

We're given the function $$F(x)=\int_{0}^{1}f(x+t)\cdot f(t)\, dt $$ and asked to prove that it is continuous at $x=0$.

I'm trying to prove that $\lim_{x\to 0} F(x)=F(0)$ or equivalently, $\lim_{x\to 0} \int_{0}^{1} f(x+t)\cdot f(t)\, dt = \int_{0}^{1}f(t)^2\,dt$, but I'm not sure how to move the limit inside the integral.

Mark
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2 Answers2

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Since continuous functions are dense in $L_2[-a,1+b]$, choose a continuous functions $f_n$ (extended as $0$ outside $[-a,1+b]$ such that $$\|f_n-f\|_{L_2[-a,1+b]} < \epsilon/3.$$ Now choose $x$ small enough such that $$|f_n(x+t)-f_n(t)| <\epsilon/3$$ for all $t\in [0,1]$.

Let $f^x(t) = f(x+t)$ and $f_{n}^x(t) = f_n(x+t)$.

Hence,

$$\|f^x -f\| \leq \| f^x -f_n^x\|+\|f_n^x -f_n\| + \| f_n -f\|<\epsilon$$

This proof follows the idea from Show that $\lim_{t \to 0} \int_{\mathbb{R}^d}|f(x)-f(x-t)|dx = 0$

Balaji sb
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  • Balaji, why did you write $f(x,.)$ and $f_n(x,.)$ in your proof ? The functions $f$ and $f_n$ are functions of a single variable, are not they ? What do you mean when you write $f(x,.)$ ? When you write it, it seems that $f$ is a function of two variables. – Angelo Aug 10 '23 at 08:29
  • You are correct. Probably i should have written $f(x+.)$ – Balaji sb Aug 10 '23 at 09:31
  • Balaji, would not it be better to write $f(x+t)$, $f(t)$, $f_n(x+t)$ and $f_n(t)$ instead of what you wrote ? Moreover, could you explain why $|f(x+t)-f(t)|<\varepsilon$ implies that $F(x)\to F(0)$ as $x\to0;?$ – Angelo Aug 10 '23 at 10:02
  • I can change notation but its $L_2$ norm and inside norm is supposed to be a function. Dont know which is a clear notation. $||f(x+.)-f(.)|| < \epsilon \implies |\int f(x+t) f(t) dt - \int f(t) f(t)| = |\int (f(x+t)-f(t)) f(t) dt| < \epsilon$ Because of Cauchy Schwartz inequality. – Balaji sb Aug 10 '23 at 14:27
  • Balaji, for a clearer notation you could define $f^+(t)=f(x+t)$ and $f_n^+(t)=f_n(x+t)$ for any $t\in[0,1].$ $\text{And then you could write :}$ $|f^+!-f|\leqslant|f^+!-f_n^+|+|f_n^+!-f_n|+|f_n-f|<\varepsilon,.;;;$ – Angelo Aug 10 '23 at 18:39
  • Changed the notation. Thanks for your inputs. – Balaji sb Aug 10 '23 at 22:50
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Thanks to the comments for pointing out the rookie error. This is fixable by using the density of continuous functions, but probably overcomplicated.

Let's first argue this for continuous $f$, for which we can use the following generalisation of the DCT (see this for a proof): let $h_n, g_n, h, g$ be integrable functions such that $h_n \to h, g_n \to g$ a.e., $|h_n| \le g_n,$ and $\int g_n \to \int g$. Then $\int h_n \to \int h$.

Let us work with the extended definition of $f$ you mention. Let $\{x_n\}$ be any sequence converging to $0$, and define $g_n = f(x_n+t)^2 + f(t)^2$. Then due to continuity of $f$, $g_n \to 2f(t)^2$, and since integrals over $\mathbb{R}$ are invariant to translation, $$ \int_{\mathbb{R}} g_n = 2\int f(t)^2 = \int g.$$ But $|f(x_n + t) f(t)| \le g_n,$ we conclude that $$ \int_{[0,1]} f(x_n+t)f(t) = \int_{\mathbb{R}} f(x_n + t) f(t) \to \int_{\mathbb{R}} f(t)^2,$$ where the equality uses that $f(t)$ is $0$ for $t \not\in [0,1]$. Since $\{x_n\}$ was arbitrary, it follows that $F$ is continuous at $0$ (although nothing really used $x_n \to 0$ in the above, so the argument works for continuity at any $x$).

Now to move to $f \in L^2,$ we can use the density of continuous functions in this space. Concretely, let $M = \|f\|_2,$ and fix $\varepsilon \in (0,1)$ and take $\varphi$ to be a continuous map such that $\|f-\varphi\|_2 \le \varepsilon.$ Then for every $n$, \begin{align} &\qquad \left| \int f(x_n +t) f(t) -\int \varphi(x_n+t)\varphi(t)\right| \\&\le \int |f(x_n+t)(f(t) - \varphi(t))| + \int |\varphi(t) (f(x_n+t) - \varphi(x_n + t)| \\ &\le 2\|f\|_2 \|f-\varphi\|_2 = 2M\varepsilon. \end{align}

It follows that $$ \limsup \int f(x_n + t) f(t) \le \limsup \int \varphi(x_n+t) \varphi(t) + 2M\varepsilon = \int \varphi^2(t) + 2M\varepsilon.$$

But $\|\varphi\|^2_2 \le \|\varphi - f\|_2^2 + \|f\|_2^2 + 2\|f\|_2 \|\varphi - f\|_2 \le \|f\|_2^2 + (2M + 1)\varepsilon$ telling us that $$ \limsup \int f(x_n+t) f(t) \le \int f(t)^2 + (4M + 1)\varepsilon.$$

Similarly, we have the lower bound $$ \liminf \int f(x_n+t) f(t) \ge \int \varphi^2 - 2M\varepsilon,$$ and $$ \|\varphi\|^2 \ge \|f\|^2 - 2\|f-\varphi \|\|\varphi\| - \|\varphi - f\|^2,$$ giving us that $$\liminf \int f(x_n +t) f(t) \ge \int f(t)^2 - (4M + 1)\varepsilon. $$

Now, since $\varepsilon$ is arbitrary, drive this to $0$ to conclude.

stochasticboy321
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    It is not necessarily true that $g_n \to 2f(t)^2$, you would need $f$ to be continuous at $t$. – copper.hat Aug 10 '23 at 03:50
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    I would suggest considering the difference of the integrals, and then using Cauchy Schwarz. Then use continuity of translations in $L^2$, in this case. Note that as mentioned above, what you did essentially assumes $f$ is continuous. – peek-a-boo Aug 10 '23 at 04:31
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    “Then clearly …” – Ted Shifrin Aug 10 '23 at 04:33
  • Stochasticboy321, why $$\int |f(x_n+t)(f(t) - \varphi(t))| + \int |\varphi(t) (f(x_n+t) - \varphi(x_n + t)| \ \le 2|f|_2 |f-\varphi|_2 = 2M\varepsilon;;?$$Obviously $;\int |f(x_n+t)(f(t) - \varphi(t))|\le|f|_2 |f-\varphi|_2;,;$ but could you explain me why$$\int |\varphi(t) (f(x_n+t) - \varphi(x_n + t)|\le|f|_2 |f-\varphi|_2;;?$$ – Angelo Aug 10 '23 at 08:05
  • Ah, I meant $|\varphi| |f- \varphi|,$ but $|\varphi| \le |f| + |f-\varphi|,$ so it's more or less the same thing up to a $+\varepsilon^2$. I guess to neaten things redefine $M = \max(1, |f|)$ and switch every $2M\varepsilon$ to $3M\varepsilon$, and $(2M + 1)\varepsilon$ to $4M\varepsilon$. I'll edit when I have the patience :P – stochasticboy321 Aug 10 '23 at 08:11