Let $g_t(x)=\dfrac{1}{2\sqrt{\pi t}}e^{-\frac{(at+x)^2}{4t}} $ and suppose that $f\in L^1$. Show that $$\lim_{t\rightarrow 0^+}\int_{-\infty}^\infty\left|\int_{-\infty}^\infty(f(x-y)-f(x))g_t(y)dy\right|dx=0$$

How can I prove this statement? The function $g$ is quite messy, and I have a double integral. Also, $f$ is just in $L^1$, so there's no assumption on continuity, just an assumption that $$\int_{|x|>n}|f(x)|dx\rightarrow 0$$ as $n\rightarrow\infty$.


$g_t$ is positive, and $\int_{-\infty}^\infty g_t(x)\,dx = 1$, so the linear operator $K_t\colon L^1(\mathbb{R}) \to L^1(\mathbb{R})$ given by convolution with $g_t$,

$$K_t(f)(x) = (f\ast g_t)(x) = \int_{-\infty}^\infty f(x-y)g_t(y)\,dy,$$

has norm $\leqslant 1$ (we don't need to know the norm is exactly $1$). Since the integral of $g_t$ is $1$, the task is to show $\lim\limits_{t\to 0^+} \lVert f\ast g_t - f\rVert_1 = 0$.

Now show that for $f \in C_c(\mathbb{R})$, continuous functions with compact support. These functions have enough regularity that the proof is easy enough:

$$\begin{align} \int_{-\infty}^\infty \left\lvert\int_{-\infty}^\infty \bigl(f(x-y)-f(x)\bigr)g_t(y)\,dy \right\rvert\,dx & \leqslant \int_{-\infty}^\infty \int_{-\infty}^\infty \left\lvert f(x-y)-f(x)\right\rvert g_t(y)\,dy \,dx\\ &= \int_{-\infty}^\infty \underbrace{\left(\int_{-\infty}^\infty \lvert f(x-y)-f(x)\rvert\,dx\right)}_{I(y)} g_t(y)\,dy\\ &= \int_{\lvert y\rvert \leqslant \delta} I(y) g_t(y)\,dy + \int_{\lvert y\rvert > \delta} I(y) g_t(y)\,dy. \end{align}$$

Given $\varepsilon > 0$, choose $\delta$ small enough to make the first integral small, then choose $t$ small enough to make the second small.

Then use the fact that $C_c(\mathbb{R})$ is dense in $L^1(\mathbb{R})$, so for every $\varepsilon > 0$, you can find a $h \in C_c(\mathbb{R})$ with $\lVert f-h\rVert_1 < \varepsilon/3$. Then

$$\lVert f\ast g_t - f\rVert_1 \leqslant \lVert f\ast g_t - h\ast g_t\rVert_1 + \lVert h\ast g_t - h\rVert_1 + \lVert h-f\rVert_1 < \lVert h\ast g_t - h\rVert_1 + 2\varepsilon/3.$$

For all small enough $t > 0$, you have $\lVert h\ast g_t -h\rVert_1 < \varepsilon/3$ by the result for continuous functions with compact support.

  • $\begingroup$ Ah thanks. For $f$ being continuous with compact support (say it's supported on $[-n,n]$), I split the integral into two parts $|x|<n+1$ and $|x|>n+1$. The first part is fine because of uniform convergence and the interval is finite, but I can't see how to do the second part. Would you mind showing a sketch? $\endgroup$ – Kunal Nov 20 '13 at 1:27
  • $\begingroup$ You split the wrong integral. I've added a sketch, it should be not too difficult to fill in the remaining parts. $\endgroup$ – Daniel Fischer Nov 20 '13 at 1:44
  • $\begingroup$ Sorry, one last bit: To show that we can choose $t$ small enough, we have to show $\lim_{t\rightarrow 0^+}\int_{|y|>\delta}g_t(y)dy=0$. I've tried for a while but the function $g_t$ is hard to work with. Any hints? $\endgroup$ – Kunal Nov 20 '13 at 2:31
  • $\begingroup$ Substitute $z = \dfrac{y}{\sqrt{t}}$. You get an integral over $\lvert z\rvert > \delta/\sqrt{t}$ of what is basically $\exp(-z^2/4)$. $\endgroup$ – Daniel Fischer Nov 20 '13 at 2:37
  • $\begingroup$ You always have elegant ways of doing things! $\endgroup$ – Kunal Nov 20 '13 at 2:48

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