# Find derivative of convolution with gaussian

Let $A(\sigma)$, $\sigma > 0$ be an operator that acts on bounded continuous functions $f$ on $\mathbb{R}$ by the rule $$(A(t)f)(x) = \int\limits_{\mathbb{R}} f(y)\frac{1}{\sqrt{2 \pi t}}\exp\left( -\frac{(x-y)^2}{2t} \right)dy,$$ and let $(A(0)f)(x) = f(x)$. Then $(A(t)f)(x) \to f(x)$ when $t \to +0$. But is it true in general that $$\left(\frac{ A(t) - A(0)}{t}f\right)(x) \to \frac{1}{2}\frac{d^2}{dx^2}f(x), \;\;\; t \to +0$$ for any $C^2(\mathbb{R})$ bounded function $f$?

If $f(x)$ is bounded and continuous on $\mathbb{R}$ then $u(t,x)=(A(t)f)(x)$ is a solution of Cauchy problem for the heat equation: $$\left\{ \begin{array}{l} u_{t} = \frac{1}{2} u_{xx}, \;\;\; t>0, \; x \in \mathbb{R} \\ u(0,x) = f(x), \;\;\; x \in \mathbb{R} \end{array} \right.$$ Then $\frac{1}{t}(A(t)-A(0))f(x) = \frac{1}{t}(u(t,x)-u(0,x))$. So the question is equivalent to a question: is it true that for bounded $f(x) \in C^2(\mathbb{R})$ there exists a right derivative $$\lim\limits_{t\to +0}\frac{u(t,x)-u(0,x)}{t} = u_{t}(+0,x) = \frac{1}{2}u_{xx}(0,x)$$ so that the equation $u_t = \frac{1}{2}u_{xx}$ is valid for limit value of $t$.

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## 2 Answers

Applying a Fourier Transform to your first equation and taking the Derivative gives $$\frac{d}{dt} F(\omega) e^{-2\pi^2\omega^2 t} = F(\omega) e^{-2\pi^2\omega^2 t} (-2\pi^2\omega^2)$$ which is transformed back

$$\left(\frac{1}{2}\frac{d^2}{dx^2} f(x)\right) \ast N(x,t) \xrightarrow{t\rightarrow +0} \frac{1}{2}\frac{d^2}{dx^2} \ ,$$ where $N(x,t)$ is the Gaussian and '$\ast$' the convolution.

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Welcome to Math.SE! Thank you for answering this question. A remark: since $f''$ can be arbitrarily large at infinity, the convergence of $f''*N(\cdot,t)$ is not immediately obvious; see my comment to the other answer. – user53153 Jan 7 '13 at 14:29

Let $N(t,x) = \frac{1}{\sqrt{2 \pi t}} \exp \left( -\frac{x^2}{2t} \right)$, $t>0$. For any bounded $f(x) \in C(\mathbb{R})$ we have $$\int f(y)N(t,x-y)dy \to f(x), \;\;\; t \to +0$$ Indeed, fix $x \in \mathbb{R}$, $t>0$ and $\varepsilon > 0$ and take $\delta = \delta(\varepsilon) > 0$ such that $|f(x)-f(y)|< \varepsilon$ when $|x-y|<\delta$, then $$\int f(y)N(t,x-y)dy = \int\limits_{|x-y|< \varepsilon} + \int\limits_{|x-y| \geqslant \varepsilon}$$ We have $$f(x)-\varepsilon \leqslant \int\limits_{|x-y|< \varepsilon}f(y)N(x-y)dy \leqslant f(x)+\varepsilon,$$ because $\int N(t,x-y) = 1$ and $N(t,x-y) > 0$ for any $y$. Next let $M$ be such constant that $|f(x)| \leqslant M$ for any $x$. Then we can write $$\left| \int\limits_{|x-y|\geqslant \varepsilon}f(y)N(t,x-y) dy \right| \leqslant M \int\limits_{|y| \geqslant \varepsilon} N(t,y) dy \leqslant \frac{M}{\varepsilon^2} \int\limits y^2 N(t,y) dy = \frac{M}{\varepsilon^2} t \to 0$$ when $t \to +0$. Then $$\varlimsup\limits_{t \to +0} \int f(y) N(t,x-y)dy \leqslant f(x) + \varepsilon \\ \varliminf\limits_{t \to +0} \int f(y) N(t,x-y)dy \geqslant f(x) - \varepsilon$$ for an arbitrary $\varepsilon > 0$. From this we can derive that $$\int f(y) N(t,x-y) \to f(x), \;\;\; t \to +0$$ for any bounded continuous $f$.

Now let $f$ be a $C^2$ bounded function but not necessary with bounded second derivative. Consider a cut-off function $\eta \in C^2_{c}(\mathbb{R})$ such that $\eta(y) = 1$ for such $y$ that $|x-y|<1$ and $\eta(y) = 0$ for $|x-y|>2$. Then it can be shown that $|f(\cdot)*N(\cdot,y) - \eta(\cdot)f(\cdot)*N(t,\cdot)| = o(t)$ when $t \to +0$. Then we can deal with $\eta f$ instead of $f$ if second derivative of $f$ is not bounded. Without loss of generality we consider $f$ with bounded second derivatives.

We have $u(t,x) = f(\cdot)*N(t,\cdot) (x)$ and for any $t>0$ we have $$u_{t}(t,x) = \frac{1}{2}u_{xx}(t,x) = \frac{1}{2}f''(\cdot)*N(t,\cdot) (x) \to \frac{1}{2}f''(x), \;\;\; t \to + 0$$ Function $t \mapsto u(t,x)$ is continuous on $[0,T]$, $u_{t}(t,x)$ exists on $(0,T)$ and exists limit $u_{t}(+0,x)$. Then $$u_{t}(+0,x) = \lim\limits_{t\to+0} \frac{u(t,x)-u(0,x)}{t}$$ (for this statement see Demidovich, Problems in Mathematical Analysis). Hence $$\lim\limits_{t \to +0} \frac{u(t,x) - u(0,x)}{t} = \frac{1}{2} f''(x).$$

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I think there's a gap here. How do you conclude that $f''*N(t,\cdot)\to f''$ as $t\to 0$? The preceding convergence argument was for bounded continuous functions, but $f''$ is not known to be bounded. // This can be fixed by introducing a cut-off function $\eta\in C^\infty_c(\mathbb R)$ is such that $\eta=1$ on $[-1,1]$. Using the boundedness of $f$ one can show that at $x=0$ we have $|f*N(t,\cdot)-(\eta f)*N(t,\cdot)|=o(t)$ as $t\to 0$. Then the rest of computation goes through with $\eta f$ in place of $f$. – user53153 Jan 7 '13 at 14:27
I'm confused by the reference to Demidovich. What you wrote for $u_t(+0,x)=\dots$ looks like the definition of one-sided derivative. What you actually use here is L'Hospital rule: if $g'(t)\to L$ as $t\to 0+$, then $(g(t)-g(0))/t\to L$. – user53153 Jan 7 '13 at 14:33
@PavelM aha, thank you. I mean example 1258 from Russian edition. – Nimza Jan 7 '13 at 14:36