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This is a question from some old masters exams.

Let $\phi_{n}$ be a sequence of continuous, real functions on $\mathbb{R}$ such that

$\phi_{n}(x) = 0$ for all $|x|\geq 1/n$ and $\phi_{n}(x)\geq 0$ for all $x\in\mathbb{R}$, and


For each continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$, let


Prove that $f_{n}(x)$ converges pointwise to $f(x)$ and prove that if $f(x)=0$ for $|x|\geq 10$, then $f_{n}$ converges uniformly to $f$.

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Do you have any thoughts on the problem? –  Kevin Carlson Aug 21 '12 at 2:49
This is a standard result about "approximations to the identity". I suggest looking up that term in your text. –  Potato Aug 21 '12 at 2:59
What must be the general behavior of a series of continuous functions supported on intervals with vanishing length, whose integrals don't vanish (and are positive)? Then, note that if $\phi_n$ has a certain behavior near $0$, then $\phi_n(x-y)$ has that same behavior whenever $x$ is near $y$. Think of the selection property of the Dirac delta, and ... –  Eugene Shvarts Aug 21 '12 at 3:18
Suppose $m<n$: $|\int_{\mathbb{R}}{\phi_{n}(x-y)f(y)dy}-\int_{mathbb{R}}{\phi_{m}(x-y)f(y)dy}| = |\int_{-1/m}^{1/m}{\phi_{n}(y)f(x-y)dy}-\int_{-1/m}^{1/m}{\phi_{m}(y)f(x-y)dy}|=‌​|\int_{-1/m}^{1/m}{(\phi_{n}(y)-\phi_{m}(y))f(x-y)dy}|$. Since $f$ is continuous on $[x-1/m,x+1/m]$, $f$ is bounded. Thus, $|\int_{-1/m}^{1/m}{(\phi_{n}(y)-\phi_{m}(y))f(x-y)dy}|\leq M\times|\int_{-1/m}^{1/m}{(\phi_{n}(y)-\phi_{m}(y))dy}|=M\times|\int_{-1/m}^{1/m‌​}{\phi_{n}(y)}-\int_{-1/m}^{1/m}{\phi_{m}(y)dy}|= 0$ since the integrals of $\phi_{n}$ and $\phi_{m}$ are both $1$ in the bounds given. This seems odd? –  neelp Aug 21 '12 at 3:24

2 Answers 2

up vote 3 down vote accepted

For pointwise convergence, note that it suffices to show that $$\int_{-\infty}^\infty \phi_n(x-y)(f(y)-f(x))dy\to 0$$ for all $x$. Fix some $x$. Since $f$ is continuous, for any $\epsilon>0$ we have some $\delta>0$ such that $$|x-y|<\delta\implies |f(x)-f(y)|<\epsilon.$$ Choose $N\in\mathbb N$ such that $N> 1/\delta$. Then for $n\geq N$ we have $$\begin{align} \left|\int_{-\infty}^\infty \phi_n(x-y)(f(y)-f(x))dy\right| &=\left|\int_{x-1/n}^{x+1/n} \phi_n(x-y)(f(y)-f(x))dy\right|\\ &\leq \int_{x-1/n}^{x+1/n} \phi_n(x-y)|f(y)-f(x)|dy\\ &< \epsilon \int_{x-1/n}^{x+1/n} \phi_n(x-y)dy = \epsilon \end{align}$$ hence $f_n\to f$ pointwise. For uniform convergence, note that if $f(x)=0$ when $|x|\geq 10$ then $f$ has compact support, hence we can choose this $\epsilon$ independently of $x$ hence the convergence is uniform in $x$.

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Our solutions are identical! –  user38268 Aug 21 '12 at 3:25
@BenjaLim Not quite, but definitely the same approach. The question is, is this the natural approach or the one we've been trained to follow? –  Alex Becker Aug 21 '12 at 3:26
My solution below is motivated by what I've been reading about good kernels and fourier series. –  user38268 Aug 21 '12 at 3:28
@Alex,could you please explain me first line why it suffices to prove that when we are asked to show its point wise limit is $f(x)$. I am sorry i might have misunderstood –  Lusin May 19 at 3:32

Here is a hint on how to do the problem. Because $\phi_n(x) = 0$ for $|x| \geq 1/n$, you can say

$$\int_{-1}^1 \phi_n(x) dx = \int_{-\infty}^{\infty} \phi_n(x) dx.$$

Now choose $x \in\Bbb{R}$. Then

$$\begin{eqnarray*} |f_n(x) - f(x) | &=& \left|\lim_{N \to \infty} \int_{-N}^N \phi_n(x-y)(f(y) - f(x)) dy \right| \\ &=&\lim_{N \to \infty} \left| \int_{-N}^N \phi_n(x-y)(f(y) - f(x)) dy \right| \\ &\leq&\lim_{N \to \infty} \int_{-N}^N |\phi_n(x-y)||f(y) - f(x)| dy \\ &=& \lim_{N \to \infty} \int_{x-N}^{x+N} |\phi_n(u)||f(x) - f(x-u)| du \\ &=& \int_{x-\frac{1}{n}}^{x + \frac{1}{n}} |\phi_n(u)||f(x) - f(x-u)| du. \end{eqnarray*}$$

Facts I used above: The absolute value function is continuous, and the integrals above are continuous functions in $x$. Also, I used that the $\phi_n(x)$ are zero for $n$ sufficiently large, and there is a change of variables above. We can now complete your problem as follows. By continuity of $f$ we know that given any $\epsilon > 0$ , there is $\delta > 0$ such that $|u| < \delta$ will imply that $|f(x) - f(x-u)| < \epsilon$. Choose $N$ so large that $1/N < \delta$. Then for all $n\geq N$, we would have that $|u| < \frac{1}{n}$ will imply that $|f(x) - f(x-u)| < \epsilon$. The integral in the last line is now less than

$$ \epsilon \int_{x-\frac{1}{n}}^{x + \frac{1}{n}} |\phi_n(u)| du = C\epsilon$$

for some constant $C$. It follows that $f_n \to f$ pointwise.

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