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Let $ f:\mathbb{R}^m \rightarrow (-\infty,\infty] $ be lower semicontinuous and bounded from below. Set $f_k(x) = \inf\{f(y)+k d( x,y ): y\in \mathbb{R}^m\} $ , where $d(x,y)$ is a metric. It is easy to see that each $f_k$ is continuous and $f_1 \leq f_2\leq ...\leq f \\$. However, I don't know how to prove that $ \lim_{k \rightarrow \infty}f_k(x) = f(x) $ for every $x\in\mathbb{R}^m $.

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4 Answers 4

I assume you are interested in point-wise convergence.

Fix $x$. Since $f$ is bounded from below, if $d(x,y)>r$, then if

$$k>\frac{f(x)-f(y)}{r}$$

we have that $f(y)+kd(x,y)>f(x)$. That is, outside a ball of radius $r$ centered at $x$, after $k$ terms of the sequence (where $k$ is given by the inequality above) the value of the quantity in brackets is larger than $f(x)$.

It follows that the infimum has to be achieved in the closed ball $d(x,y)\leq r$ (and it's then a minimum). Call this point $z_k$, that is $f_k(x) = f(z_k)$. For what we said $d(z_k,x)<r$. Although it might not be clear, $r$ and $k$ are somehow related to each other: if you increase $k$, the region where the infimum has to be achieved, shrinks.

If we let $k\to\infty$, we get $z_k\to x$. Hence

$$\lim_{k\to\infty}f_k(x)=\lim_{k\to\infty}f(z_k)\geq f(x)$$ because of the lower semicontinuity. On the other hand, $f_k(x)\leq f(x)$ $\forall k$, because we are taking the infimum on a neighborhood of $x$. Hence

$$f_k(x)\leq f(x)\Rightarrow \lim_{k\to\infty}f_k(x)\leq f(x)$$

which together with the previous inequality gives

$$\lim_{k\to\infty}f_k(x)\leq f(x) \leq \lim_{k\to\infty}f_k(x)$$ that is

$$\lim_{k\to\infty}f_k(x)=f(x)$$

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Fix $\epsilon > 0$. WLOG, $f\ge0$

There is, by lower semicontinuity, $\delta>0$ such that $f(y)>f(x)-\epsilon$, $\forall y \in B(x,\delta)$.

If $k$ is large enough, then $f(y)+kd(x,y)\ge k\delta > f(x)$ for $y$ outside the former ball. Thus, the inf required must lie inside it. But- for every $y$ in this ball, we must have that $f(y)+kd(x,y)\ge f(x)-\epsilon$.

By the observations above, taking inf on the left hand side, we get that $f_k(x) \ge f(x)-\epsilon$ for $k$ big enough, that completes the proof.

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This is where the lower semicontinuity comes in. When $y$ is close to $x$, you have a lower bound $f(y)\ge f(x)-\epsilon$. When $y$ is far from $x$, the term $kd(x,y)$ is large. Put these together and you have a lower bound on $f_k(x)$.

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Since $f$ is lower semicontinuous and the distance function is continuous it follows that $ y \mapsto f(y)+kd(x,y)$ is lower semicontinuous and bounded from below. Therefore there exists $x_k$ such that $f_k(x)\geq f(x_k)+kd(x,x_k)-\varepsilon$, where $\varepsilon$ is small enough.

First note that $f_k(x) \leq f(x)$ from definition. Let $m$ be a lower bound for $f$. Then we have $$ f(x) \geq f_k(x)\geq f(x_k)+kd(x,x_k) -\varepsilon \geq m+kd(x,x_k)-\varepsilon $$

Taking $k \to \infty$ we can easily see that we must have $d(x,x_k) \to 0$. But then by the lower semicontinuity of $f$ we have $$ \liminf f_k(x)\geq \liminf[f(x_k)+kd(x,x_k)]-\varepsilon\geq \liminf f(x_k)-\varepsilon\geq f(x)-\varepsilon.$$

For $\varepsilon \to 0$ we obtain $\liminf f_k(x) \geq f(x)$

Since the inequality $\limsup f_k(x) \leq f(x)$ is obvious we can conclude that $$ \lim f_k(x)=f(x)$$

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