It can be proved that $\forall\varepsilon>0$ there exists $C(\epsilon)>0$ such that for all $f,g\in L^{1,\infty}(\Bbb R^n)$ we have that $$ ||f+g||_{1,\infty}\le(1+\varepsilon)||f||_{1,\infty}+C(\varepsilon)||g||_{1,\infty}$$ for example $C(\epsilon)=1+\frac1{\varepsilon}$ works.

I have some problem in proving this inequality fails for $\varepsilon=0$. I should prove that for every $C>0$ there exist $f_C,g_C\in L^{1,\infty}(\Bbb R^n)$ such that

$$ ||f+g||_{1,\infty}>||f||_{1,\infty}+C||g||_{1,\infty} $$ but it seems really hard. I thought I can take some sequence of functions but I can't put this idea into concrete computations.

Can someone help me?

EDIT: Given $f:\Bbb R^n\to\Bbb R$ measurable we define, for $\alpha>0$, $\lambda_f(\alpha):=|\{x\in\Bbb R^n\;:\;|f(x)|>\alpha\}|$ (given a subset $E\subseteq\Bbb R^n$, we set |E| to denote its Lebesgue measure).

Then we set $$ L^{1,\infty}(\Bbb R^n):=\{f:\Bbb R^n\to\Bbb R\;\mbox{measurable}\;:\exists C>0\;\;\mbox{s. t.}\;\; \lambda_f(\alpha)\le\frac C{\alpha}\;\forall \alpha>0\}\;\;\;. $$

Finally we set $||f||_{1,\infty}$ as the infimum of such $C$'s.

  • 1
    $\begingroup$ The functions f, g can depend on C: at the end of your question, the quantifiers are wrong. $\endgroup$ – Clement C. Apr 22 '16 at 15:52
  • $\begingroup$ Where do you mean I'm wrong? It seems correct to me $\endgroup$ – Joe Apr 22 '16 at 21:45
  • $\begingroup$ The way you wrote it, you want to finda single pair (f,g) that work for all C. You actually need to find (possibly different) f,g for each C. $\endgroup$ – Clement C. Apr 22 '16 at 21:55
  • $\begingroup$ Yes, you are right, thanks! I've edited $\endgroup$ – Joe Apr 23 '16 at 0:15
  • $\begingroup$ Hmmm... if I'm not mistaken, I think there is still an issue: it's $f,g$ = $f(C)$, $g(C)$, not $C=C(f,g)$. $\endgroup$ – Clement C. Apr 23 '16 at 4:04

Obviously, $||f||_{1,\infty} = \sup_{\alpha>0} \alpha \lambda_f(\alpha)$. For $n=1$ and a monotonic $f$ this is the area of the largest rectangle under the graph of $f$. The idea is thus the following. Take $f$ such that this rectangle has a large horizontal side. We add a small $g$, which bumps $f$ at the point where the rectangle touches the graph, which increases the area considerably.

Let $n=1$, $K>0$ be small, and $$f(x) = \frac1x \mathbf{1}_{(0,1)}(x),\quad g(x) = \frac{K}{1-x}\mathbf{1}_{(0,1)}(x).$$ Clearly, $||f||_{1,\infty} = 1$, $||g||_{1,\infty} = K$. It is easy to see that the minimum of $f+g$ on $(0,1)$ is attained at $x=(1+\sqrt{K})^{-1}$ and is equal to $m = (1+\sqrt{K})^2$. Therefore, $$ ||f+g||_{1,\infty} \ge m \lambda_{f+g}(\alpha)(m) =m\ge 1+\sqrt{K}. $$ Taking $K<1/C^2$, we get $$ ||f+g||_{1,\infty} > 1+CK = ||f||_{1,\infty}+C||g||_{1,\infty}, $$ as required.


Here's an example for $\mathbb{R}$, though it can likely be generalized to higher dimensions.

Fix an integer $n\geqq 2$, and define $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=\frac{n}{i}$ if $x\in [\frac{i-1}{n},\frac{i}{n})$, $1\leq i\leq n$, and $f(x)=0$ otherwise. Then $||f||_{1,\infty}=1$.

Define a map $T:\mathbb{R}\to\mathbb{R}$ by $T(x)=x+\frac{1}{n}$ (mod 1) if $x\in[0,1)$, and $T(x)=x$ otherwise. For $1\leq j\leq n$, let $f_j(x)=f(T^jx)$.

Then $||f_j||_{1,\infty}=1$ for $1\leq j\leq n$, but $\sum_{j=1}^nf_j$ is equal to the constant function $n\sum_{j=1}^n\frac{1}{j}$ on $[0,1)$, hence $$ \Big|\Big|\sum_{j=1}^nf_j\Big|\Big|_{1,\infty}=n\sum_{j=1}^n\frac{1}{j} $$

There is a constant $c>0$ such that $n\sum_{j=1}^n\frac{1}{j}\geq cn\log n$ for all $n\geq 2$, hence we have shown that for each $n\geq 2$ there exist $f_1,\dots,f_n$ with $||f_j||_{1,\infty}=1$ such that $$ \Big|\Big|\sum_{j=1}^nf_j\Big|\Big|_{1,\infty}\geq cn\log n$$

If the triangle inequality holds with some constant $C$, then for any functions $g_1,\dots,g_n$ with $||g_j||_{1,\infty}=1$ we have $$||g_1+\dots+g_n||_{1,\infty}\leq ||g_1+\dots+g_{n-1}||_{1,\infty}+C\leq\dots\leq 1+(n-1)C$$

But for any constant $C$ we can choose an $n$ sufficiently large that $$ cn\log n>1+(n-1)C$$ hence the triangle inequality cannot hold.


I found an answer by my own.

We want to prove that the statement $$ \exists C>0\;\;:||f+g||_{1,\infty}\le||f||_{1,\infty}+C||g||_{1,\infty}\;\;\;\forall f,g\in L^{1,\infty}(\Bbb R^n) $$ is false (we should in fact expect this: looking at PART(A), we see that $C(\varepsilon)>0$ gets bigger and bigger as $\varepsilon$ comes closer to $0$, ). Thus we must prove that $$ \forall C>0\;\;\exists f_C,\;g_C\in L^{1,\infty}(\Bbb R^n)\;\;:\;\; ||f_C+g_C||_{1,\infty}>||f_C||_{1,\infty}+C||g_C||_{1,\infty}\;\;. $$ I took $n=1$ and I searched for two sequences of functions $\{f_k\}_k,\{g_k\}_k\subseteq L^{1,\infty}(\Bbb R)$ such that \begin{equation}\label{4} ||f_k+g_k||_{1,\infty}>||f_k||_{1,\infty}+C||g_k||_{1,\infty} \end{equation} be true for some $k=k(C)$; in this way I would have finished. Observe now that the above inequality is equivalent to $$ \frac{||f_k+g_k||_{1,\infty}}{||g_k||_{1,\infty}}>\frac{||f_k||_{1,\infty}}{||g_k||_{1,\infty}}+C $$ (we search for a $\{g_k\}_k$ such that $||g_k||_{1,\infty}\neq0\;\;\;\forall k$; otherwise if $||g_k||_{1,\infty}=0$ for a finite number of $k$'s or for infinitely many $k$'s but not definitively, we just drop these $g_k$ from the sequence; if otherwise $g_k=0$ a.e. definitively, then the above inequality becomes $||f_k||_{1,\infty}>||f_k||_{1,\infty}\;\;k\ge\bar k$, which is false). Thus finding $\{f_k\}_k,\{g_k\}_k\subseteq L^{1,\infty}(\Bbb R)$ such that

\begin{equation}\label{5} \left\{ \begin{array}{ll} \limsup_k\frac{||f_k+g_k||_{1,\infty}}{||g_k||_{1,\infty}}=+\infty\\ \limsup_k\frac{||f_k||_{1,\infty}}{||g_k||_{1,\infty}}<+\infty\\ \end{array} \right. \end{equation} would allow us to conclude. \newline \newline I began to try with different functions on different domains, until the role of every component I was using became clear; then I could make an heuristic argument which finally gave two possible candidates for the desired sequences.

So let's $n=1$ and consider $$ f_k(x):=-ke^x\chi_{]0,\sqrt k[}(x),\;\;\;\;\;\;\;g_k(x):=ke^x\chi_{]0,k[}\;\;,\;\;k\ge1 $$ from which $$ f_k(x)+g_k(x)=ke^x\chi_{[\sqrt k,k[}(x)\;\;. $$ Observe now, that, for a measurable function $f:\Bbb R^n\to\Bbb R$ we have that $$ ||f||_{1,\infty}:=\inf\left\{D>0\;:\;\lambda_f(\alpha)\le\frac D{\alpha}\;\forall\alpha>0\right\}=\sup_{\alpha>0}\{\alpha\lambda_f(\alpha)\}\;. $$ We will use this last characterization of the $1$-weak norm in the following.

Now: \begin{align*} \lambda_{f_k}(\alpha) &= \left\{ \begin{array}{lll} 0\\ \sqrt k-\log\left(\frac{\alpha}{k}\right)\\ \sqrt k \end{array} \right. \left. \begin{array}{ccc} \alpha\ge ke^{\sqrt k}\\ k\le\alpha<ke^{\sqrt k}\\ 0<\alpha<k \end{array} \right.\\ &=\left(\sqrt k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[k,ke^{\sqrt k}[}(\alpha)+\sqrt k\chi_{]0,k[}(\alpha) \end{align*} from which we get $$ ||f_k||_{1,\infty}=\sup_{\alpha>0}\left\{\alpha\left(\sqrt k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[k,ke^{\sqrt k}[}(\alpha)+\alpha\sqrt k\chi_{]0,k[}(\alpha)\right\}=k\sqrt k $$ in fact if we set $h(\alpha):=\alpha\left(\sqrt k-\log\left(\frac{\alpha}{k}\right)\right)$ we have $h'(\alpha)=\sqrt k-\log\left(\frac{\alpha}{k}\right)-k$ which is negative when $\alpha\in [k,ke^{\sqrt k}[$, thus here $h$ decreases, thus the $\sup$ above follows. Let's now look at $g_k$: \begin{align*} \lambda_{g_k}(\alpha) &= \left\{ \begin{array}{lll} 0\\ k-\log\left(\frac{\alpha}{k}\right)\\ k \end{array} \right. \left. \begin{array}{ccc} \alpha\ge ke^{k}\\ k\le\alpha<ke^{k}\\ 0<\alpha<k \end{array} \right.\\ &=\left(k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[k,ke^{k}[}(\alpha)+k\chi_{]0,k[}(\alpha) \end{align*} from which we get $$ ||g_k||_{1,\infty}=\sup_{\alpha>0}\left\{\alpha\left(k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[k,ke^{k}[}(\alpha)+\alpha k\chi_{]0,k[}(\alpha)\right\}=k^2 $$ because, with the same argument as above, $h(\alpha):=\alpha\left(k-\log\left(\frac{\alpha}{k}\right)\right)$ is decreasing in $[k,ke^k[$.\ Finally let's work with the sum $f_k+g_k$: \begin{align*} \lambda_{f_k+g_k}(\alpha) &= \left\{ \begin{array}{lll} 0\\ k-\log\left(\frac{\alpha}{k}\right)\\ k-\sqrt k \end{array} \right. \left. \begin{array}{ccc} \alpha\ge ke^{k}\\ ke^{\sqrt k}\le\alpha<ke^{k}\\ 0<\alpha<ke^{\sqrt k} \end{array} \right.\\ &=\left(k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[ke^{\sqrt k},ke^{k}[}(\alpha)+(k-\sqrt k)\chi_{]0,ke^{\sqrt k}[}(\alpha) \end{align*} from which we get $$ ||f_k+g_k||_{1,\infty}=\sup_{\alpha>0}\left\{\alpha\left(k-\log\left(\frac{\alpha}{k}\right)\right)\chi_{[ke^{\sqrt k},ke^{k}[}(\alpha)+\alpha(k-\sqrt k)\chi_{]0,ke^{\sqrt k}[}(\alpha)\right\}=ke^{\sqrt k}(k-\sqrt k) $$ again observing that $h(\alpha):=\alpha\left(k-\log\left(\frac{\alpha}{k}\right)\right)$ is decreasing in $[ke^{\sqrt k},ke^k[$. Thus, such $f_k$'s and $g_k$'s verify the above $\limsup$'s, so we have done.

  • $\begingroup$ I did not read too carefully, but it seems that $ g_k\ge f_k+g_k\ge 0$, whence $||f_k+g_k|| \le ||g_k||$, so $||f_k+g_k||/||g_k||$ cannot diverge to $\infty$. $\endgroup$ – zhoraster Apr 28 '16 at 19:53
  • 1
    $\begingroup$ I think $\|g_k\|_{1,\infty}=\|g_k+f_k\|_{1,\infty}=ke^{k-1}=\alpha \lambda_f(\alpha)$ for $\alpha=ke^{k-1}$. $\endgroup$ – san Apr 30 '16 at 17:12
  • $\begingroup$ @san: why do you think that? Where does my error is in my computation? $\endgroup$ – Joe Apr 30 '16 at 22:25
  • $\begingroup$ You have to use Zhoraster's method for the calculation of the norm: The area of the greatest rectangle under the curve. When you compute the supremum in both cases you make a mistake, since your result is lower than mine for $\alpha=k e^{k-1}$. The supremum in your case is also attained at that $\alpha$. $\endgroup$ – san Apr 30 '16 at 23:19
  • 1
    $\begingroup$ $h(\alpha)$ is not decreasing, in fact it attains a maximum at $\alpha=ke^{k-1}$. $\endgroup$ – san Apr 30 '16 at 23:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.