# Triangle inequality fails in $L^{1,\infty}$

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.

• The functions f, g can depend on C: at the end of your question, the quantifiers are wrong. – Clement C. Apr 22 '16 at 15:52
• Where do you mean I'm wrong? It seems correct to me – Joe Apr 22 '16 at 21:45
• 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. – Clement C. Apr 22 '16 at 21:55
• Yes, you are right, thanks! I've edited – Joe Apr 23 '16 at 0:15
• Hmmm... if I'm not mistaken, I think there is still an issue: it's $f,g$ = $f(C)$, $g(C)$, not $C=C(f,g)$. – 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 $$\label{4} ||f_k+g_k||_{1,\infty}>||f_k||_{1,\infty}+C||g_k||_{1,\infty}$$ 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

$$\label{5} \left\{ \begin{array}{ll} \limsup_k\frac{||f_k+g_k||_{1,\infty}}{||g_k||_{1,\infty}}=+\infty\\ \limsup_k\frac{||f_k||_{1,\infty}}{||g_k||_{1,\infty}}<+\infty\\ \end{array} \right.$$ 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.

• 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 $\infty$. – zhoraster Apr 28 '16 at 19:53
• I think $\|g_k\|_{1,\infty}=\|g_k+f_k\|_{1,\infty}=ke^{k-1}=\alpha \lambda_f(\alpha)$ for $\alpha=ke^{k-1}$. – san Apr 30 '16 at 17:12
• @san: why do you think that? Where does my error is in my computation? – Joe Apr 30 '16 at 22:25
• 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 $\alpha$. – san Apr 30 '16 at 23:19
• $h(\alpha)$ is not decreasing, in fact it attains a maximum at $\alpha=ke^{k-1}$. – san Apr 30 '16 at 23:23