Complex Lp Spaces and Test-Function Conventions: Examples

Direct multiplication gives ${\omega }^2=(-1-i\sqrt{3})\mathrm{/}2$, ${\omega }^3=1$, $1+\omega +{\omega }^2=0$ and $\mathrm{\mid }\omega \mathrm{\mid }=1$. Thus all three values of $g$ have modulus one, so F3 gives $\mathrm{\parallel }g{\mathrm{\parallel }}_1=(1+1+1)\mathrm{/}3=1$. The measure is a probability measure: disjoint sets just partition three atoms, and their weighted cardinalities add to one on $X$.

For a real test $t=(t_0,t_1,t_2)\in [-1,1{]}^3$, set $w_{\sigma }={\prod }_{j=0}^2(1+{\sigma }_j{t}_j)\mathrm{/}2$ for $\sigma \in \{-1,1{\}}^3$. Each weight is nonnegative, summing the product over all signs gives ${\sum }_{\sigma }{w}_{\sigma }=1$, and summing ${\sigma }_j{w}_{\sigma }$ gives $t_j$. Thus $t={\sum }_{\sigma }{w}_{\sigma }\sigma$. For the linear expression $L(t)=(t_0+\omega t_1+{\omega }^2{t}_2)\mathrm{/}3=\int gt$, F2 implies $\mathrm{\mid }L(t)\mathrm{\mid }\le {\sum }_{\sigma }{w}_{\sigma }\mathrm{\mid }L(\sigma )\mathrm{\mid }\le {\max }_{\sigma }\mathrm{\mid }L(\sigma )\mathrm{\mid }$.

The conjugated test has modulus one on each atom, so its essential infinity norm is one. Coordinatewise $gs=(1,1,1)$, and F3–F4 give $\int gs=1$. Every complex test of norm at most one has integral modulus at most $\mathrm{\parallel }g{\mathrm{\parallel }}_1=1$ by F1; all tests here have finite-measure support. Hence this test attains the full complex supremum.

At the two equal-sign vertices, $L(\sigma )=0$ because $1+\omega +{\omega }^2=0$. Every other sign vertex has one exceptional sign, at some coordinate $j$, so $L(\sigma )=\pm 2{\omega }^j\mathrm{/}3$ and has modulus $2\mathrm{/}3$. For instance $t=(1,-1,-1)$ gives $L(t)=2\mathrm{/}3$. This proves both the real upper bound and its attainment. Finally F4 gives $\int g^2=(1+{\omega }^2+{\omega }^4)\mathrm{/}3=(1+{\omega }^2+\omega )\mathrm{/}3=0$, proving the claimed failure of the unconjugated phase.

F2 gives measure one to both intervals; they are disjoint, and both functions vanish elsewhere. Their squared moduli are each ${\mathbf{1}}_{[0,1)}+{\mathbf{1}}_{[1,2)}$, so F3 gives $\mathrm{\parallel }f{\mathrm{\parallel }}_2^2=\mathrm{\parallel }g{\mathrm{\parallel }}_2^2=1+1=2$. Thus both are integrable $L^2$ representatives.

On the first interval $f\bar{g}=1\cdot (-i)=-i$ and on the second it is $i\cdot 1=i$. F1 and F4 therefore give $⟨f,g⟩=(-i)\cdot 1+i\cdot 1=0$. The bilinear product instead has value $i$ on each interval, so $\int fg=i\cdot 1+i\cdot 1=2i$.

The coefficients of $if$ are $i,-1$ and those of $\bar{f}$ are $1,-i$, so $(if)\bar{f}$ has values $i,i$. Hence $⟨if,f⟩=2i$. The coefficients of $\overline{if}$ are $-i,-1$, so $f\overline{if}$ has values $-i,-i$ and $⟨f,if⟩=-2i$. These computations agree with first-variable linearity and second-variable conjugate-linearity and show concretely why the bilinear expression cannot replace the inner product.

The input is bounded and supported in $[0,2]$, so F3 makes it integrable and locally integrable. In the convolution integral, $f(x-t)$ has real part one exactly for $t\in [x-1,x]$ and imaginary part one exactly for $t\in [x-2,x-1]$. Integrating the two components gives the displayed formula. Their common endpoint has measure zero by F3. F1 makes the output smooth. Since ${\rho }_{\epsilon }$ vanishes outside $[-\epsilon ,\epsilon ]$, both integrals vanish for $x<-\epsilon$ and for $x>2+\epsilon$, proving the stated support inclusion.

For $0<\epsilon <1\mathrm{/}4$, put $U={\bigcup }_{a\in \{0,1,2\}}[a-\epsilon ,a+\epsilon ]$. Outside $U$, the entire interval $[x-\epsilon ,x+\epsilon ]$ stays in a constant region of $f$, so the unit kernel mass from F1 implies $({\rho }_{\epsilon }\ast f)(x)=f(x)$. Each component of the convolution is between zero and one, because ${\rho }_{\epsilon }\ge 0$ and its integral is one. The same is true for each component of $f$, including its value at the shared endpoint. Consequently the modulus error is at most $\sqrt{2}\le 2$ everywhere and is zero outside $U$. F3–F4 give $\mu (U)\le 6\epsilon$. Since $\mathrm{\mid }{\rho }_{\epsilon }\ast f-f{\mathrm{\mid }}^p\le 2^p{\mathbf{1}}_U$, F5–F6 and the definition of $N_p$ in F2 give $\mathrm{\parallel }{\rho }_{\epsilon }\ast f-f{\mathrm{\parallel }}_p^p\le 2^p\mu (U)\le 2^{p}6\epsilon$. Taking p-th roots proves the bound, and its limit is zero for each fixed finite $p$.

Let $h$ be any continuous complex function. If $\mathrm{\parallel }h-f{\mathrm{\parallel }}_{\mathrm{\infty }}<1\mathrm{/}2$, choose a real essential bound $a<1\mathrm{/}2$ for the error. On $(-1\mathrm{/}2,0)$ the function $f$ is zero, so $\mathrm{\mid }h(x)\mathrm{\mid }\le a$ a.e.; by continuity this inequality holds throughout that interval, since any failure persists on an open interval of positive measure by F3. Similarly $\mathrm{\mid }h(x)-1\mathrm{\mid }\le a$ throughout $(0,1\mathrm{/}2)$. Taking the respective limits at zero gives $\mathrm{\mid }h(0)\mathrm{\mid }\le a$ and $\mathrm{\mid }h(0)-1\mathrm{\mid }\le a$, whence $1\le \mathrm{\mid }h(0)\mathrm{\mid }+\mathrm{\mid }h(0)-1\mathrm{\mid }\le 2a<1$, a contradiction. Thus every such $h$ has error at least $1\mathrm{/}2$, in particular the smooth function from step 1.1.