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Proof of Elementary Properties of the p-Seminorm

lemmalem:lp-seminorm-basic-2026a
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· 5,148 chars · 8 deps · depth 17 Reason: First version. Each property from the exponent laws together with monotonicity and homogeneity of the integral.

Each claim follows from the exponent laws for powers with nonnegative base together with the monotonicity, homogeneity and almost-everywhere properties of the integral.

Proof

Each result cited is universally quantified over the data appearing in its own statement, and is applied here to the data named in the statement above. Throughout, powers with nonnegative base are those of Real Power of a Nonnegative Real Number §power, and we refer to the claims of Properties of Real Powers of Nonnegative Real Numbers for their properties; for fLpf\in\mathcal{L}^{p} the number Xfpdμ\int_{X}|f|^{p}\,d\mu is a nonnegative real number by Power-Integrable Functions and the p-Seminorm §space.

Claim 1. Put I=XfpdμI=\int_{X}|f|^{p}\,d\mu, a nonnegative real number. Then, by Properties of Real Powers of Nonnegative Real Numbers §exponents and Properties of Real Powers of Nonnegative Real Numbers §agreement,

(fp)p=(I1/p)p=I(1/p)p=I1=I.\bigl(\lVert f\rVert_{p}\bigr)^{p}=\bigl(I^{1/p}\bigr)^{p}=I^{(1/p)\cdot p}=I^{1}=I .

Claim 2. The map cfcf is measurable by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. For every xXx\in X we have (cf)(x)=cf(x)|(cf)(x)|=|c|\,|f(x)| by claim 4 of Properties of the Absolute Value in an Ordered Field, so by Properties of Real Powers of Nonnegative Real Numbers §product,

((cf)(x))p=(cf(x))p=cp(f(x))p,\bigl(|(cf)(x)|\bigr)^{p}=\bigl(|c|\,|f(x)|\bigr)^{p}=|c|^{p}\bigl(|f(x)|\bigr)^{p},

that is cfp=cpfp|cf|^{p}=|c|^{p}\,|f|^{p} pointwise. The number cp|c|^{p} is a nonnegative real number by Properties of Real Powers of Nonnegative Real Numbers §values, so the homogeneity in claim 1 of Linearity and Monotonicity of the Lebesgue Integral gives Xcfpdμ=cpI\int_{X}|cf|^{p}\,d\mu=|c|^{p}I, which is finite; hence cfLpcf\in\mathcal{L}^{p}. Using Properties of Real Powers of Nonnegative Real Numbers §product and then Properties of Real Powers of Nonnegative Real Numbers §exponents and Properties of Real Powers of Nonnegative Real Numbers §agreement,

cfp=(cpI)1/p=(cp)1/pI1/p=cp(1/p)I1/p=cfp.\lVert cf\rVert_{p}=\bigl(|c|^{p}I\bigr)^{1/p}=\bigl(|c|^{p}\bigr)^{1/p}I^{1/p}=|c|^{p\cdot(1/p)}\,I^{1/p}=|c|\,\lVert f\rVert_{p}.

Claim 3. Let NN be the null set of points at which f(x)g(x)|f(x)|\le|g(x)| fails. For xNx\notin N, Properties of Real Powers of Nonnegative Real Numbers §monotone gives (f(x))p(g(x))p(|f(x)|)^{p}\le(|g(x)|)^{p}; hence fpgp|f|^{p}\le|g|^{p} almost everywhere. Both are measurable by Power-Integrable Functions and the p-Seminorm §measurable-power, so The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison yields

XfpdμXgpdμ<,\int_{X}|f|^{p}\,d\mu\le\int_{X}|g|^{p}\,d\mu<\infty ,

so fLpf\in\mathcal{L}^{p}. Applying Properties of Real Powers of Nonnegative Real Numbers §monotone with the positive exponent 1/p1/p to these two nonnegative real numbers gives fpgp\lVert f\rVert_{p}\le\lVert g\rVert_{p}.

Claim 4. The set where f|f| and g|g| differ is contained in the set where ff and gg differ, hence is null by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §null-union. So fg|f|\le|g| almost everywhere and gf|g|\le|f| almost everywhere. The first, with claim 3, gives gLpg\in\mathcal{L}^{p} once we know fLpf\in\mathcal{L}^{p}; more precisely claim 3 applied with the roles of ff and gg as stated there gives gLpg\in\mathcal{L}^{p} and gpfp\lVert g\rVert_{p}\le\lVert f\rVert_{p}, and applied in the opposite direction gives fpgp\lVert f\rVert_{p}\le\lVert g\rVert_{p}. Hence gp=fp\lVert g\rVert_{p}=\lVert f\rVert_{p}.

Claim 5. Write I=XfpdμI=\int_{X}|f|^{p}\,d\mu. By Properties of Real Powers of Nonnegative Real Numbers §values the power I1/pI^{1/p} vanishes exactly when II vanishes, so fp=0\lVert f\rVert_{p}=0 if and only if I=0I=0. By The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §vanishing, I=0I=0 if and only if (f(x))p=0(|f(x)|)^{p}=0 for almost every xx. By Properties of Real Powers of Nonnegative Real Numbers §values again, (f(x))p=0(|f(x)|)^{p}=0 holds exactly when f(x)=0|f(x)|=0, that is exactly when f(x)=0f(x)=0. Hence fp=0\lVert f\rVert_{p}=0 if and only if f=0f=0 almost everywhere.

Claim 6. Measurability of hr|h|^{r} is proved exactly as in Power-Integrable Functions and the p-Seminorm §measurable-power: the map h|h| is measurable with nonnegative values by claim 4 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions; for a real c<0c<0 the set {x:c<(h(x))r}\{x:c<(|h(x)|)^{r}\} is XX, and for 0c0\le c it equals {x:c1/r<h(x)}\{x:c^{1/r}<|h(x)|\} by Properties of Real Powers of Nonnegative Real Numbers §inverse, a member of F\mathcal{F}; the criterion in Measure Spaces and the Lebesgue Integral: Standing Notation §measurable applies.

The map hr|h|^{r} is nonnegative, so its absolute value is itself, and by Properties of Real Powers of Nonnegative Real Numbers §exponents, for every xXx\in X,

(hr(x))s=((h(x))r)s=(h(x))rs.\Bigl(\bigl|\,|h|^{r}(x)\bigr|\Bigr)^{s}=\Bigl(\bigl(|h(x)|\bigr)^{r}\Bigr)^{s}=\bigl(|h(x)|\bigr)^{rs}.

Hence the two functions hrs\bigl|\,|h|^{r}\bigr|^{s} and hrs|h|^{rs} are equal, so their integrals over XX coincide as members of [0,][0,\infty], and one is finite exactly when the other is. Since hr|h|^{r} and hh are both measurable, this says hrLs|h|^{r}\in\mathcal{L}^{s} if and only if hLrsh\in\mathcal{L}^{rs}.

Suppose this holds and write I=XhrsdμI=\int_{X}|h|^{rs}\,d\mu, a nonnegative real number. Then hrs=I1/s\bigl\lVert\,|h|^{r}\bigr\rVert_{s}=I^{1/s} by Power-Integrable Functions and the p-Seminorm §seminorm, while by Properties of Real Powers of Nonnegative Real Numbers §exponents

(hrs)r=(I1/(rs))r=Ir/(rs)=I1/s,\bigl(\lVert h\rVert_{rs}\bigr)^{r}=\bigl(I^{1/(rs)}\bigr)^{r}=I^{r/(rs)}=I^{1/s},

the exponents being equal because r1rs=1sr\cdot\frac{1}{rs}=\frac{1}{s}. The two sides therefore agree.

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