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A strictly convex function satisfies that inequality strictly whenever $x\ne y$ and $0<t<1$.

For necessity of the strict supporting-tangent property, convexity gives a nonnegative gap along a segment from $z_0$ to $z$. If the gap at its other endpoint were zero, convexity of the gap and its zero initial value would force it to be zero all along the segment. That would make $F$ affine on this nontrivial segment, contrary to strict convexity. Conversely let $z=(1-t)x+ty$ with distinct endpoints. Apply the strict supporting inequality at $z$ to both endpoints, multiply by $1-t$ and $t$, and add. The linear terms cancel, yielding strict convexity. ThusFor strict Jensen's inequality, put $m=(b-a{)}^{-1}{\int }_a^{b}u$. The supporting gap $F(u)-F(m)-DF(m)\cdot (u-m)$ is nonnegative everywhere and strictly positive where $u\ne m$. That set has positive measure because $u$ is not almost everywhere constant. Its integral is therefore strictly positive. The linear term integrates to zero, giving the requested strict integral Jensen inequality.

In this special case the endpoints are both zero, overriding the previous strict ordering. The affine function is zero and $I(0)=1$. Since $0<f(v)\le 1$, every admissible energy is at most one, proving zero is a global maximizer.

For $u_N(x)=N\min \{x,1-x\}$, the endpoint values vanish and its slopes are $\pm N$ almost everywhere. HenceEvery admissible $u$ has essentially bounded derivative, so its integrand is bounded below by the positive constant $1/(1+\Vert u^{\prime }{\Vert }_{\infty }^{2008})$. Thus no admissible function has zero energy and there is no global minimizer.

For the unlabelled Euler-Lagrange assertion, use its integrated form $f^{\prime }(u^{\prime })=c$ almost everywhere. This follows from the first variation $\int f^{\prime }(u^{\prime }){\varphi }^{\prime }=0$ for all compactly supported tests. If $u\in C^1$, the left side is continuous and so equals $c$ everywhere. Its derivative values lie in the finite fiber $(f^{\prime }{)}^{-1}(c)$, while the continuous image of the interval under $u^{\prime }$ is connected. Thus $u^{\prime }$ is constant. Endpoint values fix it to $m=(\beta -\alpha )/(b-a)$, giving exactly the affine $\bar{u}$. Conversely that affine function has constant $f^{\prime }({\bar{u}}^{\prime })$, so is a solution. This proves uniqueness among $C^1$ solutions; the scalar integrand is the one-dimensional case of the notation in the source.

If $f$ is strictly convex, Jensen's inequality applied to $u^{\prime }$ givesEquality in the strict inequality can occur only when $u^{\prime }$ is almost everywhere constant. Absolute continuity and the endpoints then give $u=\bar{u}$. Hence the affine function is the unique global minimizer.

The monotone convergence theorem states that if $f_n$ are measurable, nonnegative functions on a measure space with $f_n\le f_{n+1}$ almost everywhere and $f_n\uparrow f$ almost everywhere, thenThe equality is valid in $[0,\infty ]$, so finiteness is not required. Changes on null sets do not affect the conclusion.

Let $S_N={\sum }_{k=0}^N{a}_k$. These are nonnegative measurable functions increasing to the extended-valued sum $S$. Monotone convergence givesA nonnegative function with finite integral cannot be infinite on a set of positive measure. Hence $S(x)<\infty$ almost everywhere, which is exactly almost-everywhere convergence of the nonnegative series. One may redefine the sum arbitrarily on the exceptional null set to obtain an ordinary integrable representative.

On the product of two sigma-finite measure spaces, Tonelli's theorem states that a nonnegative measurable function has measurable iterated-integral functions andwith values allowed to be infinite. Fubini's theorem states that if $f$ is measurable and ${\int }_{X\times Y}|f|<\infty$, then its sections are integrable almost everywhere, both iterated integrals exist as finite integrals, and the same equality holds, also for real or complex signed functions. Tonelli applied to $|f|$ can first be used to verify the absolute-integrability hypothesis of Fubini.

Interpret the weighted function space as measurable functions modulo equality almost everywhere, so the integral form is a genuine inner product. Assume $q\in {\Pi }_k\subseteq V$. If the residual is orthogonal to ${\Pi }_k$, then for $p\in {\Pi }_k$,Thus $q$ is a best approximation. Conversely, if it is best, take any $r\in {\Pi }_k$ and vary through $q+tr$. The polynomialhas a minimum at $t=0$, so its derivative there vanishes. Hence $\boxed{⟨f-q,r⟩=0}$ for every $r\in {\Pi }_k$. The same Pythagorean identity proves uniqueness of the best approximation.

For product measure spaces, Tonelli's theorem states that for a nonnegative measurable function the product-space integral equals either iterated integral, allowing the value $+\infty$. Fubini's theorem states that if a measurable function has finite integral of its absolute value, then its sections are integrable almost everywhere, both iterated integrals exist and equal the product-space integral. The absolute-integrability hypothesis is what permits this conclusion for a signed function. These statements apply to the Lebesgue product spaces used here, which are sigma-finite.

The monotone convergence theorem states that if nonnegative measurable functions $h_n$ increase almost everywhere to $h$, then $\int h={\lim }_n\int h_n$, allowing infinity. The dominated convergence theorem states that if measurable $h_n\to h$ almost everywhere and $|h_n|\le g$ for an integrable $g$, then $h$ is integrable and $\int h_n\to \int h$.

Apply monotone convergence to $A_n={\sum }_{k=1}^n|a_k|$. Its limit $A={\sum }_{k\ge 1}|a_k|$ has integral ${\sum }_k\int |a_k|<\infty$, so $A<\infty$ almost everywhere. Thus the original series converges absolutely almost everywhere. Its partial sums are bounded in absolute value by the integrable function $A$, and their limit $s$ satisfies $|s|\le A$. Dominated convergence givesAssigning any value on the null exceptional set defines an integrable representative of $s$.

False. Let $k$ be the indicator of the positive integers. It is measurable, bounded by one and zero almost everywhere, so its Lebesgue integral on $(0,\infty )$ is zero. Yet $k(n)=1$ for every positive integer $n$, so it does not tend to zero at infinity. Lebesgue integrability is unchanged by modifications on a null set, whereas a pointwise limit can change.

Use the nonnegative parts $h^{+}=\max (h,0)$ and $h^{-}=\max (-h,0)$, so $h=h^{+}-h^{-}$. The zero-mean assumption gives $H:={\int }_0^{\pi }{h}^{+}={\int }_0^{\pi }{h}^{-}<\infty$. On the $k$-th period, each of the positive and negative weighted integrals lies between $H/((k+1)\pi )$ and $H/(k\pi )$. Their difference therefore satisfies the stronger estimateThe sum of the absolute values of these block integrals converges, proving the existence of the finite limit of the integrals from $\pi$ to $n\pi$.

For genuine Lebesgue integrability we instead need the integral of the absolute value. Periodicity givesIf $H>0$, summation diverges by the harmonic series. If $H=0$, then $h=0$ almost everywhere and the function is integrable. ThusThe finite signed block limit therefore need not be a Lebesgue integral.

The partial sums $S_n={\sum }_{k=1}^n{a}_k$ are nonnegative measurable functions and increase pointwise to the extended-valued sum $S={\sum }_{k=1}^{\infty }{a}_k$. The monotone convergence theorem and finite linearity giveA nonnegative measurable function with finite integral is finite almost everywhere; if it were infinite on a positive-measure set, its integral would be infinite. Define it arbitrarily on the null exceptional set if a real-valued representative is desired. Hence the series converges almost everywhere to an integrable function and

The absolute continuity of measures condition isThe Radon-Nikodym theorem says that if both positive measures are sigma-finite measures, then this condition is equivalent to the existence of a nonnegative measurable function $h$, unique up to equality $\kappa$-almost everywhere, such thatOne writes $h=d\nu /d\kappa$, the Radon-Nikodym derivative. Under these hypotheses $h$ is finite $\kappa$-almost everywhere. If $\nu$ is finite, then $h\in L^1(\kappa )$. A finite signed measure absolutely continuous with respect to a sigma-finite positive measure likewise has an integrable signed Radon-Nikodym derivative.

The sigma-finiteness hypothesis on the dominating measure cannot simply be omitted. On the Borel subsets of $[0,1]$, let $\kappa$ be counting measure and $\nu$ Lebesgue measure. Absolute continuity holds because the only $\kappa$-null set is empty. A Radon-Nikodym derivative would satisfy $h(x)=\nu (\{x\})=0$ at every point, contradicting $\nu ([0,1])=1$. Here counting measure is not sigma-finite.

Use normalized Haar measure $dm=dt/(2\pi )$. A homogeneous Banach space on the circle is a Banach subspace $B\subseteq L^1(\mathbb{T})$ with $\Vert f{\Vert }_1\le \Vert f{\Vert }_B$, such that translations ${\tau }_{s}f(t)=f(t-s)$ preserve $B$, satisfy $\Vert {\tau }_{s}f{\Vert }_B=\Vert f{\Vert }_B$, and obey $\Vert {\tau }_{s}f-f{\Vert }_B\to 0$ as $s\to 0$.

For continuous $g$, the map $s\mapsto g(s){\tau }_{s}f$ is continuous with values in the complete space $B$. Define its Banach-valued Riemann integral, equivalently its Bochner integral, byThe triangle inequality for Riemann sums and passage to their norm limit give the canonical estimateThe continuous inclusion into $L^1$ carries the same Riemann sums to the ordinary convolution, so this definition agrees almost everywhere with $\int f(t-s)g(s)dm(s)$. Density of continuous functions in $L^1$, completeness of $B$, and the estimate extend the operation uniquely to all $g\in L^1$. The ordinary $L^1$ convolution estimate identifies that extension with the same integral formula.

Let $Df(x)$ denote the upper limit of the displayed mean oscillation as $r\downarrow 0$. For a continuous integrable approximation $g$, continuity gives $Dg(x)=0$. With $h=f-g$, the triangle inequality yieldsTherefore the maximal inequality and Chebyshev's inequality giveDensity makes the right side arbitrarily small. For each positive rational $\epsilon$ the exceptional set is null, proving $Df=0$ almost everywhere, the Lebesgue differentiation theorem. Values where the chosen representative is not finite already form a null set.

For the fundamental theorem define $F(x)={\int }_a^{x}f(t)dt$, with oriented integration for $x<a$. This function is absolutely continuous: for disjoint intervals the sum of $|F(b_j)-F(a_j)|$ is at most the integral of $|f|$ over their union, which is small when its measure is small by absolute continuity of the integral. At a Lebesgue point of $f$, for $h\ne 0$,Thus $F^{\prime }=f$ almost everywhere and $F(b)-F(a)={\int }_a^{b}f$. Differentiability at every point is not claimed for an arbitrary integrable function.

A distribution has period $T>0$ if the displayed equality holds for every test function $\varphi$. This agrees with $f(x+T)=f(x)$ almost everywhere for a regular distribution. If $u$ is a tempered distribution, density of test functions in the Schwartz space extends the identity to all Schwartz functions.

For a test function $\Psi$ with ${\sum }_k\Psi (x+kT)=1$, one hasIndeed, expand $\varphi ={\sum }_k\Psi (x+kT)\varphi (x)$ in the Schwartz space and move each translation to $\varphi$ by periodicity. All Schwartz seminorms of the localized terms decay faster than any power of $|k|$, justifying the sum of pairings. This localization permits rigorous Fourier coefficients without multiplying a distribution by a discontinuous window.

On continuous functions on a compact interval, evaluation at an interior point is not continuous in the integral norm. Narrow triangular peaks have fixed point value and integral tending to zero. Subtracting such a peak from the constant function one gives functions vanishing at the point and converging in the integral norm to one. Thus the evaluation kernel is not closed, although it is closed in the uniform norm.

Here is a fully computed endpoint version, now using the L2 norm on the actual continuous functions $C([0,1])$. For each integer $n\ge 1$, set $f_n(x)=\max (1-nx,0)$. Its point evaluation at zero is $E(f_n)=f_n(0)=1$, whileConsequently $f_n\to 0$ in the L2 norm but $E(f_n)\nrightarrow 0$: point evaluation is discontinuous on this normed space. Quantitatively $|E(f_n)|/\Vert f_n{\Vert }_2=\sqrt{3n}$ is unbounded. At $n=100$, the L2 norm is approximately $0.057735$, the integral is $0.005$, and the point value is still $1$. In the supremum norm, $\Vert f_n{\Vert }_{\infty }=1$ and $|E(f)|\le \Vert f{\Vert }_{\infty }$, so $E$ has operator norm one instead.

Set $g_n=1-f_n$. Then $E(g_n)=0$ and $\Vert g_n-1{\Vert }_2=1/\sqrt{3n}\to 0$, although $E(1)=1$. Thus the kernel of $E$ fails to be a closed set. On the full L2 space, a further distinction is essential: its elements are equivalence classes under equality almost everywhere, and changing a representative at zero does not change its class. Arbitrary point evaluation is therefore not even well-defined there. Restricting first to continuous functions makes the discontinuity statement precise.^[ref]

A probability density function represents probabilities as areas under a nonnegative measurable function. For a real random variable with an absolutely continuous probability distribution, it satisfiesfor every Borel set $A$. Its height is probability per unit of the outcome variable, not the probability of that exact outcome. Thus a probability density function may exceed one, while the probability of a singleton is zero. NIST describes the interval-probability interpretation through integration. ^[ref].

For the example $f_Y(y)=2y$ on $[0,1]$, $f_Y(0.8)=1.6$, but $\mathbb{P}(Y=0.8)=0$ and $\mathbb{P}(0.7<Y<0.9)=0.9^2-0.7^2=0.32$. The cumulative distribution function is $F_Y(y)=y^2$ within that interval, connecting density height to the local accumulation rate.

Changing the scale changes the density height. If $Z=100Y$, then $f_Z(z)=f_Y(z/100)/100=z/5000$ on $[0,100]$. At the corresponding value $z=80$, its height is $0.016$, while $\mathbb{P}(70<Z<90)=0.32$ is unchanged. The factor $1/100$ compensates for the wider integration variable. This is a change of variables, not a change in the underlying probabilities.

A probability density function is determined only almost everywhere: changing its values on a Lebesgue measure zero set changes no probabilities. The Cantor distribution has no such representation, and a mixed probability distribution can have both a point mass and a density component, so differentiating its cumulative distribution function does not recover all its mass.

Away from the origin, the radial derivative is the directional derivative along the outward radial unit vector. For a Sobolev space function it is the same scalar product with the weak gradient, defined almost everywhere. On an annulus, it has no singular direction at the origin.