A convex function satisfiesIts value on a line segment never lies above the chord joining its endpoint values.

The finite-energy comparison $I_{\epsilon }(u_{\epsilon })\le I_{\epsilon }(x)<\infty$ gives $\epsilon \int |u_{\epsilon }^{\prime }{|}^{4/3}<\infty$. Together with the endpoint value and absolute continuity, this proves $u_{\epsilon }\in W^{1,4/3}(0,1)$.

For $\alpha >1/4$, the power competitor $u_{\alpha }=x^{\alpha }$ is admissible and has finite derivative energy. Direct integration givesChoose $\delta ={\epsilon }^{1/3}$ and $\alpha =1/4+\delta$. The first three terms tend to zero as $\alpha \downarrow 1/4$, while the last equals $\frac{3}{4}{\epsilon }^{2/3}(1/4+{\epsilon }^{1/3}{)}^{4/3}\to 0$. By minimality and nonnegativity,One also obtains the quantitative bound $I_{\epsilon }(u_{\epsilon })=O({\epsilon }^{2/3})$: for $\alpha =1/4+\delta$, the mean value theorem gives $|x^{4\alpha }-x|\le 4\delta x|\log x|$, and ${\int }_0^1{x}^2{\log }^{2}xdx=2/27$. Thus the squared mismatch is at most $32{\delta }^2/27$, matching the order of the derivative penalty with this choice. This is a power-law comparison for vanishing derivative regularization.

A convenient form of the Tonelli existence theorem is: on a finite interval with prescribed endpoint values, if a continuous integrand $F(x,z,v)$ is convex in $v$ and satisfies $F(x,z,v)\ge c|v{|}^p-C$ for some $p>1$, then, provided a finite-energy admissible competitor exists, its integral attains a minimum in the corresponding $W^{1,p}$ class. The lower bound may more generally subtract an integrable function of $x$. Compact one-dimensional Sobolev embedding controls $z$, while convexity gives weak lower semicontinuity in the derivative.

Here $p=4/3$ and $F_{\epsilon }=(z^4-x{)}^2+\epsilon |v{|}^{4/3}$ is continuous, convex in $v$, and bounded below by $\epsilon |v{|}^{4/3}$. The competitor $u(x)=x$ has finite energy. Thus for every $\epsilon >0$ the Tonelli existence theorem gives an admissible minimizer in $W^{1,4/3}(0,1)$.

Any finite-energy member of the larger stated $W^{1,1}$ class necessarily has derivative in $L^{4/3}$ and bounded function values by absolute continuity; it therefore belongs to $W^{1,4/3}$. All other competitors have infinite energy. Hence the same function $u_{\epsilon }$ is a global minimizer in the full stated class.

The function $u_{\ast }(x)=x^{1/4}$ has weak derivative $u_{\ast }^{\prime }(x)=\frac{1}{4}{x}^{-3/4}\in L^1(0,1)$ and the required endpoint values, so it belongs to the admissible Sobolev space. Its fourth power is $x$, and henceSince the integrand $(u^4-x{)}^2$ is nonnegative, zero is a global lower bound. Thus $u_{\ast }$ is a global minimizer. Its derivative does not belong to $L^{4/3}$, since its $4/3$ power is proportional to $x^{-1}$; positive regularization therefore needs different competitors.

Use $u_A(x)=A\sin (\pi x/b)$, which is admissible for every real $A$. Direct integration givesThe coefficient is strictly negative, so the energy tends to minus infinity as $|A|\to \infty$. Therefore

Fix an admissible $u$. The preceding result gives ${\int }_0^b{u}^{\prime 2}dx\ge {\lambda }^2{\int }_0^b{u}^{2}dx$ for every ${\lambda }^2<{\pi }^2/b^2$. Letting ${\lambda }^2$ increase to ${\pi }^2/b^2$ provesEquality is attained byIndeed the derivative and function squared integrals are $A^2{\pi }^2/(2b)$ and $A^{2}b/2$. These are all equality cases: an equality function minimizes the nonnegative threshold quadratic form, so its first variation vanishes. The resulting weak equation $u^{\prime \prime }+(\pi /b{)}^{2}u=0$ makes $u$ smooth and, with the endpoint conditions, yields precisely these multiples. This is the Wirtinger inequality.

Here $P=2$, $Q=-2{\lambda }^2$ and the Jacobi equation is $w^{\prime \prime }+{\lambda }^{2}w=0$. With $w(0)=0$ and $w^{\prime }(0)=1$, its solution is $w(x)=\sin (|\lambda |x)/|\lambda |$ for $\lambda \ne 0$, and $w(x)=x$ for $\lambda =0$. Its first positive zero is $\pi /|\lambda |$ when $\lambda \ne 0$.

If ${\lambda }^2<{\pi }^2/b^2$, no conjugate point occurs in $(0,b]$. The Jacobi positivity criterion givesSince $I_{\lambda }$ itself is this quadratic form divided by two, $I_{\lambda }(u)>0=I_{\lambda }(0)$ for every nonzero admissible $u$. Thus zero is the unique global minimizer. This argument establishes the result before using the sharp Poincaré inequality.

Zero is a strict strong local minimizer, and therefore also a strict weak local minimizer. Here $I(0)=-2/2010$, andFor $|u|\le 1$, Taylor's theorem gives $\cosh u-1\le \frac{1}{2}\cosh (1)u^2$. Also $u(0)=0$ and the Cauchy-Schwarz inequality give ${\int }_0^1{u}^{2}dx\le \frac{1}{2}{\int }_0^1{u}^{\prime 2}dx$. ThereforeThe coefficient is positive. Thus every nonzero admissible $u$ with $\Vert u{\Vert }_{\infty }\le 1$ has strictly greater energy, proving that zero is a strict strong local minimizer of a variational integral.

Zero is a strict weak local minimizer, but is not a strong local minimizer. Put $D={\int }_0^1|u^{\prime }{|}^{2}dx$. Since $u(0)=0$, the Cauchy-Schwarz inequality gives $|u(x){|}^2\le xD$, and thereforeFor sufficiently small $\Vert u^{\prime }{\Vert }_{\infty }$, $\sin (u^{\prime 2})\ge \frac{1}{2}{u}^{\prime 2}$ pointwise. Hencewith equality only when $u=0$. Thus zero is a strict weak local minimizer of a variational integral.

For the strong claim, define a Lipschitz triangular waveIt has zero endpoint values, $\Vert u_N{\Vert }_{\infty }=\sqrt{\pi }/(2N)\to 0$, and $u_N^{\prime }=\pm \sqrt{\pi }$ almost everywhere. Thus $\sin (u_N^{\prime 2})=\sin \pi =0$, while $u_N$ is nonzero on a set of positive measure. ConsequentlyThese uniformly small competitors disprove the strong local minimizer of a variational integral property.

A strong local minimizer of a variational integral is an admissible $\bar{u}$ for which some $\rho >0$ satisfiesOnly the function values must be close; admissible derivatives may be large. Every strong local minimizer is therefore also a weak local minimizer of a variational integral.

A point of convexity of $g\in C^1(\mathbb{R})$ is a point $v_0$ where its tangent line supports the entire graph from below:The Weierstrass necessary condition says that, at almost every $x$, ${\bar{u}}^{\prime }(x)$ is a point of convexity of $v\mapsto f(x,\bar{u}(x),v)$. In terms of the Weierstrass excess function,

Now use the stated derivative coercivity. If $\bar{u}$ were not a local minimizer in $L^1$, there would be $v_n\in \mathcal{A}$ with $\Vert v_n-\bar{u}{\Vert }_1\to 0$ and $I(v_n)<I(\bar{u})$. The energy bound givesThe common value $v_n(a)=\alpha$ controls their function norms as well, by $|v_n(x)|\le |\alpha |+{\int }_a^b|v_n^{\prime }|$. Thus $(v_n)$ is bounded in $W^{1,2010}(a,b)$. The one-dimensional compact Sobolev embedding into $C([a,b])$ supplies a uniformly convergent subsequence. Its limit must be $\bar{u}$, since the same subsequence converges to $\bar{u}$ in $L^1$. Eventually $\Vert v_n-\bar{u}{\Vert }_{\infty }<\rho$, contradicting strong local minimality. Hence $\bar{u}$ is also a local minimizer in $L^1$. This is coercivity transfers strong minimality to L1 minimality.

A weak local minimizer of a variational integral is an admissible $\bar{u}$ for which there is $\rho >0$ such thatThe Sobolev norm controls both the function and its first weak derivative. The boundary values remain fixed because both functions are admissible.

Zero is a strict weak local minimizer, but is not a strong local minimizer. The derivative integrand has expansion $\cosh v\cos (v^2)=1+v^2/2+O(v^4)$, so for small $|v|$ it is at least $1+3v^2/8$. Concavity gives $(1+u^2{)}^{1/4}\le 1+u^2/4$ for every real $u$. Therefore, when $\Vert u^{\prime }{\Vert }_{\infty }$ is small,using $\int u^2\le \frac{1}{2}\int u^{\prime 2}$. This proves strict weak minimality.

For the strong claim, let $v_N(x)=\sqrt{\pi }\mathrm{dist}(x,N^{-1}\mathbb{Z})$. Its amplitude tends to zero and its derivative is $\pm \sqrt{\pi }$ almost everywhere, so the derivative integrand is the constant $-\cosh \sqrt{\pi }$. The other integrand term is at most $-2$. HenceThis disproves strong minimality. The same obstruction is detected by the Weierstrass necessary condition, since the supporting-tangent inequality at slope zero fails at $v=\sqrt{\pi }$.

Zero is neither a weak nor a strong local minimizer. For any nonzero smooth zero-trace $\phi$, Taylor expansion givesThe mixed integral $\int \phi {\phi }^{\prime }=0$ by the endpoint values. Writing $D=\int {\phi }^{\prime 2}$ and $U=\int {\phi }^2$, the quadratic coefficient is $-3D/8+U/2$. The elementary estimate $U\le D/2$ on $[0,1]$ makes this at most $-D/8<0$. Thus arbitrarily small variations decrease the energy. The separate coefficient one on $u^2$ is retained in this calculation.

Zero is neither a weak nor a strong local minimizer. Take $u_{\epsilon }(x)=-\epsilon \sin (\pi x)$. It tends to zero in $W^{1,\infty }$ and has zero endpoint values. Since $\sqrt{1+t}\le 1+t/2$ for $t\ge 0$,for sufficiently small positive $\epsilon$. Thus even weakly close admissible competitors decrease the energy.

A strong local minimizer of a variational integral is an admissible $\bar{u}$ for which the same conclusion holds whenever $\Vert v-\bar{u}{\Vert }_{\infty }<\rho$. Every strong local minimizer is weak, because Sobolev closeness implies uniform closeness. A global minimizer is both, and the converse implication from weak to strong is not generally valid.

If the integrand is independent of $u$, a weak local minimizer has vanishing first variation:The bounded function $h(x)=f_v(x,{\bar{u}}^{\prime }(x))$ therefore has zero distributional derivative, so is constant almost everywhere. For an elementary test-function argument, any smooth compactly supported $\psi$ of integral zero is the derivative of a compactly supported smooth $\phi$. Hence $\int h\psi =0$; subtracting a fixed unit-integral test function from arbitrary tests shows that $h$ agrees with one constant against every test. The fundamental lemma then givesThis is the first integral for a variational integrand independent of the function value.

A point of convexity $v_0$ of a differentiable real function $g$ has a globally supporting tangent: $g(v)\ge g(v_0)+g^{\prime }(v_0)(v-v_0)$ for every $v$. The Weierstrass necessary condition for a strong minimum says that ${\bar{u}}^{\prime }(x)$ is such a point for $v\mapsto f(x,\bar{u}(x),v)$ for almost every $x$. Equivalently, the Weierstrass excess function is nonnegative at every trial slope.

A weak local minimizer of a variational integral is an admissible $\bar{u}$ for which some $\rho >0$ makes $I(v)\ge I(\bar{u})$ whenever $v\in \mathcal{A}$ and $\Vert v-\bar{u}{\Vert }_{W^{1,\infty }}<\rho$. Both the function and its derivative are required to be close.

Choose any nonzero admissible $u_0$ and set $u=\gamma u_0$. ThenThe coefficient of $-{\gamma }^6$ is strictly positive, so the energy tends to minus infinity as $|\gamma |\to \infty$. Hence

For every $0<\lambda <1$, positivity in the preceding part gives $\int u^{\prime 2}\ge {\lambda }^2\int u^2$ for each fixed admissible $u$. Let $\lambda \uparrow 1$ to obtainThe sharp constant is attained by $u(x)=\sin x$, consistently with the first conjugate point occurring at the endpoint for $\lambda =1$.

At zero, the sixth-power term has zero second derivative, soFor $0<\lambda <1$, the Jacobi solution $z(x)=\cos (\lambda (x-\pi /2))$ is strictly positive on $[0,\pi ]$. The preceding identity gives $Q_{\lambda }(u)=\int z^2[(u/z{)}^{\prime }{]}^2>0$ for nonzero zero-trace $u$.

To control the nonlinear remainder, a uniform lower bound follows directly. Let $m=\min z>0$, $K=\max |z^{\prime }|$, and $v=u/z$. Since $\max z=1$ and $\int v^2\le ({\pi }^2/2)\int v^{\prime 2}$ by the elementary endpoint estimate,Thus $Q_{\lambda }(u)\ge c_{\lambda }\int u^{\prime 2}$ with $c_{\lambda }=m^2/(K^2{\pi }^2+2)>0$. Also $\int u^6\le \Vert u{\Vert }_{\infty }^4({\pi }^2/2)\int u^{\prime 2}$, soThis is positive for every nonzero admissible $u$ sufficiently close to zero in $W^{1,\infty }$. Therefore zero has positive second variation and is a strict weak local minimizer. No sharp Poincaré inequality has been assumed in this proof.

If an admissible $u$ attained the infimum, nonnegativity and $I(u)=0$ would force $u^4=0$ almost everywhere, so its continuous Sobolev representative would be identically zero. Its derivative would then vanish almost everywhere, but the other integrand term would equal $16^4>0$. This is impossible. Equivalently, zero energy would simultaneously require $u=0$ and $|u^{\prime }|=2$ almost everywhere. Thus