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Suppose $h$ is bounded by one in absolute value on a punctured neighbourhood of zero. Suppose also that there are positive sequences $p_n,q_n\to 0$ with $h(p_n)=1$ and $h(q_n)=0$. Extend $F_a(x)=|x{|}^{a}h(x)$ by $F_a(0)=0$. ThenWhen the derivative exists, it equals zero. For continuity, the bound $|F_a(x)|\le |x{|}^a$ gives the epsilon-delta definition of continuity with $\delta ={\epsilon }^{1/a}$ when $a>0$. If $a\le 0$, the values $F_a(p_n)=p_n^a$ remain at least one for sufficiently large $n$, contradicting continuity at zero.

For differentiability, the difference quotient is bounded in absolute value by $|x{|}^{a-1}$, so it tends to zero when $a>1$. Conversely its values along $q_n$ are zero, forcing any possible derivative to be zero. Its values along $p_n$ equal $p_n^{a-1}$ and fail to tend to zero when $a\le 1$.

For example, $h(x)=\sin (\pi \sin (1/x))$ has the required positive sequences $p_n=1/(2\pi n+\pi /6)$ and $q_n=1/(2\pi n)$. This separates the suppression needed for continuity from the stronger suppression needed for differentiability.

For a smooth function $f:U\to \mathbb{R}$ on an open set $U\subseteq {\mathbb{R}}^n$, a critical point is an input $x_{\ast }$ where the gradient vanishes: $\nabla f(x_{\ast })=0$. Equivalently, every first partial derivative is zero. Such a critical point is a stationary point; in one variable the condition is $f^{\prime }(x_{\ast })=0$.

There is a broader convention in elementary one-variable calculus: a "critical number" is an interior input where the derivative is zero or does not exist, provided the function itself is defined there. Thus a corner can be a critical number without being a stationary point. Here the smooth meaning of critical point is used unless the broader convention is explicitly stated. ^[ref]

For the polynomial $f(x)=x^4-2x^2$ on the real line,The stationary points at $x=\pm 1$ are strict local minima and attain the global minimum; the stationary point at $x=0$ is a strict local maximum. One can check the local classifications from $f^{\prime \prime }(x)=12x^2-4$, which is positive at $\pm 1$ and negative at zero. The identity $f(x)=(x^2-1{)}^2-1\ge -1$ proves the global minimum directly. Since $f(x)\to +\infty$ as $|x|\to \infty$, there is no global maximum on $\mathbb{R}$.

A zero derivative alone does not guarantee an extremum. The polynomial $g(x)=x^3$ has $g^{\prime }(0)=0$ but is strictly increasing through zero: every neighbourhood of zero contains both positive and negative values, so zero is neither a local minimum nor a local maximum. Conversely, the absolute value function $h(x)=|x|$ has a strict local minimum and a unique global minimum at zero, but its left and right derivatives are $-1$ and $+1$. It is not differentiable there, so zero is a critical number under the broader convention and is not a stationary point.

The usefulness of critical points is that they reduce the search for smooth interior extrema to equations. Every local minimum or local maximum of a differentiable function at an interior point has zero gradient; this condition is necessary, and $x^3$ shows that it is not sufficient. The striking feature is that all the first directional derivatives must vanish together, even though the surrounding values can form a valley, a peak, or a saddle point of a scalar function. ^[ref]

Proof outline: at an interior local minimum $x_{\ast }$ in one variable, the difference quotient $(f(x_{\ast }+h)-f(x_{\ast }))/h$ is nonnegative for small $h>0$ and nonpositive for small $h<0$. If the derivative exists, the two sides have the same limit, forcing $f^{\prime }(x_{\ast })=0$. Reverse the inequalities for a local maximum. In several variables, apply this argument along each coordinate line to obtain zero partial derivatives and hence zero gradient. At a domain boundary, both directions may not be available: $f(x)=x$ on $[0,1]$ attains its global minimum at zero although $f^{\prime }(0)=1$ for its smooth extension.

The derivative measures the instantaneous rate at which a function changes. For a real function $f$ defined near $a$, it exists when the difference quotient has a finite two-sided limit:The quotient is the slope of the straight line joining two nearby points of the graph. Its limit is the slope of the tangent line. Thus a derivative describes a local rate without choosing a particular finite measurement interval.

For $f(x)=x^2$, expanding the numerator givesAt $a=1$, the increments $h=1$, $1/2$ and $1/10$ give secant line slopes $3$, $2.5$ and $2.1$. The limit is $2$, and the tangent line is $y=1+2(x-1)$. The picture makes the limiting process visible.

The same calculation shows why the derivative is useful for approximation:This is a local linear approximation. Near $a=1$, replacing $(1+h{)}^2$ by $1+2h$ makes the exact error $h^2$, smaller than the leading change when $h$ is small. A curved graph can therefore be studied locally using a straight line.

If the function records position against time, its derivative is velocity. If it records an objective function, its derivative tests local changes that matter in optimization; in several variables, the gradient collects these directional rates and guides gradient descent. The interpretation comes from the function and its units, rather than from a new rule for each application.

Existence matters. For the absolute value function $f(x)=|x|$, the one-sided derivatives at zero are $-1$ and $1$, so there is no single tangent line described by a finite derivative there. Nor does a zero derivative alone identify a maximum or minimum: $f(x)=x^3$ has $f^{\prime }(0)=0$ but keeps increasing through zero.

Let $I$ be a real interval, and let $f:I\to \mathbb{R}$ be differentiable in its interior and continuous at any endpoints belonging to $I$. If the derivative satisfies $f^{\prime }(x)\ge 0$ throughout the interior, then $f$ is a nondecreasing function on $I$. For $x<y$ in $I$, the mean value theorem supplies $c\in (x,y)$ withChanging $\ge 0$ to $\le 0$, $>0$, or $<0$ gives a nonincreasing function, strictly increasing function, or strictly decreasing function, respectively. The endpoint continuity assumption lets the mean value theorem apply even when $x$ or $y$ is an endpoint. ^[ref].

Conversely, if a nondecreasing function on a real interval is differentiable at an interior point $x$, then $f^{\prime }(x)\ge 0$: the difference quotient $(f(x+h)-f(x))/h$ is nonnegative for either sign of $h$. A differentiable nonincreasing function has derivative $f^{\prime }(x)\le 0$. Strict monotonicity still only forces these weak signs: the strictly increasing function $x^3$ has derivative zero at zero. The sign of the derivative helps locate intervals of monotonicity when sketching a graph. ^[ref].

The real interval hypothesis matters because a connected subset of the real line contains every point between its inputs. On a disconnected domain of a function, positive derivatives need not give a nondecreasing function. For example, takeHere $f^{\prime }(x)=1$ everywhere in $D$, but $-3/2<3/2$ while $f(-3/2)=-3/2>-17/2=f(3/2)$. Each component has a strictly increasing function, yet their outputs are ordered the wrong way across the gap. This function is neither a nondecreasing function nor a nonincreasing function on all of $D$; the mean value theorem cannot cross the missing interval.

Suppose $h$ is continuously differentiable on $[0,L]$. For $0<x\le L$, the improper integral $g(x)={\int }_0^{x}h(t)(x-t{)}^{-1/2}dt$ exists, but substituting directly into the ordinary Leibniz integral rule produces an undefined endpoint term. First use integration by parts to obtainThe new endpoint term at $t=x$ vanishes, and the differentiated kernel is locally integrable. Differentiating gives the displayed formula. One rigorous justification rewrites the second term as ${\int }_0^x[{\int }_0^s{h}^{\prime }(t)(s-t{)}^{-1/2}dt]ds$ by Fubini's theorem; the inner integral is continuous for $s>0$, so the fundamental theorem of calculus applies. Moreover $g(x)=2h(0)\sqrt{x}+O(x^{3/2})$: if $h(0)\ne 0$, the right difference quotient at zero diverges rather than defining a finite derivative.

Extend the displayed function by $f(0)=0$. Its difference quotient at zero is $h\sin (1/h)$, which tends to zero. Away from zero its derivative is $2x\sin (1/x)-\cos (1/x)$. At $x_k=1/(2\pi k)$ this derivative equals $-1$, whereas at $y_k=1/((2k+1)\pi )$ it equals $+1$. Thus differentiability implies continuity of the function but does not imply continuity of its derivative.

If a real function has a finite derivative at $x$, its difference quotient tends to $f^{\prime }(x)$ and is bounded for sufficiently small nonzero increments. Multiplying it by the increment shows $f(x+h)-f(x)\to 0$, proving continuity at $x$. The converse fails: the absolute value function is continuous but not differentiable at zero.

Differentiation computes the derivative of a function. A difference quotient approximates a derivative using finite increments; its limit is the derivative when the function is differentiable.

The directional derivative tests a function along a chosen straight line, including lines that follow neither coordinate axis. For a fixed vector $v$ and a point $a$ in an open subset of ${\mathbb{R}}^n$, defineif the limit exists. With a unit vector $v$, the parameter $t$ measures signed distance in the input Euclidean space, so the directional derivative is change per unit distance. For a general $v$, it is change per unit parameter; the norm of $v$ also sets the rate of travel.^[ref]

For a differentiable real-valued function, the multivariable chain rule givesThus one gradient predicts the local response in every direction. This formula requires differentiability, rather than merely separate partial derivatives.^[ref]

Return to $f(x,y)=x^2+xy+2y^2$ at $a=(1,1)$. Its gradient is $(3,5)$. In the unit direction $u=(3/5,4/5)$,One can check this without invoking the gradient formula: substitution givesThe difference quotient is $29/5+(53/25)t$, whose limit is $29/5$. This calculation separates the exact change along a finite path from its first-order rate at the starting point.

Changing the travel speed changes the parameter rate. Taking $v=(3,4)=5u$ gives $D_{v}f(a)=29$, because the input travels five distance units per parameter unit. For a differentiable function, $D_{cv}f(a)=c{D}_{v}f(a)$ for any real $c$: negative $c$ reverses the orientation as well as scaling the rate. Normalizing $v$ before comparing different slopes avoids confusing direction with speed.

The Cauchy-Schwarz inequality bounds the rate in any unit direction by $\Vert \nabla f(a)\Vert$. Here the largest rate is $\sqrt{34}\approx 5.831$, attained in the direction $(3,5)/\sqrt{34}$; the chosen $(3/5,4/5)$ gives the slightly smaller rate $5.8$. The perpendicular unit direction $(-5,3)/\sqrt{34}$ gives zero. That means no first-order height change, rather than no change at all: along this direction the example has $f(a+tu)=4+(28/34)t^2$.

This is useful for following a height or temperature scalar field, choosing a local search direction in mathematical optimization, and distinguishing a coordinate sensitivity from a response along an actual path. For a curved path, derivative along a curve uses its instantaneous tangent vector. Even existence of all directional derivatives does not guarantee one common linear approximation; directional derivatives do not imply differentiability gives a concrete counterexample.

With ${\delta }_{h}f(x)=[f(x+h{e}_{\ell })-f(x)]/h$, a change of variables givesThe compact support of $f$ and its translates must remain inside the integration domain, or one must integrate over the whole space with convergent integrals. In particular ${\delta }_{-h}g(x)=[g(x)-g(x-h{e}_{\ell })]/h$: the denominator's sign matters. This is the difference quotient analogue of integration by parts.

For $n=1$, differentiability at $a$ means that the difference quotient has a finite two-sided limit of a function. For $n\ge 2$, require derivatives through order $n-1$ throughout some open neighbourhood of $a$, and require the final difference quotientto exist as a finite real number. The last derivative need not exist elsewhere or be continuous at $a$. In particular, existence of $f^{\prime }(a)$ by itself does not justify taking a second derivative: the function $f^{\prime }$ must first be defined in a neighbourhood of $a$.

LetThe Fourier coefficients of $\theta$ are absolutely summable, so $a$ belongs to the Wiener algebra, as does $a^{-1}$. The identity $\theta (2t)=\theta (t)+\theta (t+\pi )$ implies $a(2t)=a(t)a(t+\pi )$. Starting with a compactly supported Daubechies wavelet having $p\ge 3$ vanishing moments, the periodic phase change of a scaling function gives $m(t)=a(t+\pi )m_0(t)$, while the canonical high-pass construction leaves $\hat{\psi }(2t)=e^{-it}\overline{m_0(t+\pi )}{\hat{\phi }}_0(t)$ unchanged.

At a dyadic point $t_0=2\pi k/2^j$, the terms of the difference quotient with $n\ge j$ and $2^n|h|\le 1$ each contribute $1+O((2^{n}h{)}^2)$. Their number tends to infinity, their total error is bounded, and the remaining tail contributes a bounded amount. Hence $(\theta (t_0+h)-\theta (t_0))/h={\log }_2(1/|h|)+O(1)$, so neither $\theta$ nor $a$ has a finite derivative there. These points are dense. Away from the isolated zero of $m_0$ near $\pi$, multiplication by its nonzero smooth value cannot remove this nondifferentiability. Thus the new MRA low-pass filter is not differentiable throughout any neighborhood of $\pi$, despite integrability of the scaling function and unchanged vanishing moments. Ordinary higher derivatives at $\pi$ cannot be inferred, although the corresponding Peano zero survives.

The right and left one-sided derivatives are the limits of the difference quotient restricted respectively to positive and negative increments:At an interior point, differentiability is equivalent to existence and equality of these two finite limits. For the absolute value function, $f_{+}^{\prime }(0)=1$ and $f_{-}^{\prime }(0)=-1$, so it is continuous at zero but not differentiable there.

A partial derivative measures the local response of a function to one input while its other inputs stay fixed. For a function $f$ on an open subset of ${\mathbb{R}}^2$, its partial derivatives at $(a,b)$ are the limitswhen the respective limits exist. Each is an ordinary derivative of a one-variable slice: $x\mapsto f(x,b)$ or $y\mapsto f(a,y)$. This is why the other variable is treated as a constant during partial differentiation.^[ref]

For an original example, take the polynomial functionHolding $y$ fixed gives $f_x=2x+y$, while holding $x$ fixed gives $f_y=x+4y$. At $(1,1)$, the height is $f=4$ and the two partial derivatives are $f_x=3$ and $f_y=5$. The values answer different experiments on the same mathematical model: change $x$ alone, or change $y$ alone.

The two exact slices through that point areDividing either height change by $t\ne 0$ gives $3+t$ or $5+2t$; taking the limit $t\to 0$ recovers the two slopes. At $t=0.1$, the finite height changes are $0.31$ and $0.52$, although their linear approximations are $0.3$ and $0.5$. A derivative measures the limiting response, rather than an exact finite change.

For a function of $n$ variables, ${\partial }_{j}f(a)$ varies only its $j$th coordinate: it is the same difference quotient along $e_j$, the $j$th standard basis vector. This gives a practical sensitivity analysis: one can compare how a mathematical model responds to changes in its individual inputs, provided the chosen system of units and the quantities held fixed are stated. Comparing the numerical sizes of two partial derivatives in different units does not by itself identify a physically preferable input.

Existence of the partial derivatives at one point is weaker than differentiability. The example under partial derivatives do not imply continuity shows how both coordinate tests can miss a bad diagonal approach. When the first partial derivatives are continuous in a neighborhood, they do give differentiability there, allowing the total differential to combine all small input changes.

A real function is differentiable at an interior point $x$ when the finite limitexists, with $h$ restricted so $x+h$ lies in its domain. In quantified form, for every $\epsilon >0$ there is $\delta >0$ such that $0<|h|<\delta$ implies that the difference quotient differs from $f^{\prime }(x)$ by less than $\epsilon$.

To prove differentiability implies continuity, the convergent difference quotient is bounded by $|f^{\prime }(x)|+1$ for sufficiently small nonzero $h$. Thus $|f(x+h)-f(x)|\le (|f^{\prime }(x)|+1)|h|\to 0$. This is exactly continuity at $x$.

At zero, the particular function has difference quotient $h\sin (1/h)$. The bound $|h\sin (1/h)|\le |h|$ and the squeeze theorem giveFor $x\ne 0$, the product rule and chain rule giveAt $x_k=1/(2\pi k)$ this equals $-1$, although $x_k\to 0$ and $f^{\prime }(0)=0$. At $y_k=1/((2k+1)\pi )$ it equals $+1$. Hence the derivative has no limit at zero and is not continuous there. This demonstrates that differentiability does not imply continuity of the derivative, even though the function itself is continuous.

For $r>0$, define the forward and backward one-sided interval averaging operatorsand set $A_0^{\pm }f=f$. Each positive-radius average is bounded in absolute value by $m_u(f)(x)$. Although $x$ is an endpoint of its averaging interval, enlarge that interval slightly so that $x$ is interior, apply the maximal bound and let the enlargement shrink to zero.

To dominate the limiting identity too without assuming the differentiation result we are proving, use the sublinear operatorThe maximal estimate and Chebyshev inequality implyThus $S$ has weak type (1,1) and dominates every $A_r^{\pm }$, including $r=0$.

Smooth compactly supported functions are dense in $L^1$, obtained by truncating and then smoothing with a compactly supported approximate identity. For each such continuous function both averages tend to its value at every point. Applying part (ii) to the two families, and intersecting their full-measure convergence sets, gives $A_r^{\pm }f(x)\to f(x)$ almost everywhere.

The forward difference quotient of the primitive is $A_r^{+}f(x)$, and the backward quotient is $A_r^{-}f(x)$:They have the same limit, proving differentiability almost everywhere, with the stronger conclusionThis proves differentiation of an indefinite Lebesgue integral directly from the maximal estimate and density, without assuming $m_u(f)\ge |f|$ as an unproved prerequisite.

For smooth $u$, the fundamental theorem along the segment in direction $e_k$ gives, for either sign of $h$,The entire segment lies in $\Omega$ under the distance hypothesis. The Cauchy-Schwarz inequality on $[0,1]$, followed by integration and translation, yieldsFor general $u\in W^{1,2}(\Omega )$, extend $u$ and its weak gradient separately by zero as $L^2$ functions on ${\mathbb{R}}^n$, and mollify $u$ with radius $\epsilon$. At points farther than $\epsilon$ from the boundary the mollified gradient equals the convolution of the zero-extended gradient, because the convolution kernel is supported wholly inside $\Omega$. Choose $\epsilon <\mathrm{dist}({\Omega }^{\prime },\partial \Omega )-|h|$. All the segments then lie in this region, and the preceding calculation, followed by the $L^2$ contraction bound for convolution, gives the same right side $\Vert D_{k}u{\Vert }_{L^2(\Omega )}$. The mollifications converge locally in $L^2$, as do their fixed-$h$ translated differences, so passage to the limit preserves the estimate. HenceThe argument proves the sharper estimate with $D_{k}u$ on the right. It is the basic Sobolev difference quotient bound.

We prove the Interior second-derivative estimate for the Poisson equation using difference quotients, rather than differentiating the $L^2$ forcing. Fix a coordinate $e_k$ and write ${\delta }_{h}v(x)=(v(x+h{e}_k)-v(x))/h$. Choose a smooth cutoff function $\eta$ equal to one on ${\Omega }^{\prime }$ with support compactly contained in $\Omega$, leaving room for all translations with sufficiently small $|h|$. We can arrange $|D\eta |\le C(n)/d$, where $d=\mathrm{dist}({\Omega }^{\prime },\partial \Omega )$.

The standard difference quotient lemmas give discrete integration by parts, $\int a{\delta }_{-h}b=-\int ({\delta }_{h}a)b$, and $\Vert {\delta }_{h}v{\Vert }_2\le \Vert D_{k}v{\Vert }_2$ on the relevant enlarged set. In the weak formulation $\int Du\cdot D\phi =-\int f\phi$, use the admissible test function $\phi =-{\delta }_{-h}({\eta }^2{\delta }_{h}u)$. Moving a difference quotient onto $Du$ givesSet $X=\Vert \eta D{\delta }_{h}u{\Vert }_2$ and $Y=\Vert D\eta {\delta }_{h}u{\Vert }_2$. The cross term has absolute value at most $2XY$. Since $0\le \eta \le 1$, the difference quotient lemma on the right givesThe Young inequality now implies $X^2\le C(Y^2+\Vert f{\Vert }_2^2)$. Moreover $Y\le C(n)d^{-1}\Vert Du{\Vert }_{L^2(\Omega )}$, independently of $h$. Hence, for every coordinate $k$,By the Sobolev characterization by bounded difference quotients, applied to each first weak derivative on smaller interior sets, these uniform bounds give all second weak derivatives. Their bounds pass to the weak limit as $h\to 0$. Thus $u\in W_{\mathrm{loc}}^{2,2}(\Omega )$ andNo derivative of $f$ is used, and no boundary condition is needed for this interior elliptic regularity argument.

Since $f(0)=0$, its difference quotient isThe absolute value of cosine is at most $1$, so this quotient lies between $0$ and $|h|$. The squeeze theorem proves

Use the usual fractional Dirichlet domain scale: $A=-{\Delta }_D$ on $H=L^2(\Omega )$, and $H_{\rho }=D(A^{\rho })$. Its Sobolev embedding theorem givesSuch a positive $\beta$ exists precisely because the assumed $\delta$ is greater than $n/4$. Consequently $v=u_t$ is continuous in time with values in this Hölder space on every compact positive-time interval. This embedding alone gives no two spatial derivatives for $v$; those must come from the equation.

First establish a bound that will also justify the nonlinear regularity steps. Put $F(s)=\lambda s-s^3$ andThe constants $K$ and $-K$ enclose both the initial data and the zero Dirichlet boundary data. Moreover $F(K)\le 0$ and $F(-K)\ge 0$, so they are a supersolution and a subsolution for the parabolic comparison principle. A direct energy proof avoids assuming pointwise continuity at time zero. Let $h=(u-K{)}_{+}$. This truncation has zero trace under the Sobolev trace operator; testing the evolution with $h$ and using integration by parts givesFor $u>K$, the mean value theorem and $F^{\prime }(s)=\lambda -3s^2\le \lambda$ imply $F(u)\le F(K)+\lambda (u-K)$. ThereforeThe Gronwall inequality gives $h=0$, since its initial L2 norm vanishes. For the initially rough datum, apply the inequality from $\epsilon >0$ and let $\epsilon \downarrow 0$, using $u(\epsilon )\to u_0$ in L2 space. The same argument for $-u$, which satisfies the same equation because $F$ is odd, provesThis is the invariant interval for a cubic reaction-diffusion equation; the energy calculation is valid by the Sobolev chain rule and approximation even before the stronger pointwise regularity is established.

Now fix $0<\tau <t_0$ and, if necessary, reduce the positive Hölder exponent $\beta$. Rearranging the evolution gives the elliptic identityIts right side is bounded on $[\tau ,t_0]\times \Omega$, because $v\in C([\tau ,t_0],C^{0,\beta })$ and $|u|\le K$. Boundary elliptic regularity first gives $u(t)\in W^{2,p}(\Omega )$ for every finite $p$, uniformly on this time interval. For $p>n$, the Sobolev embedding theorem makes $u(t)$ spatially continuously differentiable, with a positive Hölder exponent. Thus $F(u(t))$ lies in $C^{0,\beta }$ after reducing $\beta$ if necessary. The global Schauder estimate then gives $u(t)\in C^{2,\beta }(\overline{\Omega })$.

These are continuous-in-time, not just separate-time, regularity statements. Since $v\in C(C^{0,\beta })$, the identity $u(t)-u(s)={\int }_s^{t}v(r)dr$, initially an identity in L2 space, holds in $C^{0,\beta }$ once $u(s)$ is known to belong to that space. Hence $u$ is locally Lipschitz in time into $C^{0,\beta }$. Apply the global Schauder estimate to the difference of the two elliptic identities. Continuity of $v$ and of $F(u)$ into $C^{0,\beta }$ proves $u\in C([\tau ,t_0],C^{2,\beta })$.

To gain the spatial regularity of the time derivative itself, take time difference quotients of the evolution. If $d_h(t)=[u(t+h)-u(t)]/h$, their equations arewhereThe coefficients have uniformly bounded spatial Hölder norms and temporal Hölder norms on compact positive-time cylinders: this follows from the spatial regularity just proved and the local time Lipschitz bound in $C^{0,\beta }$. Also $d_h={\int }_0^{1}v(t+sh)ds$ is uniformly bounded and tends to $v$ in $C^{0,\beta }$. Positive-time parabolic Schauder estimates, applied away from the lower time edge, give uniform $C^{1+\beta /2,2+\beta }$ bounds for $d_h$, with a slightly smaller exponent if needed. Passing to the limit proves that $v$ is continuous into $C^2(\overline{\Omega })$ and satisfiesThe Dirichlet boundary condition for $v$ follows also by differentiating the identically zero boundary value in $C^0$. Since $\tau$ was arbitrary, this proves the requested, genuinely stronger regularity:This is positive-time smoothing for a semilinear heat equation; it is the extra parabolic step for $v$, not just the Sobolev embedding theorem, that gives the $C^1$ topology asserted here.

Finally, a uniform bound excludes a finite maximal existence time for this semilinear partial differential equation. Restart at any positive time $s$, using the heat semigroup on the Banach space $X=\{g\in C(\overline{\Omega }):g{|}_{\partial \Omega }=0\}$ with the supremum norm. Its variation-of-constants formula isThe heat semigroup is a contraction semigroup on $X$. On the ball of radius $K+1$, $F$ is Lipschitz with constant at most $|\lambda |+3(K+1{)}^2$, and bounded by $|\lambda |(K+1)+(K+1{)}^3$. Choose a time length $\eta >0$ so that the integral is at most $1$ and its Lipschitz constant is less than $1$. The Banach fixed-point theorem then gives a local solution on $[s,s+\eta ]$, with $\eta$ independent of $s$, because $\Vert u(s){\Vert }_{\infty }\le K$. If a finite maximal time existed, choosing $s$ less than $\eta$ before it would extend the same solution past it, a contradiction. Uniqueness identifies this continuation with the original solution, and positive-time smoothing restores the classical solution on each extended interval. The solution is globally defined and remains between $-K$ and $K$.