nyuuzyou/wikis huggingface dataset:
Continuous Mapping (Metric Space)/Examples/Identity Function with Discontinuity

== Example of Non-Continuous Mapping ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \begin {cases} x & : x \ne 0 \\ 1 & : x = 0 \end {cases}$
Then $f$ is not continuous.
== Proof ==
Let $U \subseteq \R$ be the open set $\openint 0 \to$.
Then:
:$f^{-1} \sqbrk U = \hointr 0 \to$ 
Note that we have $0 \in f^{-1} \sqbrk U$.
Let $\epsilon \in \R_{>0}$ be arbitrary.
Then:
and so $f^{-1} \sqbrk U$ is not open in $\R$.
Thus an open set does not necessarily have an open preimage.
Hence $f$ is not continuous by definition.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.15$
Category:Examples of Continuous Mappings on Metric Spaces
Continuous Mapping from Compact Space to Hausdorff Space Preserves Local Connectedness

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ be a compact topological space.
Let $T_2 = \struct {S_2, \tau_2}$ be a $T_2$ (Hausdorff) space.
Let $f: T_1 \to T_2$ be a continuous mapping.
Let $T_1$ be locally connected.
Then $T_2$ is also locally connected.
== Proof ==
Let $H$ be a component of an open set $U$ of $T_2$.
By definition of continuous mapping, $f^{-1} \sqbrk U$ is an open set of $T_1$.
Let $G$ be a component of $f^{-1} \sqbrk U$.
Thus by Continuous Image of Connected Space is Connected, $f \sqbrk G$ is connected in $T_2$.
Thus either:
:$f \sqbrk G \subseteq H$
or:
:$f \sqbrk G \cap H = \O$
Thus every component of $f^{-1} \sqbrk H$ is a component of $f^{-1} \sqbrk U$.
From Component of Locally Connected Space is Open, $f^{-1} \sqbrk H$ is open.
Let $S_1 \setminus G$ denote the complement of $f^{-1} \sqbrk H$ relative to $S_1$.
By definition of closed set, $S_1 \setminus f^{-1} \sqbrk H$ is closed in $T_1$.
From Closed Subspace of Compact Space is Compact, $S_1 \setminus f^{-1} \sqbrk H$ is also compact in $T_1$.
From Continuous Image of Compact Space is Compact, $f \sqbrk {S_1 \setminus f^{-1} \sqbrk H} = S_2 \setminus H$ is compact in $T_2$.
From Compact Subspace of Hausdorff Space is Closed, $S_2 \setminus H$ is closed in $T_2$.
Thus by definition of closed set, $H$ is open in $T_2$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Functions and Products
Category:Locally Connected Spaces
Category:Compact Spaces
Category:Hausdorff Spaces
Category:Continuous Mappings
Continuous Mapping from Compact Space to Hausdorff Space is Closed Mapping

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ be a compact topological space.
Let $T_2 = \struct {S_2, \tau_2}$ be a $T_2$ (Hausdorff) space.
Let $f: T_1 \to T_2$ be a continuous mapping.
Then $f$ is a closed mapping.
== Proof ==
Let $C$ be a closed subspace of $T_1$.
By Closed Subspace of Compact Space is Compact, $C$ is compact.
By Continuous Image of Compact Space is Compact, $f \sqbrk C$ is compact in $T_2$.
By Compact Subspace of Hausdorff Space is Closed, $f \sqbrk C$ is closed in $T_2$.
The result follows by definition of closed mapping.
Category:Compact Spaces
Category:Hausdorff Spaces
Category:Closed Mappings
Continuous Mapping is Continuous on Induced Topological Spaces

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\tau_{d_1}$ and $\tau_{d_2}$ be the topologies induced by the metrics $d_1$ and $d_2$.
Let $T_1 = \struct {A_1, \tau_{d_1} }$ and $T_2 = \struct {A_2, \tau_{d_2} }$ be the resulting topological spaces.
Let $f: A_1 \to A_2$ be a mapping.
Then $f$ is $\tuple {d_1, d_2}$-continuous $f$ is $\tuple {\tau_{d_1}, \tau_{d_2} }$-continuous.
== Proof ==
Follows directly from:
:the open set definition of continuity on a metric space
:the definition of continuity on a topological space.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces: Proposition $3.1.11$
Category:Topology
Category:Metric Spaces
Category:Continuous Mappings
Continuous Mapping is Measurable

== Theorem ==
Let $\struct {X, \tau}$ and $\struct {X', \tau'}$ be topological spaces.
Denote with $\map \BB {X, \tau}$ and $\map \BB {X', \tau'}$ their respective Borel $\sigma$-algebras.
Let $f: X \to X'$ be a continuous mapping.
Then $f$ is $\map \BB {X, \tau} \, / \, \map \BB {X', \tau'}$-measurable.
== Proof ==
As $f$ is a continuous mapping, by definition, it holds that:
:$\forall S' \in \tau': \map {f^{-1} } {S'} \in \tau$
Now, by definition, $\map \BB {X, \tau} = \map \sigma \tau$, and so $\tau \subseteq \map \BB {X, \tau}$.
Also, $\map \BB {X', \tau'} = \map \sigma {\tau'}$.
Hence Mapping Measurable iff Measurable on Generator applies.
Therefore, $f$ is $\map \BB {X, \tau} \, / \, \map \BB {X', \tau'}$-measurable.
== Sources ==
* : $7.3$
* : $\S 8$: Problem $8$
Category:Measurable Functions
Category:Continuous Mappings
Continuous Mapping is Sequentially Continuous

== Theorem ==
Let $X$ and $Y$ be topological spaces.
Let $x \in X$.
Let $f: X \to Y$ be a mapping that is continuous at $x$.
Then $f$ is sequentially continuous at $x$.
=== Corollary ===
== Proof ==
Let $\sequence {x_n}_{n \mathop \ge 1}$ be a sequence in $X$ converging to $x$.
Let $V$ be a neighborhood of $\map f x$ in $Y$.
We are required to show that there exists $N \in \N$ such that $\map f {x_n} \in V$ for all $n \ge N$.
By continuity of $f$, choose a neighborhood $U$ of $x$ in $X$ such that $\map f U \subseteq V$.
Since $\sequence {x_n}_{n \mathop \ge 1}$ converges, there exists $N \in \N$ such that $x_n \in U$ for all $n \ge N$.
Therefore we must have that $\map f {x_n} \in \map f U \subseteq V$ for all $n \ge N$, as required.
Category:Continuous Mappings
Category:Sequential Continuity
Continuous Mapping is Sequentially Continuous/Corollary

== Theorem ==
Let $X$ and $Y$ be topological spaces.
Let $f: X \to Y$ be a mapping that is continuous at $x$.
Let $f$ be continuous (everywhere) on $X$.
Then $f$ is sequentially continuous on $X$.
== Proof ==
This follows immediately from the definitions:
:$(1): \quad$ A mapping is sequentially continuous everywhere in $X$ if it is sequentially continuous at each point
:$(2): \quad$ A mapping is continuous everywhere on $X$ it is continuous at each point.
Category:Continuous Mappings
Category:Sequential Continuity
Continuous Mapping of Separation

== Theorem ==
Let $T$ and $T'$ be topological spaces.
Let $A \mid B$ be a separation of $T$.
Let $f: T \to T'$ be a mapping such that the restrictions $f \restriction_A$ and $f \restriction_B$ are both continuous.
Then $f$ is continuous on the whole of $T$.
== Proof ==
Follows directly from Pasting Lemma for Union of Open Sets.
== Sources ==
* : $6.2$: Connectedness: Proposition $6.2.5$
Category:Separations
Category:Continuous Mappings
Continuous Mapping on Finer Domain and Coarser Codomain Topologies is Continuous

== Theorem ==
Let $\struct {X, \tau_1}$ and $\struct {Y, \tau_2}$ be topological spaces.
Let $f: \struct {X, \tau_1} \to \struct {Y, \tau_2}$ be a continuous mapping.
Let $\tau'_1$ be a finer topology on $X$ than $\tau_1$, that is, $\tau_1 \subseteq \tau'_1$.
Let $\tau'_2$ be a coarser topology on $Y$ than $\tau_2$, that is, $\tau'_2 \subseteq \tau_2$.
Then:
:$f: \struct {X, \tau'_1} \to \struct {Y, \tau'_2}$ is a continuous mapping.
== Proof ==
Let $U \in \tau'_2$.
Since $\tau'_2$ is a coarser topology than $\tau_2$:
:$U \in \tau_2$
By definition of continuity:
:$\map {f^{-1}} U \in \tau_1$
Since $\tau'_1$ is a finer topology than $\tau_1$:
:$\map {f^{-1}} U \in \tau'_1$
Since $U$ was arbitrary, by definition of continuity:
:$f: \struct {X, \tau'_1} \to \struct {Y, \tau'_2}$ is a continuous mapping
Category:Topology
Continuous Mapping to Product Space

== Theorem ==
Let $T = T_1 \times T_2$ be the product space of two topological spaces $T_1$ and $T_2$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.
Let $T'$ be a topological space.
Let $f: T' \to T$ be a mapping.
Then $f$ is continuous $\pr_1 \circ f$ and $\pr_2 \circ f$ are continuous.
=== Corollary ===
=== General Result ===
== Proof ==
=== Necessary Condition ===
Let $f$ be continuous.
Then by:
:Projection from Product Topology is Continuous
:Composite of Continuous Mappings is Continuous
so are $\pr_1 \circ f$ and $\pr_2 \circ f$.
=== Sufficient Condition ===
Suppose $\pr_1 \circ f$ and $\pr_2 \circ f$ are continuous.
Let $U_1 \subseteq T_1$ and $U_2 \subseteq T_2$.
Then we have:
Let $U_1$ and $U_2$ be open in $T_1$ and $T_2$ respectively.
Then by continuity of $\pr_1 \circ f$ and $\pr_2 \circ f$:
:$\map {\paren {\pr_1 \circ f}^{-1} } {U_1}$ and $\map {\paren {\pr_2 \circ f}^{-1} } {U_2}$ are open in $T'$.
So $\map {f^{-1} } {U_1 \times U_2}$ is open in $T'$.
From Natural Basis of Product Topology of Finite Product, a basis for the product space $T$ is:
:$\BB = \set {U_1 \times U_2 : U_1 \text { is open in } T_1, U_2 \text { is open in } T_2}$
Then we have shown that:
:$\forall B \in \BB : \map {f^{-1} } B$ is open in $T'$
 
It follows from Continuity Test using Basis that $f$ is continuous.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.5$: Products: Proposition $3.5.4$
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Functions
Category:Product Topology
Category:Continuous Mappings
Continuous Mapping to Product Space/Corollary

== Corollary to Continuous Mapping to Product Space ==
Let $T = T_1 \times T_2$ be the product space of two topological spaces $T_1$ and $T_2$.
Let $T'$ be a topological space.
Let $f: T' \to T_1$ be a mapping.
Let $g: T' \to T_2$ be a mapping.
Let $f \times g : T' \to T$ be the mapping defined by:
:$\forall x \in T' : \map {\paren {f \times g}} x = \tuple {\map f x, \map g x}$
Then $f \times g$ is continuous $f$ and $g$ are continuous.
== Proof ==
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.
From Continuous Mapping to Product Space, $f \times g$ is continuous $\pr_1 \circ \paren {f \times g}$ and $\pr_2 \circ \paren {f \times g}$ are continuous.
Now:
From Equality of Mappings, $\pr_1 \circ \paren {f \times g} = f$.
Similarly:
:$\pr_2 \circ \paren {f \times g} = g$
The result follows.
Category:Product Topology
Category:Continuous Mappings
Continuous Mapping to Product Space/General Result

== Theorem ==
Let $X$ be a topological space.
Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of topological spaces for some indexing set $I$.
Let $\ds Y = \prod_{i \mathop \in I} Y_i$ be the product space of $\family {Y_i}_{i \mathop \in I}$. 
For each $i \in I$, let $\pr_i: X \to X_i$ denote the $i$th projection on $X$:
:$\forall \family {x_j}_{j \mathop \in I} \in X: \map {\pr_i} {\family {x_j}_{j \mathop \in I} } = x_i$
Let $f$ be a mapping from $X$ to $Y$. 
Then $f$ is continuous $\pr_i \circ f$ is continuous for all $i \in I$.
== Proof ==
Suppose $f$ is continuous.
From Projection from Product Topology is Continuous $\pr_i$ is continuous for each $i \in I$.
By Composite of Continuous Mappings is Continuous it follows that $\pr_i \circ f$ is continuous for each $i \in I$.
Conversely, suppose that each $\pr_i \circ f$ is continuous. 
Let $U = \pr_{i_1}^{-1} \sqbrk {U_{i_1} } \cap \cdots \cap \pr_{i_n}^{-1} \sqbrk {U_{i_n} }$ be an open set in the natural basis of the product topology for $Y$.
Note that by the definition of natural basis, $U$ is constrained to be the intersection of a finite number of sets of $\mathbb S$.
Then by Preimage of Intersection under Mapping we find:
:$f^{-1} \sqbrk U = f^{-1} \sqbrk {\pr_{i_1}^{-1} \sqbrk {U_{i_1} } } \cap \cdots \cap f^{-1} \sqbrk {\pr_{i_n}^{-1} \sqbrk {U_{i_n} } }$
Since each $\pr_{i_j} \circ f$ is continuous, each:
:$\paren {\pr_{i_j} \circ f}^{-1} \sqbrk {U_{i_j} } = f^{-1} \sqbrk {\pr_{i_j}^{-1} \sqbrk {U_{i_j} } }$ 
is open.
Therefore $f^{-1} \sqbrk U$ is open.
By the definition of basis, each open set of $Y$ can be written as a union of sets in the natural basis of the product topology for $Y$.
From Preimage of Union under Mapping, the union of the preimages of open sets of $Y$ equals the preimages of the union of those sets.
Thus the preimage of an open set of $Y$ is open in $X$.
Hence $f$ is continuous.
Category:Product Topology
Category:Continuous Mappings
Continuous Mappings into Hausdorff Space coinciding on Everywhere Dense Set coincide

== Theorem ==
Let $\struct {X, \tau_X}$ be a topological space.
Let $\struct {Y, \tau_Y}$ be a Hausdorff space. 
Let $D$ be an everywhere dense subset of $X$. 
Let $f : X \to Y$ and $g : X \to Y$ be continuous mappings such that: 
:$\map f x = \map g x$ for all $x \in D$.
Then: 
:$\map f x = \map g x$ for all $x \in X$. 
== Proof 1 ==
 
== Proof 2 ==
Category:Hausdorff Spaces
Category:Continuous Mappings
Continuous Mappings into Hausdorff Space coinciding on Everywhere Dense Set coincide/Proof 1

== Theorem ==
== Proof ==
 that there exists $x_0 \in X$ such that $\map f {x_0} \ne \map g {x_0}$.
Since $\struct {Y, \tau_Y}$ is Hausdorff, there exists an open neighborhood $U$ of $\map f {x_0}$ and an open neighborhood $V$ of $\map g {x_0}$ such that: 
:$U \cap V = \O$
We then have: 
:$x_0 \in f^{-1} \sqbrk U \cap g^{-1} \sqbrk V$
Let:
:$W = f^{-1} \sqbrk U \cap g^{-1} \sqbrk V$
Since $f$ and $g$ are continuous, $W$ is the intersection of two open sets in $\struct {X, \tau_X}$. 
So $W$ is open in $\struct {X, \tau_X}$ by the definition of a topology. 
Since $D$ is an everywhere dense subset of $X$, there therefore exists $y \in D \cap W$.
Then we have $\map f y \in U$ and $\map g y \in V$.
Since $U \cap V = \O$, it follows that $\map f y \ne \map g y$. 
But we also have $y \in D$, and by hypothesis we have that $\map f x = \map g x$ for all $x \in D$.
Hence we have a contradiction.
We conclude: 
:$\map f x = \map g x$ for all $x \in X$.
Category:Continuous Mappings into Hausdorff Space coinciding on Everywhere Dense Set coincide
Continuous Mappings into Hausdorff Space coinciding on Everywhere Dense Set coincide/Proof 2

== Theorem ==
== Proof ==
Let $x \in X$.
Since $D$ is everywhere dense, we have that $x \in \map \cl D$, where $\map \cl D$ is the topological closure of $D$. 
By Point in Set Closure iff Limit of Net, there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ in $D$ converging to $x$.
From Characterization of Continuity in terms of Nets, we have that the nets $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ and $\family {\map g {x_\lambda} }_{\lambda \in \Lambda}$ converge to $\map f x$ and $\map g x$ in $\struct {Y, \tau_Y}$ respectively.
By hypothesis, we have: 
:$\map f {x_\lambda} = \map g {x_\lambda}$ for each $\lambda \in \Lambda$.
So we have that $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ converges to both $\map f x$ and $\map g x$.
Since $\struct {Y, \tau_Y}$ is Hausdorff, from Characterization of Hausdorff Property in terms of Nets we obtain that $\map f x = \map g x$, hence the demand.
Category:Continuous Mappings into Hausdorff Space coinciding on Everywhere Dense Set coincide
Continuous Mappings preserve Compact Subsets

== Theorem ==
Let $X, Y$ be normed vector spaces.
Let $K \subseteq X$ be a compact subset.
Suppose $f : X \to Y$ is a continuous mapping at each $x \in K$.
Then $\map f K$ is a compact subset of $Y$.
== Proof ==
Let $\sequence {y_n}_{n \mathop \in \N}$ be a sequence contained in $\map f K$.
Then:
:$\forall n \in \N : \exists x_n \in K : y_n = \map f {x_n}$
$K$ is compact.
By definition, there is a convergent subsequence $\sequence {x_{n_k}}_{k \mathop \in \N}$ convergent with the limit $L \in K$.
$f$ is a continuous mapping.
By Continuous Mappings preserve Convergent Sequences, $\sequence {\map f {x_{n_k}}}_{k \mathop \in \N} = \sequence {y_{n_k}}_{k \mathop \in \N}$ is convergent with limit $\map f L \in \map f K$.
By definition, $\map f K$ is compact.
== Sources ==
* : Chapter $\S 2.2$: Continuous and linear maps. Continuous maps
Category:Compact Spaces
Category:Continuous Mappings on Normed Vector Spaces
Continuous Mappings preserve Convergent Sequences

== Theorem ==
Let $X, Y$ be normed vector spaces.
Let $c \in X$.
Let $f : X \to Y$ be a mapping.
Then $f$ is continuous at $c$ iff for every sequence $\sequence {x_n}_{n \mathop \in \N} \in X$ such that $\sequence {x_n}_{n \mathop \in \N}$ converges to $c$, $\sequence {\map f {x_n}}_{n \mathop \in \N}$ converges to $\map f c$.
== Proof ==
=== Necessary Condition ===
Let $f$ be continuous at $c$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$ such that $\sequence {x_n}_{n \mathop \in \N}$ converges to $c$.
Let $\epsilon \in \R_{\mathop > 0}$.
By definition of continuous mapping:
:$\exists \delta \in \R_{\mathop > 0} : \forall x \in X : \norm {x - c} < \delta \implies \norm {\map f x - \map f c} < \epsilon$
Since $\sequence {x_n}_{n \mathop \in \N}$ converges to $c$:
:$\exists N \in \N : \forall n \in \N : n > \N \implies \norm {x_n - c} < \delta$
By definition of continuity, and because $\sequence {x_n}_{n \mathop \in \N} \in X$:
:$\norm {\map f {x_n} - \map f c} < \epsilon$
Choice of $\epsilon$ was arbitrary.
Hence:
:$\forall \epsilon \in \R_{\mathop > 0} : \exists N \in \N : \forall n \in \N : n > \N \implies \norm {\map f {x_n} - \map f c} < \epsilon$
By definition, $\sequence {\map f {x_n}}_{n \mathop \in \N}$ converges to $\map f c$.
=== Sufficient Condition ===
Suppose $f$ is not continuous at $c$.
By definition of continuous mapping:
:$\forall \epsilon \in \R_{\mathop > 0} : \exists \delta \in \R_{\mathop > 0} : \forall x \in X : \norm {x - c} < \delta \implies \norm {\map f x - \map f c} < \epsilon$
By De Morgan's laws and negation of implication:
:$\exists \epsilon \in \R_{\mathop > 0} : \forall \delta \in \R_{\mathop > 0} : \exists x \in X : \paren {\norm {x - c} < \delta} \land \paren {\norm {\map f x - \map f c} \ge \epsilon}$
Consider a sequence $\ds \delta_n = \frac 1 n$ where $n \in \N$.
Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $X$ such that:
:$\ds \forall n \in \N : \norm {x_n - c} < \delta_n$.
At the same time we have that: 
:$\forall n \in \N : \norm {\map f {x_n} - \map f c} \ge \epsilon$
By definition of convergent sequence:
:$\forall \epsilon' \in \R_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \norm {x_n - L} < \epsilon'$
Let $N \in \N$.
Let $\epsilon' := \delta_N$
Then:
:$\forall n \in \N : n > N \implies \norm {x_n - c} < \delta_n < \delta_N$
But $N \in \N$ was arbitrary.
Hence, $\sequence {x_n}_{n \mathop \in \N}$ is convergent with limit $c$.
However, $\sequence {\map f {x_n}}_{n \mathop \in \N}$ does not converge to $\map f c$ because:
:$\forall n \in \N : \norm {\map f {x_n} - \map f c} \ge \epsilon$
In other words, if $f$ is not continuous at $c$ then not for all $\sequence {x_n}_{n \mathop \in \N}$ convergent to $c$ $\sequence {\map f {x_n}}$ converges to $\map f c$.
By rule of transposition, if for all $\sequence {x_n}_{n \mathop \in \N}$ convergent to $c$ $\sequence {\map f {x_n}}$ converges to $\map f c$ then $f$ is continuous at $c$.
== Sources ==
* : Chapter $\S 2.2$: Continuous and linear maps. Continuous maps
Category:Continuous Mappings
Category:Convergent Sequences
Continuous Midpoint-Concave Function is Concave

== Theorem ==
Let $f$ be a real function which is defined on a real interval $I$.
Let $f$ be midpoint-concave and continuous on $I$.
Then $f$ is concave.
== Proof ==
As $f$ is midpoint-concave, then $-f$ is midpoint-convex.
From Continuous Midpoint-Convex Function is Convex, $-f$ is convex. 
From Real Function is Concave iff its Negative is Convex, $f$ is concave. 
== Also see ==
* Continuous Strictly Midpoint-Concave Function is Strictly Concave
* Continuous Midpoint-Convex Function is Convex
* Continuous Strictly Midpoint-Convex Function is Strictly Convex
Category: Real Analysis
Continuous Midpoint-Convex Function is Convex

== Theorem ==
Let $f$ be a real function which is defined on a real interval $I$.
Let $f$ be midpoint-convex and continuous on $I$.
Then $f$ is convex.
== Proof ==
Let $x, y \in I$ and let $t \in \closedint 0 1$.
Let $\sequence {t_n}_{n \mathop \in \N}$ be a sequence of rational numbers in $\closedint 0 1$ converging to $t$.
It follows that:
and:
From Midpoint-Convex Function is Rational Convex, $f$ is rational convex. 
From Continuous Mapping is Sequentially Continuous, $f$ is sequentially continuous.
As $f$ is rational convex:
:$\forall n \in \N: \map f {t_n x + \paren {1 - t_n} y} \le t_n \map f x + \paren {1 - t_n} \map f y$
From sequential continuity of $f$ and Inequality Rule for Real Sequences:
which completes the proof.
== Also see ==
* Continuous Strictly Midpoint-Convex Function is Strictly Convex
* Continuous Midpoint-Concave Function is Concave
* Continuous Strictly Midpoint-Concave Function is Strictly Concave
== Sources ==
* : $\S 4.3$
Category:Real Analysis
Continuous Non-Negative Real Function with Zero Integral is Zero Function

== Theorem ==
Let $a, b$ be real numbers with $a < b$. 
Let $f : \closedint a b \to \R$ be a continuous function. 
Let: 
:$\map f x \ge 0$
for all $x \in \closedint a b$.
Let: 
:$\ds \int_a^b \map f x \rd x = 0$
Then $\map f x = 0$ for all $x \in \closedint a b$.
== Proof 1 ==
 
== Proof 2 ==
Category:Definite Integrals
Category:Continuous Non-Negative Real Function with Zero Integral is Zero Function
Continuous Non-Negative Real Function with Zero Integral is Zero Function/Proof 1

== Theorem ==
== Proof ==
From Definite Integral of Constant, if $\map f x = 0$ for all $x \in \closedint a b$, then: 
:$\ds \int_a^b \map f x \rd x = 0$
We want to show that if:
:$\ds \int_a^b \map f x \rd x = 0$
then:
:$\map f x = 0$ for all $x \in \closedint a b$.
Since $\map f x \ge 0$, by Relative Sizes of Definite Integrals:
:$\ds \int_a^b \map f x \rd x \ge 0$
It therefore suffices to show that if: 
:$\map f x > 0$ for some $x \in \closedint a b$
then:
:$\ds \int_a^b \map f x \rd x > 0$
We split this into three cases: 
:$x = a$
:$a < x < b$
:$x = b$
since continuity on endpoints is defined slightly differently.
Consider the case: 
:$a < x < b$
By continuity at $x$ we have that there exists $\delta > 0$ such that: 
:for all $y \in \openint {x - \delta} {x + \delta}$ we have $\size {\map f y - \map f x} < \dfrac {\map f x} 2$
In particular for $y \in \openint {x - \delta} {x + \delta}$ we have: 
:$0 < \dfrac {\map f x} 2 < \map f y$
Pick $\delta$ sufficiently small so that:
:$\openint {x - \delta} {x + \delta} \subseteq \closedint a b$
We then have:
In the case $x = a$, by the definition of right continuity, there exists $\delta > 0$ such that: 
:for all $x \in \openint a {a + \delta}$ we have $\size {\map f x - \map f a} < \dfrac {\map f a} 2$
That is, there exists some $x \in \openint a b$ such that: 
:$\map f x > \dfrac {\map f a} 2 > 0$
So the former proof applies in this case. 
The case $x = b$ follows similarly.
In the case $x = b$, by the definition of left continuity, there exists $\delta > 0$ such that: 
:for all $x \in \openint {b - \delta} b$ we have $\size {\map f x - \map f b} < \dfrac {\map f b} 2$
That is, there exists some $x \in \openint a b$ such that: 
:$\map f x > \dfrac {\map f b} 2 > 0$
So, again, the proof for the case $a < x < b$ applies.
We have covered all three cases, so we are done.
Category:Continuous Non-Negative Real Function with Zero Integral is Zero Function
Continuous Non-Negative Real Function with Zero Integral is Zero Function/Proof 2

== Theorem ==
== Proof ==
From Continuous Real Function is Darboux Integrable, $f$ is Darboux integrable on $\closedint a b$.
Let $F : \closedint a b \to \R$ be a real function defined by: 
:$\ds \map F x = \int_a^x \map f x \rd x$
We are assured that this function is well-defined, since $f$ is integrable on $\closedint a b$.
From Fundamental Theorem of Calculus: First Part, we have: 
:$F$ is continuous on $\closedint a b$
:$F$ is differentiable on $\openint a b$ with derivative $f$
Note that: 
for all $x \in \openint a b$. 
We therefore have, by Real Function with Positive Derivative is Increasing: 
:$F$ is increasing on $\closedint a b$.
However, by hypothesis:
So, it must be the case that: 
:$\map F x = 0$ for all $x \in \closedint a b$.
We therefore have, from Derivative of Constant: 
:$\map {F'} x = \map f x = 0$ for all $x \in \closedint a b$
as required.
Category:Continuous Non-Negative Real Function with Zero Integral is Zero Function