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Prove some continuity properties of a function

Given a function f and a closed interval [a,b] such that

    \[ |f(u) - f(v)| \leq |u-v| \]

for all u,v \in [a,b].

  1. Prove f is continuous at every point x \in [a,b].
  2. If f is integrable on [a,b] prove

        \[ \left| \int_a^b f(x) \, dx - (b-a)f(a) \right| \leq \frac{(b-a)^2}{2}. \]

  3. If c \in [a,b] is any point prove

        \[ \left| \int_a^b f(x) \, dx - (b-a) f(c) \right| \leq \frac{(b-a)^2}{2}. \]


  1. Proof. Let p be any point in [a,b]. Then, for any \varepsilon > 0 let \delta = \varepsilon. Then

        \begin{align*}   |x-p| < \delta &&\implies && |x-p| &< \varepsilon \\  && \implies && |f(x) - f(p)| &< \varepsilon. \end{align*}

    Thus, for any \varepsilon > 0, we have |f(x) - f(p)| < \varepsilon whenever |x-p| < \delta. Hence,

        \[ \lim_{x \to p} f(x) = f(p). \]

    Therefore, f is continuous at every point p \in [a,b]. \qquad \blacksquare

  2. Proof. Since f(a) is a constant (the value of the function f evaluated at the constant a), we have

        \[ (b-a) f(a) = \int_a^b f(a) \, dx. \]

    Therefore, we can compute,

        \begin{align*}  \left| \int_a^b f(x) \, dx - (b-a) f(a) \right| &= \left| \int_a^b f(x) \, dx - \int_a^b f(a) \, dx \right| \\  &= \left| \int_a^b (f(x) - f(a)) \, dx \right|\\  &\leq \int_a^b |f(x) - f(a)| \, dx. \end{align*}

    The final inequality follows since, using the monotone property of the integral,

        \begin{align*}  - |f(x)| \leq f(x) \leq |f(x)| && \implies && \int_a^b -|f(x)| \, dx \leq \int_a^b f(x) \, dx \leq \int_a^b |f(x)| \, dx \\  && \implies && - \int_a^b |f(x)| \, dx \leq \int_a^b f(x) \, dx \leq \int_a^b |f(x)| \, dx \\  && \implies && \left| \int_a^b f(x) \, dx \right| \leq \int_a^b |f(x)| \, dx. \end{align*}

    Then, since |f(x) - f(a)| \leq |x-a| by our hypothesis on f, we have

        \begin{align*}  \int_a^b |f(x) - f(a)| \, dx &\leq \int_a^b |x-a| \, dx \\  &= \int_a^b (x-a) \, dx &(x-a\geq 0 \text{ since } x \in [a,b]) \\  &= \frac{b^2}{2} - \frac{a^2}{2} - ab + a^2 \\  &= \frac{b^2-2ab+a^2}{2} \\  & = \frac{(b-a)^2}{2}. \qquad \blacksquare \end{align*}

  3. Proof. The proof proceeds similarly to that of part (b),

        \begin{align*}  \left| \int_a^b f(x) \, dx - (b-a) f(c) \right| &= \left| \int_a^b f(x) \, dx - \int_a^b f(c) \, dx \right|\\ &= \left| \int_a^b (f(x) - f(c)) \, dx \right| \\ &\leq \int_a^b | f(x) - f(c) | \, dx \\ &\leq \int_a^b |x-c| \, dx. \end{align*}

    Then, since c \in [a,b], we break the integral into two pieces (to deal with the fact that |x-c| = -(x-c) for x < c and |x-c| = x-c for x \geq c):

        \begin{align*}   \int_a^b |x-c| \, dx &= \int_a^c -(x-c) \, dx + \int_c^b (x-c) \, dx \\  &= \left. -\frac{x^2}{2} \right|_a^c + c \cdot x \biggr \rvert_a^c + \left. \frac{x^2}{2} \right|_c^b - c \cdot x \biggr \rvert_c^b \\  &= \frac{a^2-c^2}{2} + c(c-a) + \frac{b^2-c^2}{2} - c(b-c) \\  &= \frac{a^2 - 2c^2 + b^2}{2} + 2c^2-ac-bc \\  &= \frac{a^2+b^2}{2} + c^2 - c(b+a). \end{align*}

    But now, we know c \in [a,b] implies c \leq b, so we have the inequality,

        \begin{align*}    \frac{a^2+b^2}{2} + c^2 - c(b+a) &= \frac{a^2+b^2}{2} + c(c-(b+a)) \\  &\leq \frac{b^2+a^2}{2} + b(b-b-a) \\  &= \frac{b^2+a^2}{2} - ab \\  &= \frac{b^2 - 2ab + a^2}{2} \\  &= \frac{(b-a)^2}{2}. \qquad \blacksquare \end{align*}

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