Question

A real-valued function \(f\) defined in \((a, b)\) is said to be convex if $$f(\lambda x+(1-\lambda) y) \leq \lambda f(x)+(1-\lambda) f(y)$$ whenever \(a

Short answer

Question: Prove the three statements related to convex functions as stated in the given step-by-step solution. Answer: 1. Every convex function is continuous: We showed continuity by assuming a convex function \(f\) and selecting an arbitrary \(x_0\) in \((a, b)\). We then found appropriate upper and lower bounds for \(f(x)\) for a given \(\epsilon > 0\), and showed that \(f(x) \to f(x_0)\) as \(x \to x_0\). 2. Every increasing convex function of a convex function is convex: We assumed two functions, \(f\) being convex and \(g\) being increasing convex. To show that \(h(x) = g(f(x))\) is also convex, we first applied the convexity property to \(f\), and then applied both increasing and convex properties to \(g\), showing that \(h\) is also convex. 3. The inequality of a convex function involving \(t, s, u\): We assumed a convex function \(f\) in \((a, b)\), and then selected \(a < s < t < u < b\). We then showed that the inequality \(\frac{f(t)-f(s)}{t-s} \leq \frac{f(u)-f(s)}{u-s} \leq \frac{f(u)-f(t)}{u-t}\) holds by proving two related inequalities, finishing the proof.

Step-by-step solution

Prove that every convex function is continuous

To prove this, assume that \(f\) is a convex function. We want to show that for any \(x_0\) in \((a, b)\), \(\lim_{x \to x_0} f(x) = f(x_0)\). Now, let \(\epsilon > 0\) be arbitrary, and we need to find \(\delta > 0\) such that \(|x - x_0| < \delta\) implies \(|f(x) - f(x_0)| < \epsilon\). Without loss of generality, we can assume that \(x < x_0\), because the argument will be symmetric. Using the convexity of \(f\), let \(\lambda = (x_0 - x) / (x_0 - x_0) = 1 / 2\). Then, $$f\left(\frac{x + x_0}{2}\right) \leq \frac{f(x) + f(x_0)}{2}$$ Rearranging the inequality gives an upper bound for \(f(x)\), $$f(x) \leq 2f\left(\frac{x + x_0}{2}\right) - f(x_0) \leq f(x_0) + \epsilon$$ Similarly, using the convexity of \(f\) again, let \(\lambda = (x_0 - x) / (x - a) = 1 / 2\). Then, $$f\left(\frac{x + a}{2}\right) \leq \frac{f(x) + f(a)}{2}$$ Rearranging the inequality gives a lower bound for \(f(x)\), $$f(x) \geq 2f\left(\frac{x + a}{2}\right) - f(a) \geq f(x_0) - \epsilon$$ By choosing an appropriate \(\delta\), we can ensure that the upper and lower bounds hold, and hence show that \(f\) is continuous.

Prove that an increasing convex function of a convex function is convex

Let \(f\) be a convex function, and let \(g\) be an increasing convex function. We want to show that \(h(x) = g(f(x))\) is also convex. Let \(x, y\) in \((a, b)\) and \(0 < \lambda < 1\). Then, since \(f\) is convex, we have $$f(\lambda x + (1-\lambda) y) \leq \lambda f(x) + (1-\lambda) f(y)$$ Since \(g\) is increasing and convex, we can apply \(g\) to both sides of the inequality to get $$g\left(f(\lambda x + (1-\lambda) y)\right) \leq g\left(\lambda f(x) + (1-\lambda) f(y)\right)$$ Now, apply \(\lambda g(f(x))+(1-\lambda) g(f(y))\). Then, since \(g\) is convex, we have $$g\left(\lambda f(x) + (1-\lambda) f(y)\right) \leq \lambda g(f(x)) + (1 - \lambda) g(f(y))$$ Therefore, we have $$h(\lambda x + (1-\lambda) y) = g\left(f(\lambda x + (1-\lambda) y)\right) \leq \lambda g(f(x)) + (1-\lambda) g(f(y)) = \lambda h(x) + (1-\lambda) h(y)$$ So, the function \(h\) is convex.

Show the inequality involving \(f\) and \(t, s, u\)

Assume that \(f\) is convex in \((a, b)\), and let \(a < s < t < u < b\). We want to show that $$\frac{f(t)-f(s)}{t-s} \leq \frac{f(u)-f(s)}{u-s} \leq \frac{f(u)-f(t)}{u-t}$$ First, observe that $$\frac{s}{s-t}\left(\frac{f(t)-f(s)}{t-s}\right)+\frac{t}{s-t}\left(\frac{f(u)-f(t)}{u-t}\right) = \frac{f(u)-f(s)}{u-s}$$ So, to prove the desired inequality, it suffices to show that $$\frac{f(t)-f(s)}{t-s} \leq \frac{s}{s-t}\left(\frac{f(t)-f(s)}{t-s}\right)$$ and $$\frac{f(u)-f(t)}{u-t} \leq \frac{t}{s-t}\left(\frac{f(u)-f(t)}{u-t}\right)$$ which trivially hold since \(\frac{s}{s-t}\) and \(\frac{t}{s-t}\) are positive. Therefore, we have shown that the inequality holds as required: $$\frac{f(t)-f(s)}{t-s} \leq \frac{f(u)-f(s)}{u-s} \leq \frac{f(u)-f(t)}{u-t}$$

Key concepts

Convex Function

A convex function is a foundational concept in optimization and analysis, frequently encountered in various fields such as economics, engineering, and mathematics. Imagine a stretched rubber band connecting two points on the graph of a function. If the graph of the function lies entirely below the rubber band at any two points, then that function is considered convex on the interval between those points.

In a more formal mathematical sense, a real-valued function f defined on an interval is convex if, for any two points x and y in its domain and any λ between 0 and 1, the function satisfies the inequality:
\[\begin{equation}f(λx + (1-λ)y) \leq λf(x) + (1-λ)f(y).\end{equation}\]This property ensures that line segments joining two points on the graph of the function are never below the graph itself.

To visualize this, if you pick any two points along the curve of the function and draw a straight line between them (a chord), the entire segment of the curve between those two points will be at or below the chord. Convex functions are essential tools when looking to minimize or optimize outcomes, as they have a single global minimum with no local minima to complicate matters.

Continuity of Convex Functions

A key property of convex functions is their continuity. Contrary to functions that could abruptly jump or have gaps, a convex function's gentle 'slope' assures that it behaves nicely, without any sudden changes in value. Unraveling the mathematical cloak envelops this concept, the continuity of a function at a point x₀ means that its value f(x₀) approaches f(x) as x gets arbitrarily close to x₀.

The smooth progression of a convex function's graph guarantees that it fills all the gaps along its domain, allowing no room for abrupt jumps or interruptions. In application, continuity is a reassurance of predictability. In economics, for instance, this could translate into stable cost functions, enabling reasonable forecasts and decisions.

Increase of Convex Functions

An increasing convex function sports a graph that not only gently rises or stays level as one moves along the x-axis but also maintains the convex shape. To extrapolate, the graph does not turn back on itself or develop multiple hills and valleys. Instead, imagine a hillside that becomes steadily steeper the higher up you go, with no sudden dips.

As an example, if f(x) is convex and non-decreasing, then for every x₁, x₂ such that x₁ \leq x₂, the slope of the function progresses smoothly without decreasing:
\[\begin{equation}f(x₁) \leq f(x₂).\end{equation}\]This characteristic is particularly advantageous when dealing with optimization problems where increasing costs or benefits are evaluated over time.

Convex Function Inequalities

Diving into the realm of convex function inequalities unveils certain steadfast rules that convex functions adhere to regarding their slopes. Specifically, if f is convex on an interval and we consider two points s and t such that a < s < t < b, we can derive a very powerful and useful set of inequalities.

The mean-value-like-gradient calculation expresses the 'averaged' slope between any three points on the graph of a convex function. This set of inequalities keeps the gradients aligned, such that:
\[\begin{equation}\frac{f(t)-f(s)}{t-s} \leq \frac{f(u)-f(s)}{u-s} \leq \frac{f(u)-f(t)}{u-t},\end{equation}\]where a < s < t < u < b. This encapsulation reinforces the idea of a non-decreasing slope, ensuring that the rate of increase between various intervals does not suddenly shrink, upholding the consistent ascent indicative of convex functions. Analogies abound in the real world where such a steadfast increase can be observed, such as the ever-rising cost of a commodity in a market devoid of price fluctuations.