Real Analysis 29 | Combination of Continuous Functions

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This is my video series about Real Analysis. We talk about sequences, series, continuous functions, differentiable functions, and integral. I hope that it will help everyone who wants to learn about it.

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00:00 Intro
00:14 Recalling continuity
01:25 Combining 2 continuous functions
03:07 Composition of functions
05:55 Proof for composition of functions
07:27 Credits

#RealAnalysis
#Mathematics
#Calculus
#LearnMath
#Integrals
#Derivatives

I hope that this helps students, pupils and others. Have fun!

(This explanation fits to lectures for students in their first and second year of study: Mathematics for physicists, Mathematics for the natural science, Mathematics for engineers and so on)

[Independent_Man]

can you do videos on probability, statistics, estimation, learning, etc? These are hot topics in AI and machine learning; a lot of people will be interested in those.

[musiccd]

This proof is short and elegant. I used epsilo delta proof, never tried this way

[maohejiang]

I have trouble writing proofs, sir! What should I learn in order to be able to write these rigorous proofs? Thank you!

[angelmendez-rivera]

To prove that if f, g : I —> R are continuous at x0, and 0 < |g(x0)|, then f/g : I —> R is continuous at x0, you actually need the continuity product theorem, the continuity composition theorem that was proven, and you must also prove the continuity of the reciprocal function h : R\{0} —> R at g(x0).

h is continuous at x0 if and only for every real ε > 0, there exists a real δ > 0, such that for every nonzero real x, |x – x0| < δ implies |1/x – 1/x0| < ε. Notice that |1/x – 1/x0| = |x – x0|/(|x0|·|x| < δ/(|x0|·|x|). If 1/|x| can be bounded from above by an expression in terms of δ and x0, then the proof of continuity can be completed. The key here is that ||x| – |x0|| < |x – x0| or ||x| – |x0|| = |x – x0|, according to the reverse triangle inequality. Therefore, |x – x0| < δ implies ||x| – |x0|| < δ, which is equivalent to –δ < |x| – |x0| < δ. Thus |x0| – δ < |x| < |x0| + δ. If δ < |x0|, then |x0| – δ < |x| < |x0| + δ is equivalent to 1/(|x0| + δ) < 1/|x| < 1/(|x0| – δ). As such, |x – x0| < δ < |x0| implies 1/|x| < 1/(|x0| – δ). In conclusion, since |x – x0| < δ < |x0| implies |1/x – 1/x0| < δ/(|x0|·|x|), and it implies 1/|x| < 1/(|x0| – δ), it follows thus that |x – x0| < δ < |x0| implies |1/x – 1/x0| < δ/(|x0|·|x|) < δ/[|x0|·(|x0| – δ)]. Let ε = δ/[|x0|·(|x0| – δ)], so that |x0|^2·ε – |x0|·ε·δ = δ is equivalent to (1 + |x0|·ε)·δ = |x0|^2·ε, which solves with respect to δ to δ = |x0|^2·ε/(1 + |x0|·ε) = |x0| – |x0|/(1 + |x0|·ε), which satisfies 0 < δ < |x0| for every ε > 0.

As such, for every real ε > 0, there exists a δ with 0 < δ = |x0| – |x0|/(1 + |x0|·ε) < |x0|, such that for every nonzero real x, |x – x0| < |x0| – |x0|/(1 + |x0|·ε) implies |1/x – 1/x0| = |x – x0|/(|x0|·|x|) < [1 – 1/(1 + |x0|·ε)]/[|x0|/(1 + |x0|·ε)] = [(1 + |x0|·ε) – 1]/|x0| = |x0|·ε/|x0| = ε. Therefore, h is continuous at x0. Q. E. D.

[angelmendez-rivera]

If f and g are continuous at x0, then f + g and f·g are continuous at x0.

h = f + g is continuous at x0 if and only if for every real ε > 0, there exists a real δ > 0, such that for every x element in I, |x – x0| < δ implies |f(x) + g(x) – [f(x0) + g(x0)]| < ε. Notice that |f(x) + g(x) – [f(x0) + g(x0)]| = |f(x) – f(x0) + g(x) – g(x0)| = |f(x) – f(x0)| + |g(x) – g(x0)| or |f(x) + g(x) – [f(x0) + g(x0)]| < |f(x) – f(x0)| + |g(x) – g(x0)| by the triangle inequality. Since f and g are continuous, for every real ε0, ε1 > 0, there exists some real δ0, δ1 > 0, such that for every x element in I, |x – x0| < δ0 implies |f(x) – f(x0)| < ε0 and |x – x0| < δ1 implies |g(x) – g(x0)| < ε1. Thus, let δ = min(δ0, δ1), so |x – x0| < δ implies |f(x) + g(x) – [f(x0) + g(x0)]| < ε0 + ε1. Let ε = ε0 + ε1. Hence, for every real ε > 0, there exists a real δ > 0, such that for every x element in I, |x – x0| < δ implies |f(x) + g(x) – [f(x0) + g(x0)]| < ε. Therefore, if f and g are continuous at x0, then h = f + g is continuous at x0. Q. E. D.

f·g is continuous at x0 if and only for real ε > 0, there exists a δ > 0, such that for every x element in I, |x – x0| < δ implies |f(x)·g(x) – f(x0)·g(x0)| < ε. Notice that |f(x)·g(x) – f(x0)·g(x0)| = |f(x)·g(x) – f(x)·g(x0) + f(x)·g(x0) – f(x0)·g(x0)| = |f(x)·[g(x) – g(x0)] + [f(x) – f(x0)]·g(x0)| = |[f(x) – f(x0)]·[g(x) – g(x0)] + f(x0)·[g(x) – g(x0)] + g(x0)·[f(x) – f(x0)]| = |f(x) – f(x0)|·|f(x) – f(x0)| + |f(x0)|·|g(x) – g(x0)| + |g(x0)|·|f(x) – f(x0)|, or |f(x)·g(x) – f(x0)·g(x0)| < |f(x) – f(x0)|·|f(x) – f(x0)| + |f(x0)|·|g(x) – g(x0)| + |g(x0)|·|f(x) – f(x0)|. Since f and g are continuous, for every real ε0, ε1 > 0, there exist some real δ0, δ1 > 0, such that for every x element in I, |x – x0| < δ0 implies |f(x) – f(x0)| < ε0, and |x – x0| < δ1 implies |g(x) – g(x0)| < ε1. Let δ = min(δ0, δ1). Hence, |x – x0| < δ implies |f(x)·g(x) – f(x0)·g(x0)| < ε0·ε1 + |f(x0)|·ε1 + |g(x0)|·ε0. Let ε = ε0·ε1 + |f(x0)|·ε1 + |g(x0)|·ε0. Thus, for every real ε > 0, there exists a δ > 0, such that for every x element in I, |x – x0| < δ implies |f(x)·g(x) – f(x0)·g(x0)| < ε. Therefore, f·g is continuous at x0. Q. E. D.

[reconstructingmathematics]

Sir kindly upload lectures on topology and analysis of partial differential equations in English.