Problem 2. In this problem, we will show that differentiable functions are continuous. (a) Show that if M is a k x n matrix, then there is a constant c> 0 such that ||M|| ≤ c|||| for all 7 € R". ill
Problem 2. In this problem, we will show that differentiable functions are continuous. (a) Show that if M is a k x n matrix, then there is a constant c> 0 such that ||M|| ≤ c|||| for all 7 € R". ill
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Problem 2. In this problem, we will show that differentiable functions are continuous.
(a)
Show that if M is a k x n matrix, then there is a constant c> 0 such that ||M|| ≤ c|||| for
all z € R".
Hint: You will probably need the Cauchy-Schwartz inequality.
Next, consider the following theorem and its proof in the single variable case.
Theorem. Suppose that f: R→ R is differentiable at a ER. Then f is continuous at a.
Proof. Recall that if f(x) is differentiable at x = a, then we have:
f(x) [f(a) + f'(a)(x − a)]
x-a
lim
I-a
To show that f is continuous at a, we must check that:
lim f(x) = f(a), or equivalently, that
DI
Assume x a. By adding and subtracting f'(a)(x - a) we can write:
Thus, by the triangle inequality, we have
|f(x) = f(a)| = |f(x) - f(a) - f'(a)(x − a) + f'(a)(x − a)|
= f(x) [f(a) + f'(a)(x − a)] + f'(a)(x - a)|
|f(x) = f(a)| ≤ f(x) - [ƒ(a) – f'(a)(x − a)]] + [f'(a)(x − a)|
Since we are taking a limit as x→a, we may choose a close enough to a, i.e., satisfying |z − a ≤ 1.
For such , we have 1 ≤ Combining this with the inequality above we find:
z-al
Since f is differentiable at x = a, we have
|f(x) = f(a)| = |f(x) - [ƒ(a) - f'(a)(x − a)]] · 1+|f'(a)(x - a)|
≤ f(x) [f(a) - f'(a)(x-a)]].
1
|x - a|
+|f'(a)(x-a)|
= |ƒ (2) -
f(x)
[f(a)
·a)]| +
+|f'(a)(x-a).
Moreover, we also have
lim
D+Z
D+I
Thus, combining these two facts we find that
lim f(x) f(a)| = lim
x→a
= 0.
lim f(a) f(a)| = 0
I-a
f'(a)(x - a)]]
( -
xa
f(x) [f(a) f'(a)(x-a)]
D+I
= 0+0
= 0
x-a
lim f'(a)(x-a)| = |f'(a)| lim |xa| = 0.
x-a
(z - a) -
= 0.
|ƒ(x) — [ƒ(a) — ƒ'(a)(x − a)]
x-a
+|f'(a)(x − a)|
and so f is continuous at x = a.
(b)
Adapt this proof to give a similar proof of the following theorem.
Theorem. Suppose that F: R" → Rk is differentiable at a ER". Then F is continuous at a.
Hint: You will need part (a).
0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe60e14a8-856f-447c-ac90-795ae43e00b4%2Fa83206ca-8786-4862-8673-0accfa734454%2Ft7sh2pg_processed.png&w=3840&q=75)
Transcribed Image Text:Problem 2. In this problem, we will show that differentiable functions are continuous.
(a)
Show that if M is a k x n matrix, then there is a constant c> 0 such that ||M|| ≤ c|||| for
all z € R".
Hint: You will probably need the Cauchy-Schwartz inequality.
Next, consider the following theorem and its proof in the single variable case.
Theorem. Suppose that f: R→ R is differentiable at a ER. Then f is continuous at a.
Proof. Recall that if f(x) is differentiable at x = a, then we have:
f(x) [f(a) + f'(a)(x − a)]
x-a
lim
I-a
To show that f is continuous at a, we must check that:
lim f(x) = f(a), or equivalently, that
DI
Assume x a. By adding and subtracting f'(a)(x - a) we can write:
Thus, by the triangle inequality, we have
|f(x) = f(a)| = |f(x) - f(a) - f'(a)(x − a) + f'(a)(x − a)|
= f(x) [f(a) + f'(a)(x − a)] + f'(a)(x - a)|
|f(x) = f(a)| ≤ f(x) - [ƒ(a) – f'(a)(x − a)]] + [f'(a)(x − a)|
Since we are taking a limit as x→a, we may choose a close enough to a, i.e., satisfying |z − a ≤ 1.
For such , we have 1 ≤ Combining this with the inequality above we find:
z-al
Since f is differentiable at x = a, we have
|f(x) = f(a)| = |f(x) - [ƒ(a) - f'(a)(x − a)]] · 1+|f'(a)(x - a)|
≤ f(x) [f(a) - f'(a)(x-a)]].
1
|x - a|
+|f'(a)(x-a)|
= |ƒ (2) -
f(x)
[f(a)
·a)]| +
+|f'(a)(x-a).
Moreover, we also have
lim
D+Z
D+I
Thus, combining these two facts we find that
lim f(x) f(a)| = lim
x→a
= 0.
lim f(a) f(a)| = 0
I-a
f'(a)(x - a)]]
( -
xa
f(x) [f(a) f'(a)(x-a)]
D+I
= 0+0
= 0
x-a
lim f'(a)(x-a)| = |f'(a)| lim |xa| = 0.
x-a
(z - a) -
= 0.
|ƒ(x) — [ƒ(a) — ƒ'(a)(x − a)]
x-a
+|f'(a)(x − a)|
and so f is continuous at x = a.
(b)
Adapt this proof to give a similar proof of the following theorem.
Theorem. Suppose that F: R" → Rk is differentiable at a ER". Then F is continuous at a.
Hint: You will need part (a).
0
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