How big is a function

Mathematics · Calculus & Analysis · ages 22-24

Name ______________________ Date ____________

Find the exponent q that is conjugate to p = 5/4.
Answer: ______________
With p = q = 2, the 2-norm of f is 3 and the 2-norm of g is 2. What upper bound does Holder give for the integral of |fg|?
- 5
- 6
- 3
- 1.5
Which inequality bounds the integral of a product of two functions?
- Minkowski, by the sum of the two norms
- The triangle inequality applied pointwise
- Holder, by the product of the two norms
Lp treats functions that agree almost everywhere as the same element, because a nonzero difference on a null set has norm zero.
Circle one: True False
For real Lp on [0, 1] with Lebesgue measure and finite p >= 1, the given norm makes it a Hilbert space only when p = 2.
Circle one: True False
In L2 on [0, 1], the functions f = 1 and g = x minus 1/2 are orthogonal. Is this true?
Circle one: True False
Which pair of functions on [0, 1] is the same element of L2?
- The function x and the function x squared
- The constant function 1 and the constant function 2
- The indicator of [0, 1] and the indicator of (0, 1)
- The function x and the function 1 minus x on all of [0, 1]
The Lp norm of f is 5. What is the Lp norm of -2f?
- -10
- 10
- 5
For real Lp on [0, 1] with Lebesgue measure and finite p >= 1, what makes p = 2 special?
- Its norm comes from an inner product, so orthogonality makes sense
- It is the only exponent for which Minkowski's triangle inequality holds
- Every function in L2 is automatically bounded on its domain
On [0, 1], f = 2 and g = 5. With p = 3 and q = 3/2, what is the Holder bound for the integral of fg?
Answer: ______________

Answer key

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How big is a function W1-mt_Qd8jU26qgE-s1

5 · 1/q = 1 minus 4/5 = 1/5, so q = 5.
6 · Holder bounds the integral of |fg| by the product of the norms, 3 times 2, which is 6.
Holder, by the product of the two norms · Holder is the product bound, while Minkowski governs sums.
True · A difference confined to a null set integrates to 0, so definiteness demands the identification.
True · In this setting, L2 is complete and its norm comes from an inner product. The other finite exponents fail the parallelogram law.
True · Their inner product is the integral of x minus 1/2 over [0, 1], which is 1/2 minus 1/2 = 0.
The indicator of [0, 1] and the indicator of (0, 1) · The two indicators differ only at the points 0 and 1, a null set, so their difference has norm 0. The other pairs differ on sets of positive measure.
10 · Homogeneity multiplies the norm by the absolute value of the scalar: |-2| times 5 = 10.
Its norm comes from an inner product, so orthogonality makes sense · On this interval, only p = 2 satisfies the parallelogram law and gives an inner-product norm. Minkowski holds for every finite p >= 1, and L2 functions need not be bounded.
10 · On a set of measure 1 a constant has norm equal to its absolute value for every p, so the bound is 2 times 5 = 10, which here equals the integral.