Let 𝐶[0,1] denote the set of all real valued continuous functions defined on [0,1] and ‖𝑓‖_∞ = sup{|𝑓(𝑥)| ∶ 𝑥 ∈ [0,1]} for all 𝑓 ∈ 𝐶[0,1]. Let

𝑋 = { 𝑓 ∈ 𝐶[0,1] ∶ 𝑓(0) = 𝑓(1) = 0 }.

Define 𝐹 ∶ (𝐶[0,1], ‖⋅‖_∞) → ℝ by 𝐹(𝑓) = \(\rm \int_0^1f(t)dt\) for all 𝑓 ∈ 𝐶[0,1].

Denote 𝑆_𝑋 = {𝑓 ∈ 𝑋 ∶ ‖𝑓‖_∞ = 1}.

Then the set {𝑓 ∈ 𝑋 ∶ 𝐹(𝑓) = ‖𝐹‖} ∩ 𝑆_𝑋 has 

Given -

Let 𝐶[0,1] denote the set of all real valued continuous functions defined on [0,1] and ‖𝑓‖∞ = sup{|𝑓(𝑥)| ∶ 𝑥 ∈ [0,1]} for all 𝑓 ∈ 𝐶[0,1].

Let 𝑋 = { 𝑓 ∈ 𝐶[0,1] ∶ 𝑓(0) = 𝑓(1) = 0 }.

Define 𝐹 ∶ (𝐶[0,1],  ‖𝑓‖∞) → ℝ by 𝐹(𝑓) = \(\rm \int_0^1f(t)dt\)  for all 𝑓 ∈ 𝐶[0,1].

Denote 𝑆_X = {𝑓 ∈ 𝑋 ∶  ‖𝑓‖∞ = 1}.

Explanation -

The aim is to find the functions that satisfy the condition {𝑓 ∈ 𝑋 : 𝐹(𝑓) = ‖𝐹‖}.

Given ‖𝐹‖ is the supremum of |𝐹(𝑓)| over all 𝑓 ∈ 𝐶[0,1] with ‖𝑓‖∞ ≤ 1.

And also that the value of an integral is generally understood to be a "sum" of the values of 𝑓 over the interval [0,1].

Since 𝑓(0) = 𝑓(1) = 0, any such ‖𝑓‖∞ ≤ 1 that has a chance of maximizing |𝐹(𝑓)| would generally need to have positive and negative values on [0,1].

Otherwise, the integral could likely be increased by "stretching" ‖𝑓‖∞ .

But this will contradict 𝑆_X = {𝑓 ∈ 𝑋 : ‖𝑓‖∞ = 1}.

A function in 𝑋 that meets the norm condition (i.e., is in 𝑆_X) would have |𝑓(𝑡)| ≤ 1 for each 𝑡 ∈ [0,1].

Moreover, no such function can be "stretched" to have a larger value of |𝐹(𝑓)|.

Hence, it would seem that no function in 𝑆_X can satisfy ‖𝐹‖.

Therefore, it seems no element in S_X can satisfy 𝐹(𝑓) = ‖𝐹‖.

Therefore, the set {𝑓 ∈ 𝑋 ∶ 𝐹(𝑓) = ‖𝐹‖} ∩ 𝑆_X contains no elements.

Hence option (1) is correct.